Integrand size = 32, antiderivative size = 403 \[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (4 a^2-3 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 d}-\frac {2 a^3 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 d}-\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
-2/5*(a-b)*(4*a^2-3*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b) ^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b *(1+sec(d*x+c))/(a-b))^(1/2)/d+2/5*(10*a^3-4*a^2*b-4*a*b^2+3*b^3)*cot(d*x+ c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b) ^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-2* a^3*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b )/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ c))/(a-b))^(1/2)/d-2/5*b^2*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d-2/5*a*b^2*( a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(956\) vs. \(2(403)=806\).
Time = 18.69 (sec) , antiderivative size = 956, normalized size of antiderivative = 2.37 \[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=-\frac {4 (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \left (-4 a^3 b \tan \left (\frac {1}{2} (c+d x)\right )-4 a^2 b^2 \tan \left (\frac {1}{2} (c+d x)\right )+3 a b^3 \tan \left (\frac {1}{2} (c+d x)\right )+3 b^4 \tan \left (\frac {1}{2} (c+d x)\right )+8 a^3 b \tan ^3\left (\frac {1}{2} (c+d x)\right )-6 a b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )-4 a^3 b \tan ^5\left (\frac {1}{2} (c+d x)\right )+4 a^2 b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+10 a^4 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+10 a^4 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b \left (-4 a^3-4 a^2 b+3 a b^2+3 b^3\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (5 a^4-10 a^3 b-4 a^2 b^2+4 a b^3+3 b^4\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{5 d (b+a \cos (c+d x))^{3/2} \left (a^2-2 b^2+a^2 \cos (2 c+2 d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \left (-\frac {4}{5} b \left (-4 a^2+3 b^2\right ) \sin (c+d x)-\frac {8}{5} a b^2 \tan (c+d x)-\frac {4}{5} b^3 \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x)) \left (a^2-2 b^2+a^2 \cos (2 c+2 d x)\right )} \]
(-4*(a + b*Sec[c + d*x])^(3/2)*(a^2 - b^2*Sec[c + d*x]^2)*(-4*a^3*b*Tan[(c + d*x)/2] - 4*a^2*b^2*Tan[(c + d*x)/2] + 3*a*b^3*Tan[(c + d*x)/2] + 3*b^4 *Tan[(c + d*x)/2] + 8*a^3*b*Tan[(c + d*x)/2]^3 - 6*a*b^3*Tan[(c + d*x)/2]^ 3 - 4*a^3*b*Tan[(c + d*x)/2]^5 + 4*a^2*b^2*Tan[(c + d*x)/2]^5 + 3*a*b^3*Ta n[(c + d*x)/2]^5 - 3*b^4*Tan[(c + d*x)/2]^5 + 10*a^4*EllipticPi[-1, ArcSin [Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 10*a^4*Ellip ticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sq rt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(-4*a^3 - 4*a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[Ar cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/ 2]^2)/(a + b)] - (5*a^4 - 10*a^3*b - 4*a^2*b^2 + 4*a*b^3 + 3*b^4)*Elliptic F[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]* (1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d *x)/2]^2)/(a + b)]))/(5*d*(b + a*Cos[c + d*x])^(3/2)*(a^2 - 2*b^2 + a^2*Co s[2*c + 2*d*x])*Sec[c + d*x]^(7/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(-1 + Tan[(c + d*x)/2]^2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[ (c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(a^2 - b^2*Sec[c + d*x]^2)*((-4*b*...
Time = 1.63 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.531, Rules used = {3042, 4530, 25, 3042, 4406, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
| \(\displaystyle \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a^2-b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4530 |
| \(\displaystyle -\int -\left ((a-b \sec (c+d x)) (a+b \sec (c+d x))^{5/2}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int (a-b \sec (c+d x)) (a+b \sec (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4406 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (5 a^3-3 b^2 \sec ^2(c+d x) a+b \left (5 a^2-3 b^2\right ) \sec (c+d x)\right )dx-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (5 a^3-3 b^2 \sec ^2(c+d x) a+b \left (5 a^2-3 b^2\right ) \sec (c+d x)\right )dx-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 a^3-3 b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (5 a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 a^4+2 b \left (5 a^2-2 b^2\right ) \sec (c+d x) a+b^2 \left (4 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a^4+2 b \left (5 a^2-2 b^2\right ) \sec (c+d x) a+b^2 \left (4 a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a^4+2 b \left (5 a^2-2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+b^2 \left (4 a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{5} \left (b^2 \left (4 a^2-3 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {5 a^4+\left (2 a b \left (5 a^2-2 b^2\right )-b^2 \left (4 a^2-3 b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {5 a^4+\left (2 a b \left (5 a^2-2 b^2\right )-b^2 \left (4 a^2-3 b^2\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{5} \left (5 a^4 \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 a^4 \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{5} \left (b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{5} \left (b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (4 a^2-3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
(-2*b^2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((-2*(a - b)*Sqrt [a + b]*(4*a^2 - 3*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d *x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*S qrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d + (2*Sqrt[a + b]*(10*a^3 - 4*a^2 *b - 4*a*b^2 + 3*b^3)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x ]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqr t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (10*a^3*Sqrt[a + b]*Cot[c + d*x] *EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d* x]))/(a - b))])/d - (2*a*b^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/d)/5
3.8.33.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b* c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_.), x_Symbol] :> Simp[C/b^2 Int[(a + b*Csc[e + f*x])^(m + 1) *Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A*b^2 + a^2*C, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[( a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m )*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2776\) vs. \(2(364)=728\).
Time = 18.56 (sec) , antiderivative size = 2777, normalized size of antiderivative = 6.89
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2777\) |
| parts | \(\text {Expression too large to display}\) | \(3581\) |
2/5/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-b^4*sec(d*x +c)*tan(d*x+c)+2*a^2*b^2*sin(d*x+c)-3*a*b^3*sin(d*x+c)-3*a*b^3*tan(d*x+c)+ 4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+ 1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+4*(co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-3*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2 )*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-4*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ell ipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+4*(cos(d*x+c)/(c os(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt icF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-10*(cos(d*x+c)/(cos(d *x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF( cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b-3*sin(d*x+c)*b^4+4*Ellipt icE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x+c)^2+ 4*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos (d*x+c)^2-3*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+ c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2...
\[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\int { -{\left (b^{2} \sec \left (d x + c\right )^{2} - a^{2}\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
integral(-(b^3*sec(d*x + c)^3 + a*b^2*sec(d*x + c)^2 - a^2*b*sec(d*x + c) - a^3)*sqrt(b*sec(d*x + c) + a), x)
\[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\int \left (a - b \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
\[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\int { -{\left (b^{2} \sec \left (d x + c\right )^{2} - a^{2}\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\int { -{\left (b^{2} \sec \left (d x + c\right )^{2} - a^{2}\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=-\int -\left (a^2-\frac {b^2}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]