\(\int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx\) [310]

Optimal result

Integrand size = 25, antiderivative size = 610 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\frac {a e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a b^2 d}-\frac {2 \sqrt {2} \sqrt {a^2-b^2} e^2 \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} \sqrt {a^2-b^2} e^2 \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{b d \sqrt {e \tan (c+d x)}} \] Output:

1/2*a*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/b^2/d 
-1/2*(a^2-b^2)*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1 
/2)/a/b^2/d-1/2*a*e^(3/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2 
^(1/2)/b^2/d+1/2*(a^2-b^2)*e^(3/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e 
^(1/2))*2^(1/2)/a/b^2/d-1/2*a*e^(3/2)*arctanh(2^(1/2)*(e*tan(d*x+c))^(1/2) 
/(e^(1/2)+e^(1/2)*tan(d*x+c)))*2^(1/2)/b^2/d+1/2*(a^2-b^2)*e^(3/2)*arctanh 
(2^(1/2)*(e*tan(d*x+c))^(1/2)/(e^(1/2)+e^(1/2)*tan(d*x+c)))*2^(1/2)/a/b^2/ 
d-2*2^(1/2)*(a^2-b^2)^(1/2)*e^2*EllipticPi((-cos(d*x+c))^(1/2)/(1+sin(d*x+ 
c))^(1/2),b/(a-(a^2-b^2)^(1/2)),I)*sin(d*x+c)^(1/2)/a/b/d/(-cos(d*x+c))^(1 
/2)/(e*tan(d*x+c))^(1/2)+2*2^(1/2)*(a^2-b^2)^(1/2)*e^2*EllipticPi((-cos(d* 
x+c))^(1/2)/(1+sin(d*x+c))^(1/2),b/(a+(a^2-b^2)^(1/2)),I)*sin(d*x+c)^(1/2) 
/a/b/d/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)+e^2*InverseJacobiAM(c-1/4* 
Pi+d*x,2^(1/2))*sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/b/d/(e*tan(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 19.92 (sec) , antiderivative size = 548, normalized size of antiderivative = 0.90 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=-\frac {\cos (c+d x) \left (a+b \sqrt {\sec ^2(c+d x)}\right ) (e \tan (c+d x))^{3/2} \left (-5 \left (a^2-b^2\right ) \left (2 \sqrt {2} \sqrt {b} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \sqrt {b} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-(2-2 i) \sqrt [4]{a^2-b^2} \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+(2-2 i) \sqrt [4]{a^2-b^2} \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\sqrt {2} \sqrt {b} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \sqrt {b} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-(1-i) \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )+(1-i) \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )\right )+8 a b^{3/2} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \tan ^{\frac {5}{2}}(c+d x)\right )}{20 a \sqrt {b} \left (a^2-b^2\right ) d (b+a \cos (c+d x)) \tan ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[(e*Tan[c + d*x])^(3/2)/(a + b*Sec[c + d*x]),x]
 

Output:

-1/20*(Cos[c + d*x]*(a + b*Sqrt[Sec[c + d*x]^2])*(e*Tan[c + d*x])^(3/2)*(- 
5*(a^2 - b^2)*(2*Sqrt[2]*Sqrt[b]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 
2*Sqrt[2]*Sqrt[b]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] - (2 - 2*I)*(a^2 
- b^2)^(1/4)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^( 
1/4)] + (2 - 2*I)*(a^2 - b^2)^(1/4)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c 
 + d*x]])/(a^2 - b^2)^(1/4)] + Sqrt[2]*Sqrt[b]*Log[1 - Sqrt[2]*Sqrt[Tan[c 
+ d*x]] + Tan[c + d*x]] - Sqrt[2]*Sqrt[b]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x 
]] + Tan[c + d*x]] - (1 - I)*(a^2 - b^2)^(1/4)*Log[Sqrt[a^2 - b^2] - (1 + 
I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]] + (1 - 
 I)*(a^2 - b^2)^(1/4)*Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1 
/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]]) + 8*a*b^(3/2)*AppellF1[5/4, 1/ 
2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x] 
^(5/2)))/(a*Sqrt[b]*(a^2 - b^2)*d*(b + a*Cos[c + d*x])*Tan[c + d*x]^(3/2))
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4379

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}}dx}{b^2}-\frac {e^2 \int \frac {a-b \sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \int \frac {a-b \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}\)

\(\Big \downarrow \) 4372

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-b \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-b \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\right )}{b^2}\)

\(\Big \downarrow \) 3094

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {a e \int \frac {1}{\sqrt {e \tan (c+d x)} \left (\tan ^2(c+d x) e^2+e^2\right )}d(e \tan (c+d x))}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \int \frac {1}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 4380

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \tan (c+d x)}}dx}{a}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3212

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {\sqrt {e \cot (c+d x)}}{b+a \cos (c+d x)}dx}{a \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {\sqrt {-e \tan \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3208

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \int \frac {\sqrt {-\cos (c+d x)}}{(b+a \cos (c+d x)) \sqrt {\sin (c+d x)}}dx}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \int \frac {\sqrt {\sin \left (c+d x-\frac {\pi }{2}\right )}}{\sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 3386

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int -\frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}+\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int -\frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int \frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int \frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 1537

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int \frac {1}{\sqrt {1-\frac {\cos (c+d x)}{\sin (c+d x)+1}} \sqrt {\frac {\cos (c+d x)}{\sin (c+d x)+1}+1} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int \frac {1}{\sqrt {1-\frac {\cos (c+d x)}{\sin (c+d x)+1}} \sqrt {\frac {\cos (c+d x)}{\sin (c+d x)+1}+1} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

\(\Big \downarrow \) 412

\(\displaystyle \frac {e^2 \left (a^2-b^2\right ) \left (\frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{b^2}-\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {b \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{b^2}\)

Input:

Int[(e*Tan[c + d*x])^(3/2)/(a + b*Sec[c + d*x]),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 412

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* 
(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && S 
implerSqrtQ[-f/e, -d/c])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1537

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[(-a)*c, 2]}, Simp[Sqrt[-c]   Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqr 
t[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] & 
& GtQ[a, 0] && LtQ[c, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3094

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] 
:> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])   Int[ 
1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3208

Int[Sqrt[(g_.)*tan[(e_.) + (f_.)*(x_)]]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ 
)]), x_Symbol] :> Simp[Sqrt[Cos[e + f*x]]*(Sqrt[g*Tan[e + f*x]]/Sqrt[Sin[e 
+ f*x]])   Int[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*(a + b*Sin[e + f*x])) 
, x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 

rule 3212

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[g^(2*IntPart[p])*(g*Cot[e + f*x])^FracPart[p] 
*(g*Tan[e + f*x])^FracPart[p]   Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f*x]) 
^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]
 

rule 3386

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 
2]}, Simp[2*Sqrt[2]*d*((b + q)/(f*q))   Subst[Int[1/((d*(b + q) + a*x^2)*Sq 
rt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - 
 Simp[2*Sqrt[2]*d*((b - q)/(f*q))   Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 
 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x]] /; F 
reeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4372

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(e*Cot[c + d*x])^m, x], x] + Simp[b   Int[ 
(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]
 

rule 4379

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)/(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[-e^2/b^2   Int[(e*Cot[c + d*x])^(m - 2)*(a - b*Csc[ 
c + d*x]), x], x] + Simp[e^2*((a^2 - b^2)/b^2)   Int[(e*Cot[c + d*x])^(m - 
2)/(a + b*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b 
^2, 0] && IGtQ[m - 1/2, 0]
 

rule 4380

Int[1/(Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]*(csc[(c_.) + (d_.)*(x_)]*(b_.) + 
 (a_))), x_Symbol] :> Simp[1/a   Int[1/Sqrt[e*Cot[c + d*x]], x], x] - Simp[ 
b/a   Int[1/(Sqrt[e*Cot[c + d*x]]*(b + a*Sin[c + d*x])), x], x] /; FreeQ[{a 
, b, c, d, e}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1862 vs. \(2 (513 ) = 1026\).

Time = 1.87 (sec) , antiderivative size = 1863, normalized size of antiderivative = 3.05

method	result	size
default	\(\text {Expression too large to display}\)	\(1863\)

Input:

int((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

(-1/2+1/2*I)/d/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/(a^2-b^2)^(1/2) 
/a*(e*tan(d*x+c))^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2)*(2*cot(d*x+c)-2*csc 
(d*x+c)+2)^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*e*(2*(a^2-b^2)^(1/2)*Ell 
ipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b-I*(a^2 
-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1 
/2))*a^2-I*(a^2-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2 
-1/2*I,1/2*2^(1/2))*b^2+I*(a^2-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+ 
c)+1)^(1/2),-(a-b)/(-a+b+((a+b)*(a-b))^(1/2)),1/2*2^(1/2))*b^2+I*(a^2-b^2) 
^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/(a-b+((a+b)*(a-b) 
)^(1/2)),1/2*2^(1/2))*b^2-(a^2-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+ 
c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*a^2-(a^2-b^2)^(1/2)*EllipticPi((-cot(d* 
x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2-(a^2-b^2)^(1/2)*Ellipt 
icPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/(a-b+((a+b)*(a-b))^(1/2)),1/2* 
2^(1/2))*a^2+(a^2-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),( 
a-b)/(a-b+((a+b)*(a-b))^(1/2)),1/2*2^(1/2))*b^2-(a^2-b^2)^(1/2)*EllipticPi 
((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),-(a-b)/(-a+b+((a+b)*(a-b))^(1/2)),1/2*2^ 
(1/2))*a^2+(a^2-b^2)^(1/2)*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),-(a 
-b)/(-a+b+((a+b)*(a-b))^(1/2)),1/2*2^(1/2))*b^2-EllipticPi((-cot(d*x+c)+cs 
c(d*x+c)+1)^(1/2),(a-b)/(a-b+((a+b)*(a-b))^(1/2)),1/2*2^(1/2))*a^2*b-Ellip 
ticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),(a-b)/(a-b+((a+b)*(a-b))^(1/2)),...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:

integrate((e*tan(d*x+c))**(3/2)/(a+b*sec(d*x+c)),x)
 

Output:

Integral((e*tan(c + d*x))**(3/2)/(a + b*sec(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \] Input:

integrate((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")
 

Output:

integrate((e*tan(d*x + c))^(3/2)/(b*sec(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \] Input:

integrate((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="giac")
 

Output:

integrate((e*tan(d*x + c))^(3/2)/(b*sec(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{b+a\,\cos \left (c+d\,x\right )} \,d x \] Input:

int((e*tan(c + d*x))^(3/2)/(a + b/cos(c + d*x)),x)
 

Output:

int((cos(c + d*x)*(e*tan(c + d*x))^(3/2))/(b + a*cos(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right ) b +a}d x \right ) e \] Input:

int((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x)
 

Output:

sqrt(e)*int((sqrt(tan(c + d*x))*tan(c + d*x))/(sec(c + d*x)*b + a),x)*e