\(\int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx\) [619]

Optimal result

Integrand size = 25, antiderivative size = 440 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {a d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 b^{3/2} \left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 b^{3/2} \left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 b^2 \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 b^2 \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt {d \sec (e+f x)}}{b f (a+b \tan (e+f x))} \] Output:

1/2*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))^(1/2)/b^(3/2)/(a^2+b^2)^(3/4)/f/(sec(f*x+e)^2)^(1/4)+1/2*a*d^2*arctanh 
(b^(1/2)*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(1/2)/b^(3/2 
)/(a^2+b^2)^(3/4)/f/(sec(f*x+e)^2)^(1/4)+d^2*InverseJacobiAM(1/2*arctan(ta 
n(f*x+e)),2^(1/2))*(d*sec(f*x+e))^(1/2)/b^2/f/(sec(f*x+e)^2)^(1/4)-1/2*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*s 
ec(f*x+e))^(1/2)*(-tan(f*x+e)^2)^(1/2)/b^2/(a^2+b^2)/f/(sec(f*x+e)^2)^(1/4 
)-1/2*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))^(1/2)*(-tan(f*x+e)^2)^(1/2)/b^2/(a^2+b^2)/f/(sec(f*x+e) 
^2)^(1/4)-d^2*(d*sec(f*x+e))^(1/2)/b/f/(a+b*tan(f*x+e))
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.20 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.71 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {(d \sec (e+f x))^{5/2} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\frac {2 b \cos (e+f x)}{a \cos (e+f x)+b \sin (e+f x)}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)+\frac {a \left (a \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)+a \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \tan (e+f x)+\sqrt {b} \sqrt [4]{a^2+b^2} \left (\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )\right ) \sqrt {-\tan ^2(e+f x)}\right )}{\left (a^2+b^2\right ) \sqrt {-\tan ^2(e+f x)}}}{\sqrt [4]{\sec ^2(e+f x)}}\right )}{2 b^2 f (a+b \tan (e+f x))^2} \] Input:

Integrate[(d*Sec[e + f*x])^(5/2)/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Sec[e + f*x])^(5/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*((-2*b*Cos[e + 
 f*x])/(a*Cos[e + f*x] + b*Sin[e + f*x]) + (Hypergeometric2F1[1/2, 3/4, 3/ 
2, -Tan[e + f*x]^2]*Tan[e + f*x] + (a*(a*EllipticPi[-(b/Sqrt[a^2 + b^2]), 
ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Tan[e + f*x] + a*EllipticPi[b/Sqrt[a^2 
 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Tan[e + f*x] + Sqrt[b]*(a^2 + 
 b^2)^(1/4)*(ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)] + 
ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)])*Sqrt[-Tan[e + 
 f*x]^2]))/((a^2 + b^2)*Sqrt[-Tan[e + f*x]^2]))/(Sec[e + f*x]^2)^(1/4)))/( 
2*b^2*f*(a + b*Tan[e + f*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.70, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 3994, 492, 605, 229, 504, 312, 118, 25, 353, 73, 756, 218, 221, 925, 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.

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3994

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

\(\Big \downarrow \) 492

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

\(\Big \downarrow \) 605

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

\(\Big \downarrow \) 229

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 312

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

\(\Big \downarrow \) 118

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )-a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \int \frac {1}{-\tan ^4(e+f x) b^6+b^2+a^2}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right )}{2 b^2}-\frac {\sqrt [4]{\tan ^2(e+f x)+1}}{a+b \tan (e+f x)}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 925

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

\(\Big \downarrow \) 1537

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

\(\Big \downarrow \) 412

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

Input:

Int[(d*Sec[e + f*x])^(5/2)/(a + b*Tan[e + f*x])^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 118

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^( 
3/4)), x_] :> Simp[-4   Subst[Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + 
 d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e, f}, x] & 
& GtQ[-f/(d*e - c*f), 0]
 

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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 312

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[Sqrt[(-b)*(x^2/a)]/(2*x)   Subst[Int[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)* 
(c + d*x)), x], x, x^2], 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 756

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

rule 925

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 
1/(2*c)   Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 
*c)   Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), 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. 2819 vs. \(2 (378 ) = 756\).

Time = 124.40 (sec) , antiderivative size = 2820, normalized size of antiderivative = 6.41

method	result	size
default	\(\text {Expression too large to display}\)	\(2820\)

Input:

int((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*d^2/f*(cos(f*x+e)*(1+cos(f*x+e))*(a^2+b^2)^(1/2)*arctanh(1/2*((a^2+b^2 
)^(1/2)*cos(f*x+e)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^2)/ 
(1+cos(f*x+e))/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a 
^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(1+co 
s(f*x+e))^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2 
*b^3)/a^4)^(1/2)*a+(1+cos(f*x+e))*sin(f*x+e)*(a^2+b^2)^(1/2)*arctanh(1/2*( 
(a^2+b^2)^(1/2)*cos(f*x+e)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/ 
2)+b^2)/(1+cos(f*x+e))/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/(-b*((a^2+b^2) 
^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+ 
e)/(1+cos(f*x+e))^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2 
*a^2*b+2*b^3)/a^4)^(1/2)*b+cos(f*x+e)*(-cos(f*x+e)-1)*b*arctanh(1/2*((a^2+ 
b^2)^(1/2)*cos(f*x+e)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^ 
2)/(1+cos(f*x+e))/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/(-b*((a^2+b^2)^(1/2 
)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(1 
+cos(f*x+e))^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2* 
b+2*b^3)/a^4)^(1/2)*a+(-cos(f*x+e)-1)*sin(f*x+e)*arctanh(1/2*((a^2+b^2)^(1 
/2)*cos(f*x+e)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^2)/(1+c 
os(f*x+e))/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2 
*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(1+cos(f* 
x+e))^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*...
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   catd 
ef: division by zero
 

Sympy [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate((d*sec(f*x+e))**(5/2)/(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((d*sec(e + f*x))**(5/2)/(a + b*tan(e + f*x))**2, x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((d/cos(e + f*x))^(5/2)/(a + b*tan(e + f*x))^2,x)
 

Output:

int((d/cos(e + f*x))^(5/2)/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {\sqrt {d}\, d^{2} \left (-6 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )+15 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) b^{2} f +15 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a b f +21 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +21 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f +20 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +20 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f \right )}{14 a f \left (\tan \left (f x +e \right ) b +a \right )} \] Input:

int((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x)
 

Output:

(sqrt(d)*d**2*( - 6*sqrt(sec(e + f*x))*sec(e + f*x)**2*tan(e + f*x) + 15*i 
nt((sqrt(sec(e + f*x))*sec(e + f*x)**2*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 + 15*int((sqrt(sec 
(e + f*x))*sec(e + f*x)**2*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 + 21*int((sqrt(sec(e + f*x))*sec(e + f*x)**2 
*tan(e + f*x)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*ta 
n(e + f*x)*a*b*f + 21*int((sqrt(sec(e + f*x))*sec(e + f*x)**2*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 + 20*int 
((sqrt(sec(e + f*x))*sec(e + f*x)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f* 
x)*a*b + a**2),x)*tan(e + f*x)*a*b*f + 20*int((sqrt(sec(e + f*x))*sec(e + 
f*x)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*f))/(1 
4*a*f*(tan(e + f*x)*b + a))