Integrand size = 25, antiderivative size = 882 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\frac {b^3 e^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{7/2} \left (a^2-b^2\right )^{3/4} d}-\frac {2 b \sqrt [4]{a^2-b^2} e^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{7/2} d}+\frac {b^3 e^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{7/2} \left (a^2-b^2\right )^{3/4} d}-\frac {2 b \sqrt [4]{a^2-b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{7/2} d}+\frac {2 e^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 \sqrt {e \sin (c+d x)}}-\frac {5 b^2 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^4 d \sqrt {e \sin (c+d x)}}-\frac {b^4 e^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)}}{2 a^4 \left (a^2-b^2-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^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^4 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^4 e^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)}}{2 a^4 \left (a^2-b^2+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^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^4 \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {4 b e \sqrt {e \sin (c+d x)}}{a^3 d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}+\frac {b^2 e \sqrt {e \sin (c+d x)}}{a^3 d (b+a \cos (c+d x))} \]
1/2*b^3*e^(3/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2 ))/a^(7/2)/(a^2-b^2)^(3/4)/d-2*b*(a^2-b^2)^(1/4)*e^(3/2)*arctan(a^(1/2)*(e *sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/d+1/2*b^3*e^(3/2)*arct anh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/(a^2-b^2 )^(3/4)/d-2*b*(a^2-b^2)^(1/4)*e^(3/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2) /(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/d-2/3*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)*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*sin(d*x+c))^(1/2)+5*b^2*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)*EllipticF(cos(1/2*c+1/4*Pi+1/ 2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^4/d/(e*sin(d*x+c))^(1/2)+1/2*b^4*e^2*(s in(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 ^4/d/(a^2-b^2-a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-2*b^2*(a^2-b^2)*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)*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^4/d/(a^2-b^2-a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+1/2*b^4*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)*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^4/d /(a^2-b^2+a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-2*b^2*(a^2-b^2)*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)*EllipticPi(co...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 16.51 (sec) , antiderivative size = 2012, normalized size of antiderivative = 2.28 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^2*((-2*Cos[c + d*x])/(3*a^2) + b^2/(a^3*(b + a*Cos[c + d*x])))*Csc[c + d*x]*Sec[c + d*x]^2*(e*Sin[c + d*x])^(3/2))/(d*(a + b*S ec[c + d*x])^2) - ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(e*Sin[c + d*x])^ (3/2)*((2*(-2*a^2 + 3*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*S in[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*Sin[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)) + (8*a*b*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[...
Time = 2.49 (sec) , antiderivative size = 882, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle \int \left (\frac {b^2 (e \sin (c+d x))^{3/2}}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b (e \sin (c+d x))^{3/2}}{a^2 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^{3/2}}{a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e^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^4}{2 a^4 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {e^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^4}{2 a^4 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b^3}{2 a^{7/2} \left (a^2-b^2\right )^{3/4} d}+\frac {e^{3/2} \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^{7/2} \left (a^2-b^2\right )^{3/4} d}+\frac {e \sqrt {e \sin (c+d x)} b^2}{a^3 d (b+a \cos (c+d x))}-\frac {5 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} b^2}{a^4 d \sqrt {e \sin (c+d x)}}+\frac {2 \left (a^2-b^2\right ) e^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^4 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 \left (a^2-b^2\right ) e^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^4 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 \sqrt [4]{a^2-b^2} e^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{7/2} d}-\frac {2 \sqrt [4]{a^2-b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) b}{a^{7/2} d}+\frac {4 e \sqrt {e \sin (c+d x)} b}{a^3 d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}+\frac {2 e^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 \sqrt {e \sin (c+d x)}}\) |
(b^3*e^(3/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt [e])])/(2*a^(7/2)*(a^2 - b^2)^(3/4)*d) - (2*b*(a^2 - b^2)^(1/4)*e^(3/2)*Ar cTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(7/2) *d) + (b^3*e^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/ 4)*Sqrt[e])])/(2*a^(7/2)*(a^2 - b^2)^(3/4)*d) - (2*b*(a^2 - b^2)^(1/4)*e^( 3/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/ (a^(7/2)*d) + (2*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/ (3*a^2*d*Sqrt[e*Sin[c + d*x]]) - (5*b^2*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^4*d*Sqrt[e*Sin[c + d*x]]) - (b^4*e^2*EllipticPi[ (2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2 *a^4*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*b^2*(a^2 - b^2)*e^2*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2] *Sqrt[Sin[c + d*x]])/(a^4*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (b^4*e^2*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x )/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^4*(a^2 - b^2 + a*Sqrt[a^2 - b^2])*d*Sqrt[ e*Sin[c + d*x]]) + (2*b^2*(a^2 - b^2)*e^2*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^4*(a^2 - b^2 + a*Sqr t[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (4*b*e*Sqrt[e*Sin[c + d*x]])/(a^3* d) - (2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*a^2*d) + (b^2*e*Sqrt[e*Sin [c + d*x]])/(a^3*d*(b + a*Cos[c + d*x]))
3.3.44.3.1 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 = 26.56 (sec) , antiderivative size = 1350, normalized size of antiderivative = 1.53
(4*e*a*b*((e*sin(d*x+c))^(1/2)/a^4+e^2/a^4*(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^2-5*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/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*((-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))+2*cos(d*x+c)^2*sin(d*x+c))+3*b^2/a^4*(-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/a^4*b^2*(3*a^2-5*b^2)*(-1/2/(a^2-b^2)^ (1/2)/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))+1/2/(a^2-b^2)^(1/2)/a*(-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)))+2*b^4*(a^2-b^2)/a^4*(-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...
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]