Integrand size = 25, antiderivative size = 1070 \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=-\frac {7 b^3 \left (a^2-b^2\right )^{3/4} e^{9/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{13/2} d}+\frac {2 b \left (a^2-b^2\right )^{7/4} e^{9/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{13/2} d}+\frac {7 b^3 \left (a^2-b^2\right )^{3/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{13/2} d}-\frac {2 b \left (a^2-b^2\right )^{7/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{13/2} d}+\frac {7 b^4 \left (a^2-b^2\right ) e^5 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 a^7 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 b^2 \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^7 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {7 b^4 \left (a^2-b^2\right ) e^5 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 a^7 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 b^2 \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^7 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {14 e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 a^2 d \sqrt {\sin (c+d x)}}-\frac {7 b^2 \left (3 a^2-5 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^6 d \sqrt {\sin (c+d x)}}-\frac {4 b^2 \left (8 a^2-5 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^6 d \sqrt {\sin (c+d x)}}-\frac {14 e^3 \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 a^2 d}-\frac {7 b^2 e^3 (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^5 d}+\frac {4 b e^3 \left (5 \left (a^2-b^2\right )+3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 a^5 d}+\frac {4 b e (e \sin (c+d x))^{7/2}}{7 a^3 d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{7/2}}{9 a^2 d}+\frac {b^2 e (e \sin (c+d x))^{7/2}}{a^3 d (b+a \cos (c+d x))} \]
-7/2*b^3*(a^2-b^2)^(3/4)*e^(9/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2- b^2)^(1/4)/e^(1/2))/a^(13/2)/d+2*b*(a^2-b^2)^(7/4)*e^(9/2)*arctan(a^(1/2)* (e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(13/2)/d+7/2*b^3*(a^2-b^2) ^(3/4)*e^(9/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2 ))/a^(13/2)/d-2*b*(a^2-b^2)^(7/4)*e^(9/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^( 1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(13/2)/d-14/45*e^3*cos(d*x+c)*(e*sin(d*x+c ))^(3/2)/a^2/d-7/15*b^2*e^3*(5*b-3*a*cos(d*x+c))*(e*sin(d*x+c))^(3/2)/a^5/ d+4/15*b*e^3*(5*a^2-5*b^2+3*a*b*cos(d*x+c))*(e*sin(d*x+c))^(3/2)/a^5/d+4/7 *b*e*(e*sin(d*x+c))^(7/2)/a^3/d-2/9*e*cos(d*x+c)*(e*sin(d*x+c))^(7/2)/a^2/ d+b^2*e*(e*sin(d*x+c))^(7/2)/a^3/d/(b+a*cos(d*x+c))-7/2*b^4*(a^2-b^2)*e^5* (sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(c os(1/2*c+1/4*Pi+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2) /a^7/d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2*b^2*(a^2-b^2)^2*e^5*(sin (1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1 /2*c+1/4*Pi+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^7 /d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-7/2*b^4*(a^2-b^2)*e^5*(sin(1/2 *c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c +1/4*Pi+1/2*d*x),2*a/(a+(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^7/d/( a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2*b^2*(a^2-b^2)^2*e^5*(sin(1/2*c+1 /4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 15.32 (sec) , antiderivative size = 974, normalized size of antiderivative = 0.91 \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) (e \sin (c+d x))^{9/2} \left (\frac {\left (14 a^4-159 a^2 b^2+165 b^4\right ) \cos ^2(c+d x) \left (3 \sqrt {2} b \left (-a^2+b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )+8 a^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{12 a^{3/2} \left (a^2-b^2\right ) (b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (-46 a^3 b+66 a b^3\right ) \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )}{\sqrt {a} \sqrt [4]{a^2-b^2}}+\frac {b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (-a^2+b^2\right )}\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{30 a^5 d (a+b \sec (c+d x))^2 \sin ^{\frac {9}{2}}(c+d x)}+\frac {(b+a \cos (c+d x))^2 \csc ^4(c+d x) \sec ^2(c+d x) (e \sin (c+d x))^{9/2} \left (-\frac {b \left (-37 a^2+56 b^2\right ) \sin (c+d x)}{21 a^5}+\frac {a^2 b^2 \sin (c+d x)-b^4 \sin (c+d x)}{a^5 (b+a \cos (c+d x))}-\frac {\left (19 a^2-54 b^2\right ) \sin (2 (c+d x))}{90 a^4}-\frac {b \sin (3 (c+d x))}{7 a^3}+\frac {\sin (4 (c+d x))}{36 a^2}\right )}{d (a+b \sec (c+d x))^2} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(e*Sin[c + d*x])^(9/2)*(((14*a^4 - 159*a^2*b^2 + 165*b^4)*Cos[c + d*x]^2*(3*Sqrt[2]*b*(-a^2 + b^2)^(3/4)*(2*A rcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*Arc Tan[1 + (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqr t[-a^2 + b^2] - Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a* Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)* Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]]) + 8*a^(5/2)*AppellF1[3/4, -1/2, 1, 7 /4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2))* (b + a*Sqrt[1 - Sin[c + d*x]^2]))/(12*a^(3/2)*(a^2 - b^2)*(b + a*Cos[c + d *x])*(1 - Sin[c + d*x]^2)) + (2*(-46*a^3*b + 66*a*b^3)*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/ 4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d* x]] + I*a*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2 )^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]]))/(Sqrt[a]*(a^2 - b^2)^(1/4 )) + (b*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a ^2 - b^2)]*Sin[c + d*x]^(3/2))/(3*(-a^2 + b^2)))*(b + a*Sqrt[1 - Sin[c + d *x]^2]))/((b + a*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(30*a^5*d*(a + b*Sec[c + d*x])^2*Sin[c + d*x]^(9/2)) + ((b + a*Cos[c + d*x])^2*Csc[c + d* x]^4*Sec[c + d*x]^2*(e*Sin[c + d*x])^(9/2)*(-1/21*(b*(-37*a^2 + 56*b^2)...
Time = 3.22 (sec) , antiderivative size = 1070, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4360, 3042, 3391, 2009}
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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{9/2}}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{9/2}}{(-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2 (e \sin (c+d x))^{9/2}}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b (e \sin (c+d x))^{9/2}}{a^2 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^{9/2}}{a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b^2 \left (a^2-b^2\right )^2 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^5}{a^7 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {7 b^4 \left (a^2-b^2\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^5}{2 a^7 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 b^2 \left (a^2-b^2\right )^2 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^5}{a^7 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {7 b^4 \left (a^2-b^2\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^5}{2 a^7 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 b \left (a^2-b^2\right )^{7/4} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{9/2}}{a^{13/2} d}-\frac {7 b^3 \left (a^2-b^2\right )^{3/4} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{9/2}}{2 a^{13/2} d}-\frac {2 b \left (a^2-b^2\right )^{7/4} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{9/2}}{a^{13/2} d}+\frac {7 b^3 \left (a^2-b^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{9/2}}{2 a^{13/2} d}-\frac {7 b^2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)} e^4}{5 a^6 d \sqrt {\sin (c+d x)}}-\frac {4 b^2 \left (8 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)} e^4}{5 a^6 d \sqrt {\sin (c+d x)}}+\frac {14 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)} e^4}{15 a^2 d \sqrt {\sin (c+d x)}}-\frac {14 \cos (c+d x) (e \sin (c+d x))^{3/2} e^3}{45 a^2 d}-\frac {7 b^2 (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2} e^3}{15 a^5 d}+\frac {4 b \left (5 \left (a^2-b^2\right )+3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2} e^3}{15 a^5 d}-\frac {2 \cos (c+d x) (e \sin (c+d x))^{7/2} e}{9 a^2 d}+\frac {4 b (e \sin (c+d x))^{7/2} e}{7 a^3 d}+\frac {b^2 (e \sin (c+d x))^{7/2} e}{a^3 d (b+a \cos (c+d x))}\) |
(-7*b^3*(a^2 - b^2)^(3/4)*e^(9/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/(( a^2 - b^2)^(1/4)*Sqrt[e])])/(2*a^(13/2)*d) + (2*b*(a^2 - b^2)^(7/4)*e^(9/2 )*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^( 13/2)*d) + (7*b^3*(a^2 - b^2)^(3/4)*e^(9/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(2*a^(13/2)*d) - (2*b*(a^2 - b^2)^( 7/4)*e^(9/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqr t[e])])/(a^(13/2)*d) + (7*b^4*(a^2 - b^2)*e^5*EllipticPi[(2*a)/(a - Sqrt[a ^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^7*(a - Sqrt[a^ 2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (2*b^2*(a^2 - b^2)^2*e^5*EllipticPi[(2 *a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^7 *(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (7*b^4*(a^2 - b^2)*e^5*El lipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^7*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (2*b^2*(a^2 - b^2)^2*e^5*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2 ]*Sqrt[Sin[c + d*x]])/(a^7*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (14*e^4*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(15*a^2*d* Sqrt[Sin[c + d*x]]) - (7*b^2*(3*a^2 - 5*b^2)*e^4*EllipticE[(c - Pi/2 + d*x )/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^6*d*Sqrt[Sin[c + d*x]]) - (4*b^2*(8*a^2 - 5*b^2)*e^4*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^ 6*d*Sqrt[Sin[c + d*x]]) - (14*e^3*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/...
3.3.41.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (G tQ[m, 0] || IntegerQ[n])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 44.55 (sec) , antiderivative size = 1725, normalized size of antiderivative = 1.61
(4*e*a*b*(-1/21/a^6*(e*sin(d*x+c))^(3/2)*e^2*(3*cos(d*x+c)^2*a^2-10*a^2+14 *b^2)+e^4/a^6*((1/4*a^2*b^2-1/4*b^4)*(e*sin(d*x+c))^(3/2)/(-a^2*e^2*cos(d* x+c)^2+b^2*e^2)+1/4*(a^4-15/4*a^2*b^2+11/4*b^4)/a^2/(e^2*(a^2-b^2)/a^2)^(1 /4)*(2*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4))-ln(((e*sin(d *x+c))^(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d*x+c))^(1/2)-(e^2*(a^2-b^ 2)/a^2)^(1/4))))))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*e^5*(-1/45/a^6/(cos(d *x+c)^2*e*sin(d*x+c))^(1/2)*(10*a^4*cos(d*x+c)^6+42*(-sin(d*x+c)+1)^(1/2)* (2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/ 2*2^(1/2))*a^4-432*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c) ^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2*b^2+450*(-sin(d*x+ c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c) +1)^(1/2),1/2*2^(1/2))*b^4-21*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2) *sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^4+216*a^2 *b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipti cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-225*b^4*(-sin(d*x+c)+1)^(1/2)*(2*sin (d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1 /2))-34*cos(d*x+c)^4*a^4+54*cos(d*x+c)^4*a^2*b^2+24*cos(d*x+c)^2*a^4-54*co s(d*x+c)^2*a^2*b^2)+2*b^4*(a^4-2*a^2*b^2+b^4)/a^6*(-1/2*a^2/e/b^2/(a^2-b^2 )*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*a^2+b^2)+1/2 /b^2/(a^2-b^2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^...
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{9/2}}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]