Integrand size = 25, antiderivative size = 430 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=-\frac {3 a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]
-(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*Ellipti cF(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2)/(a^2+b^2)/f/( sec(f*x+e)^2)^(1/4)-3/2*a*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1 /4))*b^(1/2)*(d*sec(f*x+e))^(1/2)/(a^2+b^2)^(7/4)/f/(sec(f*x+e)^2)^(1/4)-3 /2*a*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*b^(1/2)*(d*sec( f*x+e))^(1/2)/(a^2+b^2)^(7/4)/f/(sec(f*x+e)^2)^(1/4)+3/2*a^2*cot(f*x+e)*El lipticPi((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)/(a^2+b^2)^2/f/(sec(f*x+e)^2)^(1/4)+3/2*a^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)/(a^2+b^2)^2/f/(sec(f*x+e)^2)^(1/4)-b*(d*sec(f*x+e)) ^(1/2)/(a^2+b^2)/f/(a+b*tan(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.76 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\frac {b}{a (a-i b) (a+i b)}+\frac {b^2 \sin (e+f x)}{a (a-i b) (a+i b) (a \cos (e+f x)+b \sin (e+f x))}\right )}{f (a+b \tan (e+f x))^2}+\frac {\sqrt {d \sec (e+f x)} \sec ^2(e+f x)^{3/4} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\left (\left (a^2+b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )+3 a \left (-\sqrt {b} \sqrt [4]{a^2+b^2} \left (\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )\right )+a \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}+a \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}\right )\right )}{2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^2} \]
(Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*( -(b/(a*(a - I*b)*(a + I*b))) + (b^2*Sin[e + f*x])/(a*(a - I*b)*(a + I*b)*( a*Cos[e + f*x] + b*Sin[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2) + (Sqrt[d*S ec[e + f*x]]*(Sec[e + f*x]^2)^(3/4)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*(- ((a^2 + b^2)*Hypergeometric2F1[1/2, 3/4, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x ]) + 3*a*(-(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]), ArcSin[(Se c[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2] + a*Cot[e + f*x]*EllipticP i[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x ]^2])))/(2*(a^2 + b^2)^2*f*(a + b*Tan[e + f*x])^2)
Time = 0.72 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.75, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3994, 498, 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.
| \(\displaystyle \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \int \frac {1}{(a+b \tan (e+f x))^2 \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 498 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (-\frac {\int -\frac {2 a-b \tan (e+f x)}{2 (a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{a^2+b^2}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {\int \frac {2 a-b \tan (e+f x)}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 719 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \int \frac {1}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-\int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \int \frac {1}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 504 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (a \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 312 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {-\frac {\tan (e+f x)}{b}} \left (\frac {\tan (e+f x)}{b}+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )}{2 b}-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 118 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int -\frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-\frac {1}{2} \int \frac {1}{\left (\frac {\tan (e+f x)}{b}+1\right )^{3/4} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \int \frac {1}{-\tan ^4(e+f x) b^6+b^2+a^2}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}+\frac {\int \frac {1}{\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt {a^2+b^2}}+\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {1}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 925 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \int \frac {1}{\left (1-\frac {b^3 \tan ^2(e+f x)}{\sqrt {a^2+b^2}}\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}-\frac {b^2 \int \frac {1}{\left (\frac {\tan ^2(e+f x) b^3}{\sqrt {a^2+b^2}}+1\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 1537 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \int \frac {1}{\left (1-\frac {b^3 \tan ^2(e+f x)}{\sqrt {a^2+b^2}}\right ) \sqrt {1-\sqrt [4]{\frac {\tan (e+f x)}{b}+1}} \sqrt {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}+1}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}-\frac {b^2 \int \frac {1}{\left (\frac {\tan ^2(e+f x) b^3}{\sqrt {a^2+b^2}}+1\right ) \sqrt {1-\sqrt [4]{\frac {\tan (e+f x)}{b}+1}} \sqrt {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}+1}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 412 |
| \(\displaystyle \frac {\sqrt {d \sec (e+f x)} \left (\frac {3 a \left (-\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (-\frac {b^2 \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right ),-1\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right ),-1\right )}{2 \left (a^2+b^2\right )}\right )}{b}-2 b^2 \left (\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 \sqrt {b} \left (a^2+b^2\right )^{3/4}}\right )\right )-2 b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt [4]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) (a+b \tan (e+f x))}\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\) |
(Sqrt[d*Sec[e + f*x]]*(-((b^2*(1 + Tan[e + f*x]^2)^(1/4))/((a^2 + b^2)*(a + b*Tan[e + f*x]))) + (-2*b*EllipticF[ArcTan[Tan[e + f*x]]/2, 2] + 3*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]), ArcSin[(1 + Tan[e + f*x]/b)^(1/4)], -1])/(a^2 + b^2) - (b^2*Elli pticPi[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*(a^2 + b^2))))/(b*f*(Sec[e + f*x]^ 2)^(1/4))
3.7.13.3.1 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_))^(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]
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]
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]
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]
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]
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]
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]
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + S imp[b/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p*(c*(n + 1) - d*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[n , -1] && ((LtQ[n, -1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]) || (SumSimp lerQ[n, 1] && IntegerQ[p]) || ILtQ[Simplify[n + 2*p + 3], 0])
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]
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]
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]
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]
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[((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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 12205 vs. \(2 (395 ) = 790\).
Time = 11.21 (sec) , antiderivative size = 12206, normalized size of antiderivative = 28.39
Timed out. \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\sqrt {d \sec {\left (e + f x \right )}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\sqrt {d \sec \left (f x + e\right )}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\sqrt {d \sec \left (f x + e\right )}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]