Integrand size = 37, antiderivative size = 555 \[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=-\frac {i a (g+h x)^3 \cos (e+f x)}{3 h \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a (g+h x)^2 \arctan \left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a (g+h x)^2 \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {2 i a h (g+h x) \cos (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 i a h (g+h x) \cos (e+f x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {i a h (g+h x) \cos (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {2 a h^2 \cos (e+f x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {2 a h^2 \cos (e+f x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a h^2 \cos (e+f x) \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}} \]
-1/3*I*a*(h*x+g)^3*cos(f*x+e)/h/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1 /2)-2*I*a*(h*x+g)^2*arctan(exp(I*(f*x+e)))*cos(f*x+e)/f/(a-a*sin(f*x+e))^( 1/2)/(c+c*sin(f*x+e))^(1/2)+a*(h*x+g)^2*cos(f*x+e)*ln(1+exp(2*I*(f*x+e)))/ f/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)+2*I*a*h*(h*x+g)*cos(f*x+e) *polylog(2,-I*exp(I*(f*x+e)))/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^ (1/2)-2*I*a*h*(h*x+g)*cos(f*x+e)*polylog(2,I*exp(I*(f*x+e)))/f^2/(a-a*sin( f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-I*a*h*(h*x+g)*cos(f*x+e)*polylog(2,-e xp(2*I*(f*x+e)))/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-2*a*h^2 *cos(f*x+e)*polylog(3,-I*exp(I*(f*x+e)))/f^3/(a-a*sin(f*x+e))^(1/2)/(c+c*s in(f*x+e))^(1/2)+2*a*h^2*cos(f*x+e)*polylog(3,I*exp(I*(f*x+e)))/f^3/(a-a*s in(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)+1/2*a*h^2*cos(f*x+e)*polylog(3,-ex p(2*I*(f*x+e)))/f^3/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)
Time = 2.50 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.35 \[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\frac {\sqrt {2} \left (i+e^{i (e+f x)}\right ) \left (f^2 (g+h x)^2 \left (f (g+h x)-6 i h \log \left (1+i e^{-i (e+f x)}\right )\right )+12 f h^2 (g+h x) \operatorname {PolyLog}\left (2,-i e^{-i (e+f x)}\right )-12 i h^3 \operatorname {PolyLog}\left (3,-i e^{-i (e+f x)}\right )\right ) \sqrt {a-a \sin (e+f x)}}{3 \left (-i+e^{i (e+f x)}\right ) \sqrt {-i c e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2} f^3 h} \]
(Sqrt[2]*(I + E^(I*(e + f*x)))*(f^2*(g + h*x)^2*(f*(g + h*x) - (6*I)*h*Log [1 + I/E^(I*(e + f*x))]) + 12*f*h^2*(g + h*x)*PolyLog[2, (-I)/E^(I*(e + f* x))] - (12*I)*h^3*PolyLog[3, (-I)/E^(I*(e + f*x))])*Sqrt[a - a*Sin[e + f*x ]])/(3*(-I + E^(I*(e + f*x)))*Sqrt[((-I)*c*(I + E^(I*(e + f*x)))^2)/E^(I*( e + f*x))]*f^3*h)
Time = 1.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.48, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, 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.
\(\displaystyle \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c \sin (e+f x)+c}} \, dx\) |
\(\Big \downarrow \) 5115 |
\(\displaystyle \frac {\cos (e+f x) \int (g+h x)^2 \sec (e+f x) (a-a \sin (e+f x))dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {\cos (e+f x) \int a (g+h x)^2 \sec (e+f x) (1-\sin (e+f x))dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \cos (e+f x) \int (g+h x)^2 \sec (e+f x) (1-\sin (e+f x))dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {a \cos (e+f x) \int \left ((g+h x)^2 \sec (e+f x)-(g+h x)^2 \tan (e+f x)\right )dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \cos (e+f x) \left (-\frac {2 i (g+h x)^2 \arctan \left (e^{i (e+f x)}\right )}{f}-\frac {2 h^2 \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {2 h^2 \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {h^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {2 i h (g+h x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i h (g+h x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {i h (g+h x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac {(g+h x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {i (g+h x)^3}{3 h}\right )}{\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\) |
(a*Cos[e + f*x]*(((-1/3*I)*(g + h*x)^3)/h - ((2*I)*(g + h*x)^2*ArcTan[E^(I *(e + f*x))])/f + ((g + h*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f + ((2*I)*h* (g + h*x)*PolyLog[2, (-I)*E^(I*(e + f*x))])/f^2 - ((2*I)*h*(g + h*x)*PolyL og[2, I*E^(I*(e + f*x))])/f^2 - (I*h*(g + h*x)*PolyLog[2, -E^((2*I)*(e + f *x))])/f^2 - (2*h^2*PolyLog[3, (-I)*E^(I*(e + f*x))])/f^3 + (2*h^2*PolyLog [3, I*E^(I*(e + f*x))])/f^3 + (h^2*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^ 3)))/(Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])
3.2.78.3.1 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]
\[\int \frac {\left (h x +g \right )^{2} \sqrt {a -\sin \left (f x +e \right ) a}}{\sqrt {c +c \sin \left (f x +e \right )}}d x\]
Exception generated. \[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\int \frac {\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )} \left (g + h x\right )^{2}}{\sqrt {c \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
\[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\int { \frac {{\left (h x + g\right )}^{2} \sqrt {-a \sin \left (f x + e\right ) + a}}{\sqrt {c \sin \left (f x + e\right ) + c}} \,d x } \]
\[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\int { \frac {{\left (h x + g\right )}^{2} \sqrt {-a \sin \left (f x + e\right ) + a}}{\sqrt {c \sin \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {(g+h x)^2 \sqrt {a-a \sin (e+f x)}}{\sqrt {c+c \sin (e+f x)}} \, dx=\int \frac {{\left (g+h\,x\right )}^2\,\sqrt {a-a\,\sin \left (e+f\,x\right )}}{\sqrt {c+c\,\sin \left (e+f\,x\right )}} \,d x \]