Integrand size = 23, antiderivative size = 399 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \]
2/315*(a-b)*(10*a^4-279*a^2*b^2-147*b^4)*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)/b^3/d+2/315*(a-b)*(10*a^3+165 *a^2*b-114*a*b^2+147*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)/b^2/d-2/315*(10*a^2-49*b^2)*(a+b*sec(d*x+c)) ^(3/2)*tan(d*x+c)/b/d-4/63*a*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*(a+ b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d-4/315*a*(5*a^2-57*b^2)*(a+b*sec(d*x+c)) ^(1/2)*tan(d*x+c)/b/d
Time = 13.33 (sec) , antiderivative size = 552, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 \left (10 a^5+10 a^4 b-279 a^3 b^2-279 a^2 b^3-147 a b^4-147 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b \left (-10 a^4+155 a^3 b+279 a^2 b^2+261 a b^3+147 b^4\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (10 a^4-279 a^2 b^2-147 b^4\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^2 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 \left (-10 a^4+279 a^2 b^2+147 b^4\right ) \sin (c+d x)}{315 b^2}+\frac {2}{315} \sec ^2(c+d x) \left (75 a^2 \sin (c+d x)+49 b^2 \sin (c+d x)\right )+\frac {2 \sec (c+d x) \left (5 a^3 \sin (c+d x)+163 a b^2 \sin (c+d x)\right )}{315 b}+\frac {38}{63} a b \sec ^2(c+d x) \tan (c+d x)+\frac {2}{9} b^2 \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*(10 *a^5 + 10*a^4*b - 279*a^3*b^2 - 279*a^2*b^3 - 147*a*b^4 - 147*b^5)*Sqrt[Co s[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos [c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(- 10*a^4 + 155*a^3*b + 279*a^2*b^2 + 261*a*b^3 + 147*b^4)*Sqrt[Cos[c + d*x]/ (1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x])) ]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (10*a^4 - 279*a^2 *b^2 - 147*b^4)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[( c + d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c + d*x)/2]^2]*S ec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(-10*a ^4 + 279*a^2*b^2 + 147*b^4)*Sin[c + d*x])/(315*b^2) + (2*Sec[c + d*x]^2*(7 5*a^2*Sin[c + d*x] + 49*b^2*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(5*a^3*Si n[c + d*x] + 163*a*b^2*Sin[c + d*x]))/(315*b) + (38*a*b*Sec[c + d*x]^2*Tan [c + d*x])/63 + (2*b^2*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2)
Time = 1.70 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 4327, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4327 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (7 b-2 a \sec (c+d x)) (a+b \sec (c+d x))^{5/2}dx}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (7 b-2 a \sec (c+d x)) (a+b \sec (c+d x))^{5/2}dx}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (7 b-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (39 a b-\left (10 a^2-49 b^2\right ) \sec (c+d x)\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (39 a b-\left (10 a^2-49 b^2\right ) \sec (c+d x)\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (39 a b+\left (49 b^2-10 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (55 a^2+49 b^2\right )-2 a \left (5 a^2-57 b^2\right ) \sec (c+d x)\right )dx-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (55 a^2+49 b^2\right )-2 a \left (5 a^2-57 b^2\right ) \sec (c+d x)\right )dx-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (55 a^2+49 b^2\right )-2 a \left (5 a^2-57 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (a b \left (155 a^2+261 b^2\right )-\left (10 a^4-279 b^2 a^2-147 b^4\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (a b \left (155 a^2+261 b^2\right )-\left (10 a^4-279 b^2 a^2-147 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (155 a^2+261 b^2\right )+\left (-10 a^4+279 b^2 a^2+147 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left ((a-b) \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\) |
(2*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((-4*a*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((-2*(10*a^2 - 49*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((2*(a - b)*Sqrt[a + b]*(10*a^4 - 27 9*a^2*b^2 - 147*b^4)*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))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(10* a^3 + 165*a^2*b - 114*a*b^2 + 147*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))])/(b*d))/3 - (4*a*(5* a^2 - 57*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/7)/(9*b)
3.6.45.3.1 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[(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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B 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} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(3238\) vs. \(2(361)=722\).
Time = 65.48 (sec) , antiderivative size = 3239, normalized size of antiderivative = 8.12
2/315/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(147*b^ 5*sin(d*x+c)+442*a^2*b^3*sin(d*x+c)+49*b^5*tan(d*x+c)+80*a^3*b^2*sin(d*x+c )-279*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(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^2*b^3 *cos(d*x+c)^2-261*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(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*b^4*cos(d*x+c)^2-20*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) *(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^4*b*cos(d*x+c)+558*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 ))^(1/2)*(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^2*cos(d*x+c)+558*(1/(a+b)*(b+a*cos(d*x+c))/(co s(d*x+c)+1))^(1/2)*(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^3*cos(d*x+c)-10*(1/(a+b)*(b+a*cos(d* x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(co t(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b*cos(d*x+c)^2+294*(1/(a+b)*( b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*El lipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^4*cos(d*x+c)+20*(1/ (a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(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*b*cos(d*x+c) -310*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d...
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integral((b^2*sec(d*x + c)^5 + 2*a*b*sec(d*x + c)^4 + a^2*sec(d*x + c)^3)* sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]