Integrand size = 25, antiderivative size = 422 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \]
-1/2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)/e^(1/2)+1/ 2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)/e^(1/2)-1/4*l n(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^(1/2)/e^( 1/2)+1/4*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2 ^(1/2)/e^(1/2)-2*b*EllipticPi((-cos(d*x+c))^(1/2)/(1+sin(d*x+c))^(1/2),b/( a-(a^2-b^2)^(1/2)),I)*2^(1/2)*sin(d*x+c)^(1/2)/a/d/(a^2-b^2)^(1/2)/(-cos(d *x+c))^(1/2)/(e*tan(d*x+c))^(1/2)+2*b*EllipticPi((-cos(d*x+c))^(1/2)/(1+si n(d*x+c))^(1/2),b/(a+(a^2-b^2)^(1/2)),I)*2^(1/2)*sin(d*x+c)^(1/2)/a/d/(a^2 -b^2)^(1/2)/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 38.30 (sec) , antiderivative size = 1475, normalized size of antiderivative = 3.50 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx =\text {Too large to display} \]
((b + a*Cos[c + d*x])*Sec[c + d*x]*Sqrt[Tan[c + d*x]]*((2*Sec[c + d*x]^3*( a + b*Sqrt[1 + Tan[c + d*x]^2])*(((-1/8 + I/8)*a*(2*ArcTan[1 - ((1 + I)*Sq rt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[ b]*Sqrt[Tan[c + d*x]])/(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]] - Log[Sqr t[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]]))/(Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(-a^2 + b^2)*AppellF1[1 /4, -1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[ Tan[c + d*x]]*Sqrt[1 + Tan[c + d*x]^2])/((5*(a^2 - b^2)*AppellF1[1/4, -1/2 , 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(2*b^2*Ap pellF1[5/4, -1/2, 2, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2 )] + (a^2 - b^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]^2)*(a^2 - b^2*(1 + Tan[c + d*x]^2)))))/ ((b + a*Cos[c + d*x])*(1 + Tan[c + d*x]^2)^2) + (Cos[2*(c + d*x)]*Sec[c + d*x]^3*(a + b*Sqrt[1 + Tan[c + d*x]^2])*((-20*Sqrt[2]*ArcTan[1 - Sqrt[2]*S qrt[Tan[c + d*x]]])/a + (20*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] )/a + ((10 - 10*I)*(a^2 - 2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - ((10 - 10*I)*(a ^2 - 2*b^2)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1 /4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (10*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[...
Time = 1.24 (sec) , antiderivative size = 412, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4380, 3042, 3212, 3042, 3208, 3042, 3386, 25, 1537, 412, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {1}{\sqrt {e \tan (c+d x)} (a+b \sec (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4380 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \tan (c+d x)}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {1}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\) |
| \(\Big \downarrow \) 3212 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {\sqrt {e \cot (c+d x)}}{b+a \cos (c+d x)}dx}{a \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \int \frac {\sqrt {-e \tan \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a \sqrt {e \tan (c+d x)} \sqrt {e \cot (c+d x)}}\) |
| \(\Big \downarrow \) 3208 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \int \frac {\sqrt {-\cos (c+d x)}}{(b+a \cos (c+d x)) \sqrt {\sin (c+d x)}}dx}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \int \frac {\sqrt {\sin \left (c+d x-\frac {\pi }{2}\right )}}{\sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3386 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int -\frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}+\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int -\frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int \frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int \frac {1}{\sqrt {1-\frac {\cos ^2(c+d x)}{(\sin (c+d x)+1)^2}} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1537 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \int \frac {1}{\sqrt {1-\frac {\cos (c+d x)}{\sin (c+d x)+1}} \sqrt {\frac {\cos (c+d x)}{\sin (c+d x)+1}+1} \left (a-\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \int \frac {1}{\sqrt {1-\frac {\cos (c+d x)}{\sin (c+d x)+1}} \sqrt {\frac {\cos (c+d x)}{\sin (c+d x)+1}+1} \left (a+\sqrt {a^2-b^2}+\frac {b \cos (c+d x)}{\sin (c+d x)+1}\right )}d\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}}{d}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 412 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}}dx}{a}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {e \int \frac {1}{\sqrt {e \tan (c+d x)} \left (\tan ^2(c+d x) e^2+e^2\right )}d(e \tan (c+d x))}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 e \int \frac {1}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {2 e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 e \left (\frac {\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 e \left (\frac {\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 e \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{a d}-\frac {b \sqrt {\sin (c+d x)} \left (-\frac {2 \sqrt {2} \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (a-\sqrt {a^2-b^2}\right )}-\frac {2 \sqrt {2} \left (\frac {a}{\sqrt {a^2-b^2}}+1\right ) \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{d \left (\sqrt {a^2-b^2}+a\right )}\right )}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}\) |
(2*e*((-(ArcTan[1 - Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e])) + Arc Tan[1 + Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e]))/(2*e) + (-1/2*Log [e - Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(Sqrt[2]*Sqrt[e]) + Log[e + Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sq rt[e]))/(2*e)))/(a*d) - (b*((-2*Sqrt[2]*(1 - a/Sqrt[a^2 - b^2])*EllipticPi [b/(a - Sqrt[a^2 - b^2]), ArcSin[Sqrt[-Cos[c + d*x]]/Sqrt[1 + Sin[c + d*x] ]], -1])/((a - Sqrt[a^2 - b^2])*d) - (2*Sqrt[2]*(1 + a/Sqrt[a^2 - b^2])*El lipticPi[b/(a + Sqrt[a^2 - b^2]), ArcSin[Sqrt[-Cos[c + d*x]]/Sqrt[1 + Sin[ c + d*x]]], -1])/((a + Sqrt[a^2 - b^2])*d))*Sqrt[Sin[c + d*x]])/(a*Sqrt[-C os[c + d*x]]*Sqrt[e*Tan[c + d*x]])
3.4.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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])
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]]))
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (343 ) = 686\).
Time = 3.58 (sec) , antiderivative size = 1919, normalized size of antiderivative = 4.55
1/2/d*2^(1/2)/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/(a^2-b^2)^(1/2)/ (a-b)/a*(3*I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1 /2))*(a^2-b^2)^(1/2)*a^2*b-3*I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2), 1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2-3*I*EllipticPi((csc(d*x+c)-co t(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b+3*I*Ellipti cPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2) *a*b^2-3*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2)) *(a^2-b^2)^(1/2)*a*b^2+3*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/ 2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b-3*EllipticPi((csc(d*x+c)-cot(d*x+c) +1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2+2*(a^2-b^2)^(1/2)*E llipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)), 1/2*2^(1/2))*a*b^2+2*(a^2-b^2)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^ (1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a*b^2-I*EllipticPi((c sc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*b-I*E llipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2) ^(1/2)*a^3-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1 /2))*(a^2-b^2)^(3/2)*a-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/ 2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3+I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1 )^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*a+I*EllipticPi((csc(d*x+c)- cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3+I*Ellipt...
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]
\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}} \,d x } \]
\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]