Integrand size = 25, antiderivative size = 1054 \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {5 b^3 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 \sqrt {a} \left (a^2-b^2\right )^{9/4} d e^{3/2}}+\frac {2 b \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{5/4} d e^{3/2}}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 \sqrt {a} \left (a^2-b^2\right )^{9/4} d e^{3/2}}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{5/4} d e^{3/2}}-\frac {2 \cos (c+d x)}{a^2 d e \sqrt {e \sin (c+d x)}}+\frac {b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}}+\frac {4 b (a-b \cos (c+d x))}{a^2 \left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}+\frac {b^2 \left (5 a b-\left (3 a^2+2 b^2\right ) \cos (c+d x)\right )}{a^2 \left (a^2-b^2\right )^2 d e \sqrt {e \sin (c+d x)}}-\frac {5 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 a \left (a^2-b^2\right )^2 \left (a-\sqrt {a^2-b^2}\right ) d e \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)}}{a \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {5 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 a \left (a^2-b^2\right )^2 \left (a+\sqrt {a^2-b^2}\right ) d e \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)}}{a \left (a^2-b^2\right ) \left (a+\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {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^2 d e^2 \sqrt {\sin (c+d x)}}-\frac {4 b^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^2 \left (a^2-b^2\right ) d e^2 \sqrt {\sin (c+d x)}}-\frac {b^2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a^2 \left (a^2-b^2\right )^2 d e^2 \sqrt {\sin (c+d x)}} \]
5/2*b^3*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/(a^2- b^2)^(9/4)/d/e^(3/2)/a^(1/2)+2*b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2- b^2)^(1/4)/e^(1/2))/(a^2-b^2)^(5/4)/d/e^(3/2)/a^(1/2)-5/2*b^3*arctanh(a^(1 /2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/(a^2-b^2)^(9/4)/d/e^(3/2 )/a^(1/2)-2*b*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2) )/(a^2-b^2)^(5/4)/d/e^(3/2)/a^(1/2)-2*cos(d*x+c)/a^2/d/e/(e*sin(d*x+c))^(1 /2)+b^2/a/(a^2-b^2)/d/e/(b+a*cos(d*x+c))/(e*sin(d*x+c))^(1/2)+4*b*(a-b*cos (d*x+c))/a^2/(a^2-b^2)/d/e/(e*sin(d*x+c))^(1/2)+b^2*(5*a*b-(3*a^2+2*b^2)*c os(d*x+c))/a^2/(a^2-b^2)^2/d/e/(e*sin(d*x+c))^(1/2)+5/2*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(cos(1/2*c+1/4*P i+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a/(a^2-b^2)^2 /d/e/(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)*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/(a^2-b^2)/d/e/(a-(a ^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+5/2*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(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/(a^2-b^2)^2/d/e/(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)/si n(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/(a^2-b^2)/d/e/(a+(a^2-b^2)^(1/2)...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 7.23 (sec) , antiderivative size = 772, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x)) \left (-\frac {\left (a^2-b^2\right ) \left (b+a \sqrt {\cos ^2(c+d x)}\right ) \sec ^3(c+d x) \sin ^{\frac {3}{2}}(c+d x) \left (\left (2 a^3+3 a b^2\right ) \cos (c+d x) \left (3 \sqrt {2} b \left (-a^2+b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )+8 a^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )+(1+i) a \left (6 a^2 b+4 b^3\right ) \sqrt {\cos ^2(c+d x)} \left (3 \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )-(4-4 i) \sqrt {a} b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )\right )}{a^{3/2} (a-b)^2 (a+b)^2}+24 \left (-2 (b+a \cos (c+d x)) \left (-2 a b+\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)+a b^2 \sin (c+d x)\right ) \tan ^2(c+d x)\right )}{24 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \]
((b + a*Cos[c + d*x])*(-(((a^2 - b^2)*(b + a*Sqrt[Cos[c + d*x]^2])*Sec[c + d*x]^3*Sin[c + d*x]^(3/2)*((2*a^3 + 3*a*b^2)*Cos[c + d*x]*(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)*Sq rt[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)*App ellF1[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)) + (1 + I)*a*(6*a^2*b + 4*b^3)*Sqrt[Cos[c + d*x]^2]*(3 *(a^2 - b^2)^(3/4)*(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]]) - (4 - 4*I)*Sqr t[a]*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))))/(a^(3/2)*(a - b)^2*(a + b)^2)) + 24*(-2*(b + a*Cos[c + d*x])*(-2*a*b + (a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x] + a*b^ 2*Sin[c + d*x])*Tan[c + d*x]^2))/(24*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^ 2*(e*Sin[c + d*x])^(3/2))
Time = 2.99 (sec) , antiderivative size = 1054, 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.
| \(\displaystyle \int \frac {1}{(e \sin (c+d x))^{3/2} (a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2} \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2} (-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2} \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2}{a^2 (e \sin (c+d x))^{3/2} (-a \cos (c+d x)-b)^2}+\frac {2 b}{a^2 (e \sin (c+d x))^{3/2} (-a \cos (c+d x)-b)}+\frac {1}{a^2 (e \sin (c+d x))^{3/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {5 \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 \left (a^2-b^2\right )^2 \left (a-\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {5 \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 \left (a^2-b^2\right )^2 \left (a+\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}+\frac {5 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 \sqrt {a} \left (a^2-b^2\right )^{9/4} d e^{3/2}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 \sqrt {a} \left (a^2-b^2\right )^{9/4} d e^{3/2}}-\frac {\left (3 a^2+2 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)} b^2}{a^2 \left (a^2-b^2\right )^2 d e^2 \sqrt {\sin (c+d x)}}-\frac {4 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^2 \left (a^2-b^2\right ) d e^2 \sqrt {\sin (c+d x)}}+\frac {\left (5 a b-\left (3 a^2+2 b^2\right ) \cos (c+d x)\right ) b^2}{a^2 \left (a^2-b^2\right )^2 d e \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}{a \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) d e \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}{a \left (a^2-b^2\right ) \left (a+\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}+\frac {b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}}+\frac {2 \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{\sqrt {a} \left (a^2-b^2\right )^{5/4} d e^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{\sqrt {a} \left (a^2-b^2\right )^{5/4} d e^{3/2}}+\frac {4 (a-b \cos (c+d x)) b}{a^2 \left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a^2 d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{a^2 d e \sqrt {e \sin (c+d x)}}\) |
(5*b^3*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])]) /(2*Sqrt[a]*(a^2 - b^2)^(9/4)*d*e^(3/2)) + (2*b*ArcTan[(Sqrt[a]*Sqrt[e*Sin [c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(Sqrt[a]*(a^2 - b^2)^(5/4)*d*e^( 3/2)) - (5*b^3*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*S qrt[e])])/(2*Sqrt[a]*(a^2 - b^2)^(9/4)*d*e^(3/2)) - (2*b*ArcTanh[(Sqrt[a]* Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(Sqrt[a]*(a^2 - b^2)^( 5/4)*d*e^(3/2)) - (2*Cos[c + d*x])/(a^2*d*e*Sqrt[e*Sin[c + d*x]]) + b^2/(a *(a^2 - b^2)*d*e*(b + a*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]]) + (4*b*(a - b* Cos[c + d*x]))/(a^2*(a^2 - b^2)*d*e*Sqrt[e*Sin[c + d*x]]) + (b^2*(5*a*b - (3*a^2 + 2*b^2)*Cos[c + d*x]))/(a^2*(a^2 - b^2)^2*d*e*Sqrt[e*Sin[c + d*x]] ) - (5*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*(a^2 - b^2)^2*(a - Sqrt[a^2 - b^2])*d*e*Sqrt[e*Si n[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*(a^2 - b^2)*(a - Sqrt[a^2 - b^2])*d*e*Sq rt[e*Sin[c + d*x]]) - (5*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*(a^2 - b^2)^2*(a + Sqrt[a^2 - b ^2])*d*e*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*(a^2 - b^2)*(a + Sqrt[ a^2 - b^2])*d*e*Sqrt[e*Sin[c + d*x]]) - (2*EllipticE[(c - Pi/2 + d*x)/2, 2 ]*Sqrt[e*Sin[c + d*x]])/(a^2*d*e^2*Sqrt[Sin[c + d*x]]) - (4*b^2*Ellipti...
3.3.47.3.1 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 = 22.48 (sec) , antiderivative size = 1417, normalized size of antiderivative = 1.34
(-2*e*a*b*(-2/e^2/(a^2-b^2)^2/(e*sin(d*x+c))^(1/2)-2/e^2/(a-b)^2/(a+b)^2*( 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+1/4* 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*((a^2+b^2)/(a^2-b^2)^2*(2*(-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))-(-s in(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)^2)/(cos(d*x+c)^2*e*sin(d*x+c))^ (1/2)+2*b^4/(a-b)/(a+b)*(-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)^(1/2)/(cos(d*x+c)^2*e*sin(d*x +c))^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-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)*(-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))+3/8/(a^2-b^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*s in(d*x+c))^(1/2)/(1-(a^2-b^2)^(1/2)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2)...
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]