3.7.15 \(\int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx\) [615]

3.7.15.1 Optimal result
3.7.15.2 Mathematica [C] (verified)
3.7.15.3 Rubi [A] (warning: unable to verify)
3.7.15.4 Maple [B] (warning: unable to verify)
3.7.15.5 Fricas [F(-1)]
3.7.15.6 Sympy [F]
3.7.15.7 Maxima [F]
3.7.15.8 Giac [F]
3.7.15.9 Mupad [F(-1)]

3.7.15.1 Optimal result

Integrand size = 25, antiderivative size = 520 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=-\frac {7 a b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{2 \left (a^2+b^2\right )^{11/4} f (d \sec (e+f x))^{3/2}}-\frac {7 a b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{2 \left (a^2+b^2\right )^{11/4} f (d \sec (e+f x))^{3/2}}+\frac {\left (2 a^2-5 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}+\frac {7 a^2 b^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sec ^2(e+f x)^{3/4} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^3 f (d \sec (e+f x))^{3/2}}+\frac {7 a^2 b^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sec ^2(e+f x)^{3/4} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^3 f (d \sec (e+f x))^{3/2}}+\frac {b \left (2 a^2-5 b^2\right ) \sec ^2(e+f x)}{3 \left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))} \]

output
-7/2*a*b^(5/2)*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f 
*x+e)^2)^(3/4)/(a^2+b^2)^(11/4)/f/(d*sec(f*x+e))^(3/2)-7/2*a*b^(5/2)*arcta 
nh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(3/4)/(a^2 
+b^2)^(11/4)/f/(d*sec(f*x+e))^(3/2)+1/3*(2*a^2-5*b^2)*(cos(1/2*arctan(tan( 
f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*arctan(tan 
(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^2/f/(d*sec(f*x+e))^(3/2) 
+7/2*a^2*b^2*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2) 
,I)*(sec(f*x+e)^2)^(3/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2)^3/f/(d*sec(f*x+e) 
)^(3/2)+7/2*a^2*b^2*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b/(a^2+b^2) 
^(1/2),I)*(sec(f*x+e)^2)^(3/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2)^3/f/(d*sec( 
f*x+e))^(3/2)+1/3*b*(2*a^2-5*b^2)*sec(f*x+e)^2/(a^2+b^2)^2/f/(d*sec(f*x+e) 
)^(3/2)/(a+b*tan(f*x+e))+2/3*(b+a*tan(f*x+e))/(a^2+b^2)/f/(d*sec(f*x+e))^( 
3/2)/(a+b*tan(f*x+e))
 
3.7.15.2 Mathematica [C] (verified)

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

Time = 12.23 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\frac {\sec ^4(e+f x) (a \cos (e+f x)+b \sin (e+f x))^2 \left (\frac {b \left (2 a^2-3 b^2\right )}{3 a (a-i b)^2 (a+i b)^2}+\frac {2 a b \cos (2 (e+f x))}{3 (a-i b)^2 (a+i b)^2}+\frac {b^4 \sin (e+f x)}{a (a-i b)^2 (a+i b)^2 (a \cos (e+f x)+b \sin (e+f x))}+\frac {\left (a^2-b^2\right ) \sin (2 (e+f x))}{3 (a-i b)^2 (a+i b)^2}\right )}{f (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}+\frac {\sec ^2(e+f x) \sec ^2(e+f x)^{3/4} (a \cos (e+f x)+b \sin (e+f x))^2 \left (\left (2 a^4-3 a^2 b^2-5 b^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)+21 a b^2 \left (-\sqrt {b} \sqrt [4]{a^2+b^2} \left (\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )\right )+a \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}+a \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}\right )\right )}{6 \left (a^2+b^2\right )^3 f (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \]

input
Integrate[1/((d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^2),x]
 
output
(Sec[e + f*x]^4*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*((b*(2*a^2 - 3*b^2))/( 
3*a*(a - I*b)^2*(a + I*b)^2) + (2*a*b*Cos[2*(e + f*x)])/(3*(a - I*b)^2*(a 
+ I*b)^2) + (b^4*Sin[e + f*x])/(a*(a - I*b)^2*(a + I*b)^2*(a*Cos[e + f*x] 
+ b*Sin[e + f*x])) + ((a^2 - b^2)*Sin[2*(e + f*x)])/(3*(a - I*b)^2*(a + I* 
b)^2)))/(f*(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^2) + (Sec[e + f*x]^ 
2*(Sec[e + f*x]^2)^(3/4)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*((2*a^4 - 3*a 
^2*b^2 - 5*b^4)*Hypergeometric2F1[1/2, 3/4, 3/2, -Tan[e + f*x]^2]*Tan[e + 
f*x] + 21*a*b^2*(-(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)])) + a*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), Arc 
Sin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2] + a*Cot[e + f*x]*El 
lipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[ 
e + f*x]^2])))/(6*(a^2 + b^2)^3*f*(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f* 
x])^2)
 
3.7.15.3 Rubi [A] (warning: unable to verify)

Time = 0.88 (sec) , antiderivative size = 413, normalized size of antiderivative = 0.79, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3994, 496, 27, 688, 27, 719, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \int \frac {1}{(a+b \tan (e+f x))^2 \left (\tan ^2(e+f x)+1\right )^{7/4}}d(b \tan (e+f x))}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 688

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac {b^2 \int -\frac {2 a \left (\frac {a^2}{b^2}+8\right )-\left (5-\frac {2 a^2}{b^2}\right ) b \tan (e+f x)}{2 (a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{a^2+b^2}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \int \frac {2 a \left (\frac {a^2}{b^2}+8\right )-\left (5-\frac {2 a^2}{b^2}\right ) b \tan (e+f x)}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \int \frac {1}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-\left (5-\frac {2 a^2}{b^2}\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \int \frac {1}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (a \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 312

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {-\frac {\tan (e+f x)}{b}} \left (\frac {\tan (e+f x)}{b}+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )}{2 b}-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 118

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int -\frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-\frac {1}{2} \int \frac {1}{\left (\frac {\tan (e+f x)}{b}+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \int \frac {1}{-\tan ^4(e+f x) b^6+b^2+a^2}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}+\frac {\int \frac {1}{\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}+\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 925

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \int \frac {1}{\left (1-\frac {b^3 \tan ^2(e+f x)}{\sqrt {a^2+b^2}}\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}-\frac {b^2 \int \frac {1}{\left (\frac {\tan ^2(e+f x) b^3}{\sqrt {a^2+b^2}}+1\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 1537

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \int \frac {1}{\left (1-\frac {b^3 \tan ^2(e+f x)}{\sqrt {a^2+b^2}}\right ) \sqrt {1-\sqrt [4]{\frac {\tan (e+f x)}{b}+1}} \sqrt {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}+1}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}-\frac {b^2 \int \frac {1}{\left (\frac {\tan ^2(e+f x) b^3}{\sqrt {a^2+b^2}}+1\right ) \sqrt {1-\sqrt [4]{\frac {\tan (e+f x)}{b}+1}} \sqrt {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}+1}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 412

\(\displaystyle \frac {\sec ^2(e+f x)^{3/4} \left (\frac {\frac {b^2 \left (21 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right ),-1\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right ),-1\right )}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \left (5-\frac {2 a^2}{b^2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )\right )}{2 \left (a^2+b^2\right )}+\frac {b^2 \left (2 a^2-5 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}}{3 \left (a^2+b^2\right )}+\frac {2 \left (a b \tan (e+f x)+b^2\right )}{3 \left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{3/4} (a+b \tan (e+f x))}\right )}{b f (d \sec (e+f x))^{3/2}}\)

input
Int[1/((d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^2),x]
 
output
((Sec[e + f*x]^2)^(3/4)*((2*(b^2 + a*b*Tan[e + f*x]))/(3*(a^2 + b^2)*(a + 
b*Tan[e + f*x])*(1 + Tan[e + f*x]^2)^(3/4)) + ((b^2*(2*a^2 - 5*b^2)*(1 + T 
an[e + f*x]^2)^(1/4))/((a^2 + b^2)*(a + b*Tan[e + f*x])) + (b^2*(-2*(5 - ( 
2*a^2)/b^2)*b*EllipticF[ArcTan[Tan[e + f*x]]/2, 2] + 21*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)) + 
ArcTanh[(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]), Arc 
Sin[(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)))*Sqr 
t[-Tan[e + f*x]^2])/b)))/(2*(a^2 + b^2)))/(3*(a^2 + b^2))))/(b*f*(d*Sec[e 
+ f*x])^(3/2))
 

3.7.15.3.1 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 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 496
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[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 688
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( 
(m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2))   Int[(d + 
 e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m 
 + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 719
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 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]
 
3.7.15.4 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. 14916 vs. \(2 (477 ) = 954\).

Time = 13.29 (sec) , antiderivative size = 14917, normalized size of antiderivative = 28.69

method result size
default \(\text {Expression too large to display}\) \(14917\)

input
int(1/(d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 
output
result too large to display
 
3.7.15.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]

input
integrate(1/(d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 
output
Timed out
 
3.7.15.6 Sympy [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\int \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]

input
integrate(1/(d*sec(f*x+e))**(3/2)/(a+b*tan(f*x+e))**2,x)
 
output
Integral(1/((d*sec(e + f*x))**(3/2)*(a + b*tan(e + f*x))**2), x)
 
3.7.15.7 Maxima [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

input
integrate(1/(d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 
output
integrate(1/((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a)^2), x)
 
3.7.15.8 Giac [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

input
integrate(1/(d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 
output
integrate(1/((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a)^2), x)
 
3.7.15.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2} \, dx=\int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]

input
int(1/((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^2),x)
 
output
int(1/((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^2), x)