Integrand size = 25, antiderivative size = 850 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=-\frac {3 b^3 e^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{9/2} \sqrt [4]{a^2-b^2} d}+\frac {2 b \left (a^2-b^2\right )^{3/4} e^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{9/2} d}+\frac {3 b^3 e^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{9/2} \sqrt [4]{a^2-b^2} d}-\frac {2 b \left (a^2-b^2\right )^{3/4} e^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{9/2} d}+\frac {3 b^4 e^3 \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^5 \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 ) e^3 \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^5 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 b^4 e^3 \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^5 \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 ) e^3 \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^5 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {6 e^2 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^2 d \sqrt {\sin (c+d x)}}-\frac {7 b^2 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a^4 d \sqrt {\sin (c+d x)}}+\frac {4 b e (e \sin (c+d x))^{3/2}}{3 a^3 d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d}+\frac {b^2 e (e \sin (c+d x))^{3/2}}{a^3 d (b+a \cos (c+d x))} \]
-3/2*b^3*e^(5/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/ 2))/a^(9/2)/(a^2-b^2)^(1/4)/d+2*b*(a^2-b^2)^(3/4)*e^(5/2)*arctan(a^(1/2)*( e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(9/2)/d+3/2*b^3*e^(5/2)*arc tanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(9/2)/(a^2-b^ 2)^(1/4)/d-2*b*(a^2-b^2)^(3/4)*e^(5/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2 )/(a^2-b^2)^(1/4)/e^(1/2))/a^(9/2)/d+4/3*b*e*(e*sin(d*x+c))^(3/2)/a^3/d-2/ 5*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/a^2/d+b^2*e*(e*sin(d*x+c))^(3/2)/a^3/d /(b+a*cos(d*x+c))-3/2*b^4*e^3*(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^5/d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1 /2)+2*b^2*(a^2-b^2)*e^3*(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^5/d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-3/ 2*b^4*e^3*(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)*El lipticPi(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^5/d/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2*b^2*(a^2-b^2)* e^3*(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)*Elliptic Pi(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^5/d/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-6/5*e^2*(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)*EllipticE(cos(1/2*c+1/4...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 15.51 (sec) , antiderivative size = 886, normalized size of antiderivative = 1.04 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(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))^{5/2} \left (\frac {\left (-6 a^2+35 b^2\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 {28 a b \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 )}{10 a^3 d (a+b \sec (c+d x))^2 \sin ^{\frac {5}{2}}(c+d x)}+\frac {(b+a \cos (c+d x))^2 \csc ^2(c+d x) \sec ^2(c+d x) (e \sin (c+d x))^{5/2} \left (\frac {4 b \sin (c+d x)}{3 a^3}+\frac {b^2 \sin (c+d x)}{a^3 (b+a \cos (c+d x))}-\frac {\sin (2 (c+d x))}{5 a^2}\right )}{d (a+b \sec (c+d x))^2} \]
-1/10*((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(e*Sin[c + d*x])^(5/2)*(((-6* a^2 + 35*b^2)*Cos[c + d*x]^2*(3*Sqrt[2]*b*(-a^2 + b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ( Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 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]] + 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*Sq rt[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)) + (28*a*b*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]] + L og[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])))/(a^3*d*(a + b*Sec[c + d*x])^2*Sin[c + d*x] ^(5/2)) + ((b + a*Cos[c + d*x])^2*Csc[c + d*x]^2*Sec[c + d*x]^2*(e*Sin[c + d*x])^(5/2)*((4*b*Sin[c + d*x])/(3*a^3) + (b^2*Sin[c + d*x])/(a^3*(b +...
Time = 2.37 (sec) , antiderivative size = 850, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{5/2}}{(-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2 (e \sin (c+d x))^{5/2}}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b (e \sin (c+d x))^{5/2}}{a^2 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^{5/2}}{a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 e^3 \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)} b^4}{2 a^5 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 e^3 \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)} b^4}{2 a^5 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 e^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{9/2} \sqrt [4]{a^2-b^2} d}+\frac {3 e^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{9/2} \sqrt [4]{a^2-b^2} d}+\frac {e (e \sin (c+d x))^{3/2} b^2}{a^3 d (b+a \cos (c+d x))}-\frac {7 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)} b^2}{a^4 d \sqrt {\sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) e^3 \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)} b^2}{a^5 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right ) e^3 \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)} b^2}{a^5 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2} b}{3 a^3 d}+\frac {2 \left (a^2-b^2\right )^{3/4} e^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{9/2} d}-\frac {2 \left (a^2-b^2\right )^{3/4} e^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{9/2} d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d}+\frac {6 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}\) |
(-3*b^3*e^(5/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*S qrt[e])])/(2*a^(9/2)*(a^2 - b^2)^(1/4)*d) + (2*b*(a^2 - b^2)^(3/4)*e^(5/2) *ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(9 /2)*d) + (3*b^3*e^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2 )^(1/4)*Sqrt[e])])/(2*a^(9/2)*(a^2 - b^2)^(1/4)*d) - (2*b*(a^2 - b^2)^(3/4 )*e^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e ])])/(a^(9/2)*d) + (3*b^4*e^3*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^5*(a - Sqrt[a^2 - b^2])*d*Sqrt [e*Sin[c + d*x]]) - (2*b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^5*(a - Sqrt[a^2 - b ^2])*d*Sqrt[e*Sin[c + d*x]]) + (3*b^4*e^3*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^5*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (2*b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^5*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (6*e^2*EllipticE[(c - Pi/2 + d* x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^2*d*Sqrt[Sin[c + d*x]]) - (7*b^2*e^2*E llipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(a^4*d*Sqrt[Sin[c + d*x]]) + (4*b*e*(e*Sin[c + d*x])^(3/2))/(3*a^3*d) - (2*e*Cos[c + d*x]*(e*S in[c + d*x])^(3/2))/(5*a^2*d) + (b^2*e*(e*Sin[c + d*x])^(3/2))/(a^3*d*(b + a*Cos[c + d*x]))
3.3.43.3.1 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 = 34.74 (sec) , antiderivative size = 1482, normalized size of antiderivative = 1.74
(4*e*a*b*(1/3*(e*sin(d*x+c))^(3/2)/a^4+e^2/a^4*(1/4*(e*sin(d*x+c))^(3/2)*b ^2/(-a^2*e^2*cos(d*x+c)^2+b^2*e^2)+1/4*(a^2-7/4*b^2)/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^3*(-1/5/a^2/(c os(d*x+c)^2*e*sin(d*x+c))^(1/2)*(6*(-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))-3*(-si n(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))-2*cos(d*x+c)^4+2*cos(d*x+c)^2)+3*b^2/a^4*(-si n(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e* sin(d*x+c))^(1/2)*(2*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-Elliptic F((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2)))-b^2*(3*a^2-5*b^2)/a^4*(-1/2/a^2*(-si n(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e* sin(d*x+c))^(1/2)/(1-(a^2-b^2)^(1/2)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1 /(1-(a^2-b^2)^(1/2)/a),1/2*2^(1/2))-1/2/a^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d *x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(1+(a^2- b^2)^(1/2)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(a^2-b^2)^(1/2)/a),1/2 *2^(1/2)))+2*b^4*(a^2-b^2)/a^4*(-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)*(-si n(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2...
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]