Integrand size = 14, antiderivative size = 403 \[ \int (a+b \sec (c+d x))^{7/2} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}+\frac {2 \sqrt {a+b} \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 a^3 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {26 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
-2/15*(a-b)*(58*a^2+9*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+2/15*(60*a^3-58*a^2*b+22*a*b^2-9*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+26/15* 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. \(873\) vs. \(2(403)=806\).
Time = 13.63 (sec) , antiderivative size = 873, normalized size of antiderivative = 2.17 \[ \int (a+b \sec (c+d x))^{7/2} \, dx=\frac {2 (a+b \sec (c+d x))^{7/2} \left (58 a^3 b \tan \left (\frac {1}{2} (c+d x)\right )+58 a^2 b^2 \tan \left (\frac {1}{2} (c+d x)\right )+9 a b^3 \tan \left (\frac {1}{2} (c+d x)\right )+9 b^4 \tan \left (\frac {1}{2} (c+d x)\right )-116 a^3 b \tan ^3\left (\frac {1}{2} (c+d x)\right )-18 a b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )+58 a^3 b \tan ^5\left (\frac {1}{2} (c+d x)\right )-58 a^2 b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+9 a b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-9 b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )-30 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}}-30 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 (58 a^3+58 a^2 b+9 a b^2+9 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 (15 a^4-60 a^3 b-58 a^2 b^2-22 a b^3-9 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 )}{15 d (b+a \cos (c+d x))^{7/2} \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))^{7/2} \left (\frac {2}{15} b \left (58 a^2+9 b^2\right ) \sin (c+d x)+\frac {32}{15} a b^2 \tan (c+d x)+\frac {2}{5} b^3 \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^3} \]
(2*(a + b*Sec[c + d*x])^(7/2)*(58*a^3*b*Tan[(c + d*x)/2] + 58*a^2*b^2*Tan[ (c + d*x)/2] + 9*a*b^3*Tan[(c + d*x)/2] + 9*b^4*Tan[(c + d*x)/2] - 116*a^3 *b*Tan[(c + d*x)/2]^3 - 18*a*b^3*Tan[(c + d*x)/2]^3 + 58*a^3*b*Tan[(c + d* x)/2]^5 - 58*a^2*b^2*Tan[(c + d*x)/2]^5 + 9*a*b^3*Tan[(c + d*x)/2]^5 - 9*b ^4*Tan[(c + d*x)/2]^5 - 30*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)] - 30*a^4*EllipticPi[-1, ArcSin[Tan[( c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*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)] + b*(58*a^3 + 58*a^2*b + 9*a*b^2 + 9*b^3)*EllipticE[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)] + (15* a^4 - 60*a^3*b - 58*a^2*b^2 - 22*a*b^3 - 9*b^4)*EllipticF[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)] ))/(15*d*(b + a*Cos[c + d*x])^(7/2)*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])^(7/2)*((2*b*(58*a^2 + 9 *b^2)*Sin[c + d*x])/15 + (32*a*b^2*Tan[c + d*x])/15 + (2*b^3*Sec[c + d*...
Time = 1.57 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4269, 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.
\(\displaystyle \int (a+b \sec (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx\) |
\(\Big \downarrow \) 4269 |
\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (5 a^3+13 b^2 \sec ^2(c+d x) a+3 b \left (5 a^2+b^2\right ) \sec (c+d x)\right )dx+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (5 a^3+13 b^2 \sec ^2(c+d x) a+3 b \left (5 a^2+b^2\right ) \sec (c+d x)\right )dx+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 a^3+13 b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+3 b \left (5 a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4544 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {15 a^4+2 b \left (30 a^2+11 b^2\right ) \sec (c+d x) a+b^2 \left (58 a^2+9 b^2\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 a^4+2 b \left (30 a^2+11 b^2\right ) \sec (c+d x) a+b^2 \left (58 a^2+9 b^2\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 a^4+2 b \left (30 a^2+11 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+b^2 \left (58 a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b^2 \left (58 a^2+9 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 a^4+\left (2 a b \left (30 a^2+11 b^2\right )-b^2 \left (58 a^2+9 b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b^2 \left (58 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 a^4+\left (2 a b \left (30 a^2+11 b^2\right )-b^2 \left (58 a^2+9 b^2\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^4 \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b^2 \left (58 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^4 \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^2 \left (58 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b^2 \left (58 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b^2 \left (58 a^2+9 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {30 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\) |
(2*b^2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt [a + b]*(58*a^2 + 9*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)]* Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d + (2*Sqrt[a + b]*(60*a^3 - 58*a ^2*b + 22*a*b^2 - 9*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)]* Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (30*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)/3 + (26*a*b^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]) /(3*d))/5
3.6.53.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*C ot[c + d*x]*((a + b*Csc[c + d*x])^(n - 2)/(d*(n - 1))), x] + Simp[1/(n - 1) Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) + 3* a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] / ; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]
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_)]*(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_)]*(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 = 13.71 (sec) , antiderivative size = 2777, normalized size of antiderivative = 6.89
2/15/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(18*(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))*b^4*cos(d*x+c)+15*(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)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*cos(d*x+c)^2 -30*(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)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^4*c os(d*x+c)^2+9*(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)) *b^4*cos(d*x+c)^2+30*(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^4*cos(d*x+c)-60*(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)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a- b)/(a+b))^(1/2))*a^4*cos(d*x+c)+19*a*b^3*sin(d*x+c)+9*b^4*sin(d*x+c)+15*El lipticF(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^4-30*Elliptic Pi(cot(d*x+c)-csc(d*x+c),-1,((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^4+9*EllipticE(c ot(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)*b^4+74*a^2*b^2*sin(d*...
\[ \int (a+b \sec (c+d x))^{7/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \,d x } \]
integral((b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3)*sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int (a+b \sec (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (a+b \sec (c+d x))^{7/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \,d x } \]
\[ \int (a+b \sec (c+d x))^{7/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (a+b \sec (c+d x))^{7/2} \, dx=\int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x \]