Integrand size = 25, antiderivative size = 1101 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\frac {5 b^3 \sqrt [4]{a^2-b^2} e^{7/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{11/2} d}-\frac {2 b \left (a^2-b^2\right )^{5/4} e^{7/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{11/2} d}+\frac {5 b^3 \sqrt [4]{a^2-b^2} e^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{2 a^{11/2} d}-\frac {2 b \left (a^2-b^2\right )^{5/4} e^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{11/2} d}+\frac {10 e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d \sqrt {e \sin (c+d x)}}-\frac {5 b^2 \left (a^2-3 b^2\right ) e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^6 d \sqrt {e \sin (c+d x)}}-\frac {4 b^2 \left (4 a^2-3 b^2\right ) e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a^6 d \sqrt {e \sin (c+d x)}}-\frac {5 b^4 \left (a^2-b^2\right ) e^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^6 \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 )^2 e^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)}}{a^6 \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 b^4 \left (a^2-b^2\right ) e^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^6 \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 )^2 e^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)}}{a^6 \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {10 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}-\frac {5 b^2 e^3 (3 b-a \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 a^5 d}+\frac {4 b e^3 \left (3 \left (a^2-b^2\right )+a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{3 a^5 d}+\frac {4 b e (e \sin (c+d x))^{5/2}}{5 a^3 d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 a^2 d}+\frac {b^2 e (e \sin (c+d x))^{5/2}}{a^3 d (b+a \cos (c+d x))} \]
5/2*b^3*(a^2-b^2)^(1/4)*e^(7/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b ^2)^(1/4)/e^(1/2))/a^(11/2)/d-2*b*(a^2-b^2)^(5/4)*e^(7/2)*arctan(a^(1/2)*( e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(11/2)/d+5/2*b^3*(a^2-b^2)^ (1/4)*e^(7/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2) )/a^(11/2)/d-2*b*(a^2-b^2)^(5/4)*e^(7/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1 /2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(11/2)/d+4/5*b*e*(e*sin(d*x+c))^(5/2)/a^3/d -2/7*e*cos(d*x+c)*(e*sin(d*x+c))^(5/2)/a^2/d+b^2*e*(e*sin(d*x+c))^(5/2)/a^ 3/d/(b+a*cos(d*x+c))-10/21*e^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)*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)+5/3*b^2*(a^2-3*b^2)*e^4*(sin(1/2*c+1/4*P i+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^6/d/(e*sin(d*x+c))^(1/2)+4/3*b^2*(4*a^ 2-3*b^2)*e^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) *EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^6/d/(e*si n(d*x+c))^(1/2)+5/2*b^4*(a^2-b^2)*e^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^6/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)^2*e^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^6/d/(a^2-b^2-a*(a^2-b^2)^(...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 16.73 (sec) , antiderivative size = 2095, normalized size of antiderivative = 1.90 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^2*(-1/42*((23*a^2 - 84*b^2)*Cos[c + d*x])/a^4 - (b^2 *(-a^2 + b^2))/(a^5*(b + a*Cos[c + d*x])) - (2*b*Cos[2*(c + d*x)])/(5*a^3) + Cos[3*(c + d*x)]/(14*a^2))*Csc[c + d*x]^3*Sec[c + d*x]^2*(e*Sin[c + d*x ])^(7/2))/(d*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x ]^2*(e*Sin[c + d*x])^(7/2)*((2*(50*a^4 - 273*a^2*b^2 + 105*b^4)*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]*Sq rt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sqrt[2]*Sqr t[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)*Appe llF1[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*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)) + (2*(-139*a^3*b + 210*a*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)] - ...
Time = 3.30 (sec) , antiderivative size = 1101, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{7/2}}{(-a \cos (c+d x)-b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (\frac {b^2 (e \sin (c+d x))^{7/2}}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b (e \sin (c+d x))^{7/2}}{a^2 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^{7/2}}{a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {5 b^2 \left (a^2-3 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^4}{3 a^6 d \sqrt {e \sin (c+d x)}}-\frac {4 b^2 \left (4 a^2-3 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^4}{3 a^6 d \sqrt {e \sin (c+d x)}}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right ) \sqrt {\sin (c+d x)} e^4}{21 a^2 d \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \left (a^2-b^2\right )^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)} e^4}{a^6 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 b^4 \left (a^2-b^2\right ) \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)} e^4}{2 a^6 \left (a^2-\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 b^2 \left (a^2-b^2\right )^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)} e^4}{a^6 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 b^4 \left (a^2-b^2\right ) \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)} e^4}{2 a^6 \left (a^2+\sqrt {a^2-b^2} a-b^2\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 b \left (a^2-b^2\right )^{5/4} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{7/2}}{a^{11/2} d}+\frac {5 b^3 \sqrt [4]{a^2-b^2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{7/2}}{2 a^{11/2} d}-\frac {2 b \left (a^2-b^2\right )^{5/4} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{7/2}}{a^{11/2} d}+\frac {5 b^3 \sqrt [4]{a^2-b^2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right ) e^{7/2}}{2 a^{11/2} d}-\frac {10 \cos (c+d x) \sqrt {e \sin (c+d x)} e^3}{21 a^2 d}-\frac {5 b^2 (3 b-a \cos (c+d x)) \sqrt {e \sin (c+d x)} e^3}{3 a^5 d}+\frac {4 b \left (3 \left (a^2-b^2\right )+a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)} e^3}{3 a^5 d}-\frac {2 \cos (c+d x) (e \sin (c+d x))^{5/2} e}{7 a^2 d}+\frac {4 b (e \sin (c+d x))^{5/2} e}{5 a^3 d}+\frac {b^2 (e \sin (c+d x))^{5/2} e}{a^3 d (b+a \cos (c+d x))}\) |
(5*b^3*(a^2 - b^2)^(1/4)*e^(7/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a ^2 - b^2)^(1/4)*Sqrt[e])])/(2*a^(11/2)*d) - (2*b*(a^2 - b^2)^(5/4)*e^(7/2) *ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(1 1/2)*d) + (5*b^3*(a^2 - b^2)^(1/4)*e^(7/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(2*a^(11/2)*d) - (2*b*(a^2 - b^2)^(5 /4)*e^(7/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt [e])])/(a^(11/2)*d) + (10*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a^2*d*Sqrt[e*Sin[c + d*x]]) - (5*b^2*(a^2 - 3*b^2)*e^4*Ellipt icF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*a^6*d*Sqrt[e*Sin[c + d*x ]]) - (4*b^2*(4*a^2 - 3*b^2)*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin [c + d*x]])/(3*a^6*d*Sqrt[e*Sin[c + d*x]]) - (5*b^4*(a^2 - b^2)*e^4*Ellipt icPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x] ])/(2*a^6*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*b^2 *(a^2 - b^2)^2*e^4*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x )/2, 2]*Sqrt[Sin[c + d*x]])/(a^6*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*Sqrt[e* Sin[c + d*x]]) - (5*b^4*(a^2 - b^2)*e^4*EllipticPi[(2*a)/(a + Sqrt[a^2 - b ^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*a^6*(a^2 - b^2 + a*Sqr t[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*b^2*(a^2 - b^2)^2*e^4*EllipticP i[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/ (a^6*(a^2 - b^2 + a*Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (10*e^3*...
3.3.42.3.1 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 = 45.23 (sec) , antiderivative size = 1494, normalized size of antiderivative = 1.36
(4*e*a*b*(-1/5/a^6*(e*sin(d*x+c))^(1/2)*e^2*(cos(d*x+c)^2*a^2-6*a^2+10*b^2 )+e^4/a^6*((1/4*a^2*b^2-1/4*b^4)*(e*sin(d*x+c))^(1/2)/(-a^2*e^2*cos(d*x+c) ^2+b^2*e^2)+1/16*(4*a^4-13*a^2*b^2+9*b^4)*(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^4*(-1/21/a ^6/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*(-6*a^4*cos(d*x+c)^4*sin(d*x+c)+5*(-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))*a^4-84*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)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/ 2))+105*b^4*(-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))+16*a^4*cos(d*x+c)^2*sin(d*x+c )-42*a^2*b^2*cos(d*x+c)^2*sin(d*x+c))+2*b^4*(a^4-2*a^2*b^2+b^4)/a^6*(-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/...
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]