Integrand size = 33, antiderivative size = 536 \[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=-\frac {3 a x^2}{c f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {3 i a x^2 \cos (e+f x)}{c f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {12 i a x \arctan \left (e^{i (e+f x)}\right ) \cos (e+f x)}{c f^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {6 a x \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{c f^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {6 i a \cos (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{c f^4 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {6 i a \cos (e+f x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{c f^4 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {3 i a \cos (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{c f^4 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {a x^3 \sec (e+f x)}{c f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {3 a x^2 \sin (e+f x)}{c f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a x^3 \tan (e+f x)}{c f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}} \]
-3*a*x^2/c/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-3*I*a*x^2*cos (f*x+e)/c/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-12*I*a*x*arcta n(exp(I*(f*x+e)))*cos(f*x+e)/c/f^3/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e)) ^(1/2)+6*a*x*cos(f*x+e)*ln(1+exp(2*I*(f*x+e)))/c/f^3/(a-a*sin(f*x+e))^(1/2 )/(c+c*sin(f*x+e))^(1/2)+6*I*a*cos(f*x+e)*polylog(2,-I*exp(I*(f*x+e)))/c/f ^4/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-6*I*a*cos(f*x+e)*polylog( 2,I*exp(I*(f*x+e)))/c/f^4/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-3* I*a*cos(f*x+e)*polylog(2,-exp(2*I*(f*x+e)))/c/f^4/(a-a*sin(f*x+e))^(1/2)/( c+c*sin(f*x+e))^(1/2)-a*x^3*sec(f*x+e)/c/f/(a-a*sin(f*x+e))^(1/2)/(c+c*sin (f*x+e))^(1/2)+3*a*x^2*sin(f*x+e)/c/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f* x+e))^(1/2)+a*x^3*tan(f*x+e)/c/f/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^( 1/2)
Time = 2.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.36 \[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a-a \sin (e+f x)} \left (12 i \operatorname {PolyLog}\left (2,-i e^{-i (e+f x)}\right ) (1+\sin (e+f x))+f x \left (3 i f x-f^2 x^2-3 f x \cos (e+f x)+12 \log \left (1+i e^{-i (e+f x)}\right )+3 \left (i f x+4 \log \left (1+i e^{-i (e+f x)}\right )\right ) \sin (e+f x)\right )\right )}{f^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c (1+\sin (e+f x)))^{3/2}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[a - a*Sin[e + f*x]]*((12*I)*Po lyLog[2, (-I)/E^(I*(e + f*x))]*(1 + Sin[e + f*x]) + f*x*((3*I)*f*x - f^2*x ^2 - 3*f*x*Cos[e + f*x] + 12*Log[1 + I/E^(I*(e + f*x))] + 3*(I*f*x + 4*Log [1 + I/E^(I*(e + f*x))])*Sin[e + f*x])))/(f^4*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c*(1 + Sin[e + f*x]))^(3/2))
Time = 2.95 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5115, 7292, 27, 7293, 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.
| \(\displaystyle \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c \sin (e+f x)+c)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5115 |
| \(\displaystyle \frac {\cos (e+f x) \int x^3 \sec ^3(e+f x) (a-a \sin (e+f x))^2dx}{a c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {\cos (e+f x) \int a^2 x^3 \sec ^3(e+f x) (1-\sin (e+f x))^2dx}{a c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \cos (e+f x) \int x^3 \sec ^3(e+f x) (1-\sin (e+f x))^2dx}{c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {a \cos (e+f x) \int \left (\sec ^3(e+f x) x^3+\sec (e+f x) \tan ^2(e+f x) x^3-2 \sec ^2(e+f x) \tan (e+f x) x^3\right )dx}{c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \cos (e+f x) \left (-\frac {12 i x \arctan \left (e^{i (e+f x)}\right )}{f^3}+\frac {6 i \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^4}-\frac {6 i \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^4}-\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^4}+\frac {6 x \log \left (1+e^{2 i (e+f x)}\right )}{f^3}+\frac {3 x^2 \tan (e+f x)}{f^2}-\frac {3 x^2 \sec (e+f x)}{f^2}-\frac {x^3 \sec ^2(e+f x)}{f}+\frac {x^3 \tan (e+f x) \sec (e+f x)}{f}-\frac {3 i x^2}{f^2}\right )}{c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
(a*Cos[e + f*x]*(((-3*I)*x^2)/f^2 - ((12*I)*x*ArcTan[E^(I*(e + f*x))])/f^3 + (6*x*Log[1 + E^((2*I)*(e + f*x))])/f^3 + ((6*I)*PolyLog[2, (-I)*E^(I*(e + f*x))])/f^4 - ((6*I)*PolyLog[2, I*E^(I*(e + f*x))])/f^4 - ((3*I)*PolyLo g[2, -E^((2*I)*(e + f*x))])/f^4 - (3*x^2*Sec[e + f*x])/f^2 - (x^3*Sec[e + f*x]^2)/f + (3*x^2*Tan[e + f*x])/f^2 + (x^3*Sec[e + f*x]*Tan[e + f*x])/f)) /(c*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])
3.2.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)* ((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m] *c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPa rt[m]/Cos[e + f*x]^(2*FracPart[m])) Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ[2*m] && IGeQ[n - m, 0]
\[\int \frac {x^{3} \sqrt {a -\sin \left (f x +e \right ) a}}{\left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {-a \sin \left (f x + e\right ) + a} x^{3}}{{\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral(-sqrt(-a*sin(f*x + e) + a)*sqrt(c*sin(f*x + e) + c)*x^3/(c^2*cos( f*x + e)^2 - 2*c^2*sin(f*x + e) - 2*c^2), x)
\[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\int \frac {x^{3} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}{\left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {-a \sin \left (f x + e\right ) + a} x^{3}}{{\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {-a \sin \left (f x + e\right ) + a} x^{3}}{{\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\int \frac {x^3\,\sqrt {a-a\,\sin \left (e+f\,x\right )}}{{\left (c+c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]