Integrand size = 25, antiderivative size = 224 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac {16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 \cos (c+d x)}{15 a^2 d e \sqrt {e \sin (c+d x)}}-\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 a^2 d e^2 \sqrt {\sin (c+d x)}} \]
4/9*e^3/a^2/d/(e*sin(d*x+c))^(9/2)-2/9*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c)) ^(9/2)-2/9*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(9/2)-4/5*e/a^2/d/(e*sin( d*x+c))^(5/2)+16/45*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(5/2)-4/15*cos(d*x+c )/a^2/d/e/(e*sin(d*x+c))^(1/2)+4/15*(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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*si n(d*x+c))^(1/2)/a^2/d/e^2/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.99 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (-31-40 \cos (c+d x)-19 \cos (2 (c+d x))+e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right )^4 \sqrt {1-e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 i (c+d x)}\right )+16 i \sin (c+d x)+13 i \sin (2 (c+d x))\right )}{180 a^2 d e \sqrt {e \sin (c+d x)}} \]
(Sec[(c + d*x)/2]^4*(Cos[c + d*x] + I*Sin[c + d*x])*(-31 - 40*Cos[c + d*x] - 19*Cos[2*(c + d*x)] + ((1 + E^(I*(c + d*x)))^4*Sqrt[1 - E^((2*I)*(c + d *x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + (16*I)*Sin[c + d*x] + (13*I)*Sin[2*(c + d*x)]))/(180*a^2*d*e*Sqrt [e*Sin[c + d*x]])
Time = 0.98 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4360, 3042, 3354, 3042, 3352, 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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \sin (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2 (e \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{11/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{11/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{11/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{11/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{11/2}}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \left (-\frac {4 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 d e^6 \sqrt {\sin (c+d x)}}-\frac {4 a^2 \cos (c+d x)}{15 d e^5 \sqrt {e \sin (c+d x)}}-\frac {4 a^2}{5 d e^3 (e \sin (c+d x))^{5/2}}+\frac {16 a^2 \cos (c+d x)}{45 d e^3 (e \sin (c+d x))^{5/2}}+\frac {4 a^2}{9 d e (e \sin (c+d x))^{9/2}}-\frac {2 a^2 \cos ^3(c+d x)}{9 d e (e \sin (c+d x))^{9/2}}-\frac {2 a^2 \cos (c+d x)}{9 d e (e \sin (c+d x))^{9/2}}\right )}{a^4}\) |
(e^4*((4*a^2)/(9*d*e*(e*Sin[c + d*x])^(9/2)) - (2*a^2*Cos[c + d*x])/(9*d*e *(e*Sin[c + d*x])^(9/2)) - (2*a^2*Cos[c + d*x]^3)/(9*d*e*(e*Sin[c + d*x])^ (9/2)) - (4*a^2)/(5*d*e^3*(e*Sin[c + d*x])^(5/2)) + (16*a^2*Cos[c + d*x])/ (45*d*e^3*(e*Sin[c + d*x])^(5/2)) - (4*a^2*Cos[c + d*x])/(15*d*e^5*Sqrt[e* Sin[c + d*x]]) - (4*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d* x]])/(15*d*e^6*Sqrt[Sin[c + d*x]])))/a^4
3.2.32.3.1 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] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
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 = 6.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.95
| method | result | size |
| default | \(\frac {\frac {4 e^{3} \left (9 \cos \left (d x +c \right )^{2}-4\right )}{45 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}+\frac {\frac {4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {11}{2}} \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{15}-\frac {2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {11}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{15}+\frac {4 \sin \left (d x +c \right )^{7}}{15}-\frac {38 \sin \left (d x +c \right )^{5}}{45}+\frac {46 \sin \left (d x +c \right )^{3}}{45}-\frac {4 \sin \left (d x +c \right )}{9}}{e \,a^{2} \sin \left (d x +c \right )^{5} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(213\) |
(4/45*e^3/a^2/(e*sin(d*x+c))^(9/2)*(9*cos(d*x+c)^2-4)+2/45/e*(6*(-sin(d*x+ c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(11/2)*EllipticE((-sin(d*x+c )+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin (d*x+c)^(11/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+6*sin(d*x+c)^7 -19*sin(d*x+c)^5+23*sin(d*x+c)^3-10*sin(d*x+c))/a^2/sin(d*x+c)^5/cos(d*x+c )/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (6 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} + 19 \, \cos \left (d x + c\right ) + 8\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{45 \, {\left (a^{2} d e^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} d e^{2} \cos \left (d x + c\right ) + a^{2} d e^{2}\right )} \sin \left (d x + c\right )} \]
-2/45*(3*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2)) *sqrt(-I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, c os(d*x + c) + I*sin(d*x + c))) + 3*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2 )*cos(d*x + c) - I*sqrt(2))*sqrt(I*e)*sin(d*x + c)*weierstrassZeta(4, 0, w eierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + (6*cos(d*x + c) ^3 + 12*cos(d*x + c)^2 + 19*cos(d*x + c) + 8)*sqrt(e*sin(d*x + c)))/((a^2* d*e^2*cos(d*x + c)^2 + 2*a^2*d*e^2*cos(d*x + c) + a^2*d*e^2)*sin(d*x + c))
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]