\(\int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx\) [615]

Optimal result

Integrand size = 25, antiderivative size = 451 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}+\frac {2 a E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {a b \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {a b \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}} \] Output:

b^(3/2)*arctan(b^(1/2)*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2 
)^(1/4)/(a^2+b^2)^(5/4)/f/(d*sec(f*x+e))^(1/2)-b^(3/2)*arctanh(b^(1/2)*(se 
c(f*x+e)^2)^(1/4)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(5/4)/f/ 
(d*sec(f*x+e))^(1/2)+2*a*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(s 
ec(f*x+e)^2)^(1/4)/(a^2+b^2)/f/(d*sec(f*x+e))^(1/2)-2*a*tan(f*x+e)/(a^2+b^ 
2)/f/(d*sec(f*x+e))^(1/2)-a*b*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)^(1/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2)^ 
(3/2)/f/(d*sec(f*x+e))^(1/2)+a*b*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)^(1/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2 
)^(3/2)/f/(d*sec(f*x+e))^(1/2)+2*(b+a*tan(f*x+e))/(a^2+b^2)/f/(d*sec(f*x+e 
))^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 2.54 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=-\frac {28 d \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{4},\frac {5}{4},\frac {7}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a \cos (e+f x)+b \sin (e+f x))}{5 b f (d \sec (e+f x))^{3/2} \left (5 (a+i b) \operatorname {AppellF1}\left (\frac {7}{2},\frac {5}{4},\frac {9}{4},\frac {9}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+5 (a-i b) \operatorname {AppellF1}\left (\frac {7}{2},\frac {9}{4},\frac {5}{4},\frac {9}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+14 \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{4},\frac {5}{4},\frac {7}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a+b \tan (e+f x))\right )} \] Input:

Integrate[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])),x]
 

Output:

(-28*d*AppellF1[5/2, 5/4, 5/4, 7/2, (a - I*b)/(a + b*Tan[e + f*x]), (a + I 
*b)/(a + b*Tan[e + f*x])]*(a*Cos[e + f*x] + b*Sin[e + f*x]))/(5*b*f*(d*Sec 
[e + f*x])^(3/2)*(5*(a + I*b)*AppellF1[7/2, 5/4, 9/4, 9/2, (a - I*b)/(a + 
b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])] + 5*(a - I*b)*AppellF1[7/ 
2, 9/4, 5/4, 9/2, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + 
 f*x])] + 14*AppellF1[5/2, 5/4, 5/4, 7/2, (a - I*b)/(a + b*Tan[e + f*x]), 
(a + I*b)/(a + b*Tan[e + f*x])]*(a + b*Tan[e + f*x])))
 

Rubi [A] (warning: unable to verify)

Time = 0.76 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.76, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3994, 496, 27, 25, 719, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}dx\)

\(\Big \downarrow \) 3994

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 225

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

\(\Big \downarrow \) 212

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 310

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 993

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

\(\Big \downarrow \) 1537

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

\(\Big \downarrow \) 412

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

Input:

Int[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])),x]
 

Output:

((Sec[e + f*x]^2)^(1/4)*((2*(b^2 + a*b*Tan[e + f*x]))/((a^2 + b^2)*(1 + Ta 
n[e + f*x]^2)^(1/4)) - (-(b^2*(-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 
)) + a*(-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)))/(b*f*Sqrt[d*Sec[e + f*x]])
 

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 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 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 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 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. 4861 vs. \(2 (392 ) = 784\).

Time = 9.16 (sec) , antiderivative size = 4862, normalized size of antiderivative = 10.78

method	result	size
default	\(\text {Expression too large to display}\)	\(4862\)

Input:

int(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

1/2/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^( 
1/2)*(4*(a^2+b^2)^(3/2)*(-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*((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^3*sin(f*x+e)-( 
a^2+b^2)^(1/2)*ln(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*((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*b^3+(a^2+b^2)^(3/2)*ln(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*((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*b-4*(a^2+b^2)^(1/2)*(-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*((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^3*b^2*sin(f*x+e)-2*ln(2*(2*cos(f*x+e)*(-cos 
(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos 
(f*x+e)+1)/(1+cos(f*x+e)))*(a^2+b^2)^(3/2)*(-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*((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*b+2*ln(2*(2*cos(f*x+e)*(-cos(f 
*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos(f 
*x+e)+1)/(1+cos(f*x+e)))*(a^2+b^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*((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*b^3+2*ln((2*cos(f*x+e)*(-cos(...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\int \frac {1}{\sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )}\, dx \] Input:

integrate(1/(d*sec(f*x+e))**(1/2)/(a+b*tan(f*x+e)),x)
 

Output:

Integral(1/(sqrt(d*sec(e + f*x))*(a + b*tan(e + f*x))), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right ) \tan \left (f x +e \right ) b +\sec \left (f x +e \right ) a}d x \right )}{d} \] Input:

int(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x)
 

Output:

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