Integrand size = 25, antiderivative size = 1089 \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=-\frac {7 \sqrt {a} b^3 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 \left (a^2-b^2\right )^{11/4} d e^{5/2}}-\frac {2 \sqrt {a} b \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}-\frac {7 \sqrt {a} b^3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 \left (a^2-b^2\right )^{11/4} d e^{5/2}}-\frac {2 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}-\frac {2 \cos (c+d x)}{3 a^2 d e (e \sin (c+d x))^{3/2}}+\frac {b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x)) (e \sin (c+d x))^{3/2}}+\frac {4 b (a-b \cos (c+d x))}{3 a^2 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {b^2 \left (7 a b-\left (5 a^2+2 b^2\right ) \cos (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^2 d e^2 \sqrt {e \sin (c+d x)}}+\frac {4 b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {b^2 \left (5 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^2 \left (a^2-b^2\right )^2 d e^2 \sqrt {e \sin (c+d x)}}+\frac {7 b^4 \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 \left (a^2-b^2\right )^2 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \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)}}{\left (a^2-b^2\right ) \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {7 b^4 \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 \left (a^2-b^2\right )^2 \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \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)}}{\left (a^2-b^2\right ) \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}} \]
-2/3*cos(d*x+c)/a^2/d/e/(e*sin(d*x+c))^(3/2)+b^2/a/(a^2-b^2)/d/e/(b+a*cos( d*x+c))/(e*sin(d*x+c))^(3/2)+4/3*b*(a-b*cos(d*x+c))/a^2/(a^2-b^2)/d/e/(e*s in(d*x+c))^(3/2)+1/3*b^2*(7*a*b-(5*a^2+2*b^2)*cos(d*x+c))/a^2/(a^2-b^2)^2/ d/e/(e*sin(d*x+c))^(3/2)-7/2*b^3*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2- b^2)^(1/4)/e^(1/2))*a^(1/2)/(a^2-b^2)^(11/4)/d/e^(5/2)-2*b*arctan(a^(1/2)* (e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*a^(1/2)/(a^2-b^2)^(7/4)/d/e^ (5/2)-7/2*b^3*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2) )*a^(1/2)/(a^2-b^2)^(11/4)/d/e^(5/2)-2*b*arctanh(a^(1/2)*(e*sin(d*x+c))^(1 /2)/(a^2-b^2)^(1/4)/e^(1/2))*a^(1/2)/(a^2-b^2)^(7/4)/d/e^(5/2)-2/3*(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)*EllipticF(cos(1/2*c +1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^2/d/e^2/(e*sin(d*x+c))^(1/2)- 4/3*b^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)*Elli pticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^2/(a^2-b^2)/d/ e^2/(e*sin(d*x+c))^(1/2)-1/3*b^2*(5*a^2+2*b^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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^( 1/2))*sin(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/e^2/(e*sin(d*x+c))^(1/2)-7/2*b^4* (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^2-b^2)^2/d/e^2/(a^2-b^2-a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-2*b^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)*EllipticP...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 12.52 (sec) , antiderivative size = 1320, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \]
-1/6*((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Sin[c + d*x]^(5/2)*((2*(-2*a^3 - 5*a*b^2)*Cos[c + d*x]^2*(b + a*Sqrt[1 - Sin[c + d*x]^2])*((b*(-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]]))/(4*Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(3/4)) - (5*a*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((5*(a^ 2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/ (a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (a^2*S in[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[ c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2* (-1 + Sin[c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (2 *(10*a^2*b + 4*b^3)*Cos[c + d*x]*(b + a*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[a]*(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]]))/(a^2 - b^2)^(3/...
Time = 3.15 (sec) , antiderivative size = 1089, 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.
| \(\displaystyle \int \frac {1}{(e \sin (c+d x))^{5/2} (a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2} \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle \int \left (\frac {b^2}{a^2 (e \sin (c+d x))^{5/2} (-a \cos (c+d x)-b)^2}+\frac {2 b}{a^2 (e \sin (c+d x))^{5/2} (-a \cos (c+d x)-b)}+\frac {1}{a^2 (e \sin (c+d x))^{5/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {7 \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 \left (a^2-b^2\right )^2 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {7 \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 \left (a^2-b^2\right )^2 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {7 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 \left (a^2-b^2\right )^{11/4} d e^{5/2}}-\frac {7 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 \left (a^2-b^2\right )^{11/4} d e^{5/2}}+\frac {\left (5 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{3 a^2 \left (a^2-b^2\right )^2 d e^2 \sqrt {e \sin (c+d x)}}+\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{3 a^2 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {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)} b^2}{\left (a^2-b^2\right ) \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {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)} b^2}{\left (a^2-b^2\right ) \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {\left (7 a b-\left (5 a^2+2 b^2\right ) \cos (c+d x)\right ) b^2}{3 a^2 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{3/2}}+\frac {b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x)) (e \sin (c+d x))^{3/2}}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}+\frac {4 (a-b \cos (c+d x)) b}{3 a^2 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^2 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 a^2 d e (e \sin (c+d x))^{3/2}}\) |
(-7*Sqrt[a]*b^3*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*S qrt[e])])/(2*(a^2 - b^2)^(11/4)*d*e^(5/2)) - (2*Sqrt[a]*b*ArcTan[(Sqrt[a]* Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/((a^2 - b^2)^(7/4)*d*e ^(5/2)) - (7*Sqrt[a]*b^3*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^ 2)^(1/4)*Sqrt[e])])/(2*(a^2 - b^2)^(11/4)*d*e^(5/2)) - (2*Sqrt[a]*b*ArcTan h[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/((a^2 - b^2 )^(7/4)*d*e^(5/2)) - (2*Cos[c + d*x])/(3*a^2*d*e*(e*Sin[c + d*x])^(3/2)) + b^2/(a*(a^2 - b^2)*d*e*(b + a*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2)) + (4* b*(a - b*Cos[c + d*x]))/(3*a^2*(a^2 - b^2)*d*e*(e*Sin[c + d*x])^(3/2)) + ( b^2*(7*a*b - (5*a^2 + 2*b^2)*Cos[c + d*x]))/(3*a^2*(a^2 - b^2)^2*d*e*(e*Si n[c + d*x])^(3/2)) + (2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x] ])/(3*a^2*d*e^2*Sqrt[e*Sin[c + d*x]]) + (4*b^2*EllipticF[(c - Pi/2 + d*x)/ 2, 2]*Sqrt[Sin[c + d*x]])/(3*a^2*(a^2 - b^2)*d*e^2*Sqrt[e*Sin[c + d*x]]) + (b^2*(5*a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]]) /(3*a^2*(a^2 - b^2)^2*d*e^2*Sqrt[e*Sin[c + d*x]]) + (7*b^4*EllipticPi[(2*a )/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*(a^ 2 - b^2)^2*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*e^2*Sqrt[e*Sin[c + d*x]]) + ( 2*b^2*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[ Sin[c + d*x]])/((a^2 - b^2)*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*e^2*Sqrt[e*S in[c + d*x]]) + (7*b^4*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/...
3.3.48.3.1 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 = 31.29 (sec) , antiderivative size = 1331, normalized size of antiderivative = 1.22
(4*e*a*b*(1/3/e^2/(a^2-b^2)^2/(e*sin(d*x+c))^(3/2)+1/e^2/(a-b)^2/(a+b)^2*( 1/4*(e*sin(d*x+c))^(1/2)*b^2/(-a^2*e^2*cos(d*x+c)^2+b^2*e^2)+1/16*(4*a^2+3 *b^2)*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*(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)))+2*arctan((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^2*(1/3*(a^2+b^2)/(a^2-b^2)^2/(cos(d*x+c)^2*e*s in(d*x+c))^(1/2)/(cos(d*x+c)^2-1)*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^ (1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*cos( d*x+c)^2*sin(d*x+c))+2*b^4/(a-b)/(a+b)*(-1/2*a^2/e/b^2/(a^2-b^2)*(cos(d*x+ c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*a^2+b^2)-1/4/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)^(1/2)/(cos(d*x+c)^2*e*sin (d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-1/4/b^2/(a^2-b ^2)^(3/2)*a*(-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))+5/8/(a^2-b^2)^(3/2)/a* (-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))+1/4/b^2/(a^2-b^2)^(3/2)*a*(-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/...
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]