Integrand size = 25, antiderivative size = 809 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\frac {b^3 \sqrt {e} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{5/2} \left (a^2-b^2\right )^{5/4} d}+\frac {2 b \sqrt {e} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{5/2} \sqrt [4]{a^2-b^2} d}-\frac {b^3 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{5/2} \left (a^2-b^2\right )^{5/4} d}-\frac {2 b \sqrt {e} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{5/2} \sqrt [4]{a^2-b^2} d}-\frac {2 b^2 e \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^3 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^4 e \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^3 \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 b^2 e \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^3 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^4 e \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^3 \left (a^2-b^2\right ) \left (a+\sqrt {a^2-b^2}\right ) d \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 \sqrt {\sin (c+d x)}}-\frac {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 \sqrt {\sin (c+d x)}}+\frac {b^2 (e \sin (c+d x))^{3/2}}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x))} \]
b^2*(e*sin(d*x+c))^(3/2)/a/(a^2-b^2)/d/e/(b+a*cos(d*x+c))+1/2*b^3*arctan(a ^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*e^(1/2)/a^(5/2)/(a^2- b^2)^(5/4)/d+2*b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/ 2))*e^(1/2)/a^(5/2)/(a^2-b^2)^(1/4)/d-1/2*b^3*arctanh(a^(1/2)*(e*sin(d*x+c ))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*e^(1/2)/a^(5/2)/(a^2-b^2)^(5/4)/d-2*b*ar ctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*e^(1/2)/a^(5/2 )/(a^2-b^2)^(1/4)/d+2*b^2*e*(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^3/d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2 )+1/2*b^4*e*(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^3/(a^2-b^2)/d/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2*b^ 2*e*(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^3/d/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+1/2*b^4*e*(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^3/(a^2-b^ 2)/d/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-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*Pi+1/2*d*x),2^ (1/2))*(e*sin(d*x+c))^(1/2)/a^2/d/sin(d*x+c)^(1/2)+b^2*(sin(1/2*c+1/4*P...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.87 (sec) , antiderivative size = 854, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \left (\frac {\left (-2 a^2+3 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 {4 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 )}{2 a (-a+b) (a+b) d (a+b \sec (c+d x))^2 \sqrt {\sin (c+d x)}}+\frac {b^2 (b+a \cos (c+d x)) \sec (c+d x) \sqrt {e \sin (c+d x)} \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Sqrt[e*Sin[c + d*x]]*(((-2*a^2 + 3* 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]*S qrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sq rt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]] + Lo g[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 - Si n[c + d*x]^2]))/(12*a^(3/2)*(a^2 - b^2)*(b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*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]*S qrt[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, Si n[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])))/(2*a*(-a + b)*(a + b)*d*(a + b*Sec[c + d*x])^2*Sqrt[ Sin[c + d*x]]) + (b^2*(b + a*Cos[c + d*x])*Sec[c + d*x]*Sqrt[e*Sin[c + d*x ]]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2)
Time = 2.16 (sec) , antiderivative size = 809, 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.
| \(\displaystyle \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) \sqrt {e \sin (c+d x)}}{(-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )}}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2 \sqrt {e \sin (c+d x)}}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b \sqrt {e \sin (c+d x)}}{a^2 (a \cos (c+d x)+b)}+\frac {\sqrt {e \sin (c+d x)}}{a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^4}{2 a^3 \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {e \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^4}{2 a^3 \left (a^2-b^2\right ) \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {\sqrt {e} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{5/2} \left (a^2-b^2\right )^{5/4} d}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{5/2} \left (a^2-b^2\right )^{5/4} d}+\frac {(e \sin (c+d x))^{3/2} b^2}{a \left (a^2-b^2\right ) d e (b+a \cos (c+d x))}-\frac {E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)} b^2}{a^2 \left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}-\frac {2 e \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^3 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 e \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^3 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{5/2} \sqrt [4]{a^2-b^2} d}-\frac {2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{5/2} \sqrt [4]{a^2-b^2} d}+\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a^2 d \sqrt {\sin (c+d x)}}\) |
(b^3*Sqrt[e]*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt [e])])/(2*a^(5/2)*(a^2 - b^2)^(5/4)*d) + (2*b*Sqrt[e]*ArcTan[(Sqrt[a]*Sqrt [e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(5/2)*(a^2 - b^2)^(1/4) *d) - (b^3*Sqrt[e]*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/ 4)*Sqrt[e])])/(2*a^(5/2)*(a^2 - b^2)^(5/4)*d) - (2*b*Sqrt[e]*ArcTanh[(Sqrt [a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(5/2)*(a^2 - b^ 2)^(1/4)*d) - (2*b^2*e*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^3*(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (b^4*e*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x) /2, 2]*Sqrt[Sin[c + d*x]])/(2*a^3*(a^2 - b^2)*(a - Sqrt[a^2 - b^2])*d*Sqrt [e*Sin[c + d*x]]) - (2*b^2*e*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^3*(a + Sqrt[a^2 - b^2])*d*Sqrt[e* Sin[c + d*x]]) - (b^4*e*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^3*(a^2 - b^2)*(a + Sqrt[a^2 - b^2])* d*Sqrt[e*Sin[c + d*x]]) + (2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(a^2*d*Sqrt[Sin[c + d*x]]) - (b^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(a^2*(a^2 - b^2)*d*Sqrt[Sin[c + d*x]]) + (b^2*(e* Sin[c + d*x])^(3/2))/(a*(a^2 - b^2)*d*e*(b + a*Cos[c + d*x]))
3.3.45.3.1 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 = 19.71 (sec) , antiderivative size = 1313, normalized size of antiderivative = 1.62
(-2*e*a*b*(-1/2*b^2/a^2/(a^2-b^2)*(e*sin(d*x+c))^(3/2)/(-a^2*e^2*cos(d*x+c )^2+b^2*e^2)-1/8*(4*a^2-3*b^2)/a^4/(a^2-b^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*(-1/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)*(2*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-EllipticF((-sin(d*x+c)+ 1)^(1/2),1/2*2^(1/2)))+2*b^4/a^2*(-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)/(co s(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*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)*(-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...
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sqrt {e \sin {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {e\,\sin \left (c+d\,x\right )}}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]