\(\int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx\) [618]

Optimal result

Integrand size = 25, antiderivative size = 480 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=-\frac {3 a d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {3 d^2 E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac {3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}+\frac {3 a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{2 b^3 \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {3 a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{2 b^3 \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))} \] Output:

-3/2*a*d^2*arctan(b^(1/2)*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(1/4))*(d*sec(f*x 
+e))^(3/2)/b^(5/2)/(a^2+b^2)^(1/4)/f/(sec(f*x+e)^2)^(3/4)+3/2*a*d^2*arctan 
h(b^(1/2)*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/b^(5/ 
2)/(a^2+b^2)^(1/4)/f/(sec(f*x+e)^2)^(3/4)-3*d^2*EllipticE(sin(1/2*arctan(t 
an(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(3/2)/b^2/f/(sec(f*x+e)^2)^(3/4)+3*d^2 
*cos(f*x+e)*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/b^2/f+3/2*a^2*d^2*cot(f*x+e)*E 
llipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(3/2)* 
(-tan(f*x+e)^2)^(1/2)/b^3/(a^2+b^2)^(1/2)/f/(sec(f*x+e)^2)^(3/4)-3/2*a^2*d 
^2*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b/(a^2+b^2)^(1/2),I)*(d*sec( 
f*x+e))^(3/2)*(-tan(f*x+e)^2)^(1/2)/b^3/(a^2+b^2)^(1/2)/f/(sec(f*x+e)^2)^( 
3/4)-d^2*(d*sec(f*x+e))^(3/2)/b/f/(a+b*tan(f*x+e))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.99 (sec) , antiderivative size = 1129, normalized size of antiderivative = 2.35 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:

Integrate[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^2,x]
 

Output:

(Cos[e + f*x]*(d*Sec[e + f*x])^(7/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*( 
(3*Cos[e + f*x])/(a*b) + (3*Sin[e + f*x])/b^2 - 1/(b*(a*Cos[e + f*x] + b*S 
in[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2) + (3*(d*Sec[e + f*x])^(7/2)*(a* 
Cos[e + f*x] + b*Sin[e + f*x])^2*(-((a*EllipticE[ArcSin[Tan[(e + f*x)/2]], 
 -1]*Sqrt[1 + Tan[(e + f*x)/2]^2])/Sqrt[1 - Tan[(e + f*x)/2]^2]) + (2*a*El 
lipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 + Tan[(e + f*x)/2]^2])/Sqrt[1 
 - Tan[(e + f*x)/2]^2] + (-2*Sqrt[2]*a*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[ 
Sqrt[((1 + I)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2] 
*Sqrt[-((1 + I*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] + Sqrt[2]*a^2*Sq 
rt[a^2 + b^2]*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b + 
Sqrt[a^2 + b^2]), ArcSin[Sqrt[((1 + I)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e 
 + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((1 + I*Tan[(e + f*x)/2])/(I + Tan[(e + f* 
x)/2]))] + a^2*(a + I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b 
+ Sqrt[a^2 + b^2])))/(a + b - Sqrt[a^2 + b^2]), ArcSin[Sqrt[((1 + I)*(1 + 
Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((2 + (2*I)* 
Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] - a^3*EllipticPi[((1 + I)*(a - 
I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[((1 + I)* 
(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((2 + ( 
2*I)*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] - I*a^2*b*EllipticPi[((1 + 
 I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sq...
 

Rubi [A] (warning: unable to verify)

Time = 0.70 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.68, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 3994, 492, 605, 225, 212, 504, 310, 353, 73, 827, 218, 221, 993, 1537, 412}

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.

\(\displaystyle \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \int \frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{(a+b \tan (e+f x))^2}d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 492

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \int \frac {b \tan (e+f x)}{(a+b \tan (e+f x)) \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 605

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (\int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-a \int \frac {1}{(a+b \tan (e+f x)) \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \int \frac {1}{(a+b \tan (e+f x)) \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \int \frac {1}{(a+b \tan (e+f x)) \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (a \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 310

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-\int \frac {b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-\frac {1}{2} \int \frac {1}{\sqrt [4]{\frac {\tan (e+f x)}{b}+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \int \frac {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{-\tan ^4(e+f x) b^6+b^2+a^2}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}-\frac {\int \frac {1}{\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 993

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {1}{2} b \int \frac {1}{\left (\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\tan ^2(e+f x)+1}-\frac {1}{2} b \int \frac {1}{\left (\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\tan ^2(e+f x)+1}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 1537

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {1}{2} b \int \frac {1}{\sqrt {1-b^2 \tan ^2(e+f x)} \sqrt {b^2 \tan ^2(e+f x)+1} \left (\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)\right )}d\sqrt [4]{\tan ^2(e+f x)+1}-\frac {1}{2} b \int \frac {1}{\sqrt {1-b^2 \tan ^2(e+f x)} \sqrt {b^2 \tan ^2(e+f x)+1} \left (\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}\right )}d\sqrt [4]{\tan ^2(e+f x)+1}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 412

\(\displaystyle \frac {d^2 (d \sec (e+f x))^{3/2} \left (\frac {3 \left (-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {b \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\tan ^2(e+f x)+1}\right ),-1\right )}{2 \sqrt {a^2+b^2}}-\frac {b \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\tan ^2(e+f x)+1}\right ),-1\right )}{2 \sqrt {a^2+b^2}}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )-2 b E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )+\frac {2 b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1}}\right )}{2 b^2}-\frac {\left (\tan ^2(e+f x)+1\right )^{3/4}}{a+b \tan (e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

Input:

Int[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^2,x]
 

Output:

(d^2*(d*Sec[e + f*x])^(3/2)*(-((1 + Tan[e + f*x]^2)^(3/4)/(a + b*Tan[e + f 
*x])) + (3*(-2*b*EllipticE[ArcTan[Tan[e + f*x]]/2, 2] + (2*b*Tan[e + f*x]) 
/(1 + Tan[e + f*x]^2)^(1/4) - a*(-2*b^2*(-1/2*ArcTan[(b^(3/2)*Tan[e + f*x] 
)/(a^2 + b^2)^(1/4)]/(b^(3/2)*(a^2 + b^2)^(1/4)) + ArcTanh[(b^(3/2)*Tan[e 
+ f*x])/(a^2 + b^2)^(1/4)]/(2*b^(3/2)*(a^2 + b^2)^(1/4))) + (2*a*Cot[e + f 
*x]*(-1/2*(b*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(1 + Tan[e + f*x]^2)^ 
(1/4)], -1])/Sqrt[a^2 + b^2] + (b*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(1 
+ Tan[e + f*x]^2)^(1/4)], -1])/(2*Sqrt[a^2 + b^2]))*Sqrt[-Tan[e + f*x]^2]) 
/b)))/(2*b^2)))/(b*f*(Sec[e + f*x]^2)^(3/4))
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 218

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

rule 221

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

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 310

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[2*(Sqrt[(-b)*(x^2/a)]/x)   Subst[Int[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d* 
x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

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 492

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 1))), x] - Simp[2*b*(p/(d*(n + 1)) 
)   Int[x*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, 
d, n}, x] && GtQ[p, 0] && (IntegerQ[p] || LtQ[n, -1]) && NeQ[n, -1] &&  !IL 
tQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 605

Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] 
:> Simp[1/d   Int[x^(m - 1)*(a + b*x^2)^p, x], x] - Simp[c/d   Int[x^(m - 1 
)*((a + b*x^2)^p/(c + d*x)), x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 
 0] && LtQ[-1, p, 0]
 

rule 827

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

rule 993

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> 
With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b)   Int[1/((r 
 - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
a*d, 0]
 

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 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]))   Subst[Int[(a + x)^n*(1 + 
x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

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. 12831 vs. \(2 (413 ) = 826\).

Time = 244.87 (sec) , antiderivative size = 12832, normalized size of antiderivative = 26.73

method	result	size
default	\(\text {Expression too large to display}\)	\(12832\)

Input:

int((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

integral(sqrt(d*sec(f*x + e))*d^3*sec(f*x + e)^3/(b^2*tan(f*x + e)^2 + 2*a 
*b*tan(f*x + e) + a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(7/2)/(a+b*tan(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((d/cos(e + f*x))^(7/2)/(a + b*tan(e + f*x))^2,x)
 

Output:

int((d/cos(e + f*x))^(7/2)/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {\sqrt {d}\, d^{3} \left (-10 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )+35 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) b^{2} f +35 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a b f +45 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +45 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f +28 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +28 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f \right )}{18 a f \left (\tan \left (f x +e \right ) b +a \right )} \] Input:

int((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x)
 

Output:

(sqrt(d)*d**3*( - 10*sqrt(sec(e + f*x))*sec(e + f*x)**3*tan(e + f*x) + 35* 
int((sqrt(sec(e + f*x))*sec(e + f*x)**3*tan(e + f*x)**3)/(tan(e + f*x)**2* 
b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*b**2*f + 35*int((sqrt(se 
c(e + f*x))*sec(e + f*x)**3*tan(e + f*x)**3)/(tan(e + f*x)**2*b**2 + 2*tan 
(e + f*x)*a*b + a**2),x)*a*b*f + 45*int((sqrt(sec(e + f*x))*sec(e + f*x)** 
3*tan(e + f*x)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*t 
an(e + f*x)*a*b*f + 45*int((sqrt(sec(e + f*x))*sec(e + f*x)**3*tan(e + f*x 
)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*f + 28*in 
t((sqrt(sec(e + f*x))*sec(e + f*x)**3)/(tan(e + f*x)**2*b**2 + 2*tan(e + f 
*x)*a*b + a**2),x)*tan(e + f*x)*a*b*f + 28*int((sqrt(sec(e + f*x))*sec(e + 
 f*x)**3)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*f))/( 
18*a*f*(tan(e + f*x)*b + a))