3.178 \(\int \frac{(g+h x)^2 \sqrt{a-a \sin (e+f x)}}{\sqrt{c+c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=555 \[ \frac{2 i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 a h^2 \cos (e+f x) \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{2 a h^2 \cos (e+f x) \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a h^2 \cos (e+f x) \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a (g+h x)^3 \cos (e+f x)}{3 h \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a (g+h x)^2 \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a (g+h x)^2 \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

[Out]

((-I/3)*a*(g + h*x)^3*Cos[e + f*x])/(h*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((2*I)*a*(g + h*x)
^2*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*(g + h*x)^
2*Cos[e + f*x]*Log[1 + E^((2*I)*(e + f*x))])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + ((2*I)*a*
h*(g + h*x)*Cos[e + f*x]*PolyLog[2, (-I)*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*
x]]) - ((2*I)*a*h*(g + h*x)*Cos[e + f*x]*PolyLog[2, I*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c +
c*Sin[e + f*x]]) - (I*a*h*(g + h*x)*Cos[e + f*x]*PolyLog[2, -E^((2*I)*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*
x]]*Sqrt[c + c*Sin[e + f*x]]) - (2*a*h^2*Cos[e + f*x]*PolyLog[3, (-I)*E^(I*(e + f*x))])/(f^3*Sqrt[a - a*Sin[e
+ f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (2*a*h^2*Cos[e + f*x]*PolyLog[3, I*E^(I*(e + f*x))])/(f^3*Sqrt[a - a*Sin[e
+ f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*h^2*Cos[e + f*x]*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^3*Sqrt[a - a*S
in[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.876905, antiderivative size = 555, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.27, Rules used = {4604, 6741, 12, 6742, 4181, 2531, 2282, 6589, 3719, 2190} \[ \frac{2 i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h (g+h x) \cos (e+f x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 a h^2 \cos (e+f x) \text{PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{2 a h^2 \cos (e+f x) \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a h^2 \cos (e+f x) \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a (g+h x)^3 \cos (e+f x)}{3 h \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a (g+h x)^2 \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a (g+h x)^2 \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

Antiderivative was successfully verified.

[In]

Int[((g + h*x)^2*Sqrt[a - a*Sin[e + f*x]])/Sqrt[c + c*Sin[e + f*x]],x]

[Out]

((-I/3)*a*(g + h*x)^3*Cos[e + f*x])/(h*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((2*I)*a*(g + h*x)
^2*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*(g + h*x)^
2*Cos[e + f*x]*Log[1 + E^((2*I)*(e + f*x))])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + ((2*I)*a*
h*(g + h*x)*Cos[e + f*x]*PolyLog[2, (-I)*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*
x]]) - ((2*I)*a*h*(g + h*x)*Cos[e + f*x]*PolyLog[2, I*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c +
c*Sin[e + f*x]]) - (I*a*h*(g + h*x)*Cos[e + f*x]*PolyLog[2, -E^((2*I)*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*
x]]*Sqrt[c + c*Sin[e + f*x]]) - (2*a*h^2*Cos[e + f*x]*PolyLog[3, (-I)*E^(I*(e + f*x))])/(f^3*Sqrt[a - a*Sin[e
+ f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (2*a*h^2*Cos[e + f*x]*PolyLog[3, I*E^(I*(e + f*x))])/(f^3*Sqrt[a - a*Sin[e
+ f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*h^2*Cos[e + f*x]*PolyLog[3, -E^((2*I)*(e + f*x))])/(2*f^3*Sqrt[a - a*S
in[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])

Rule 4604

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^F
racPart[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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{(g+h x)^2 \sqrt{a-a \sin (e+f x)}}{\sqrt{c+c \sin (e+f x)}} \, dx &=\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+c \sin (e+f x)}}\\ &=\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+c \sin (e+f x)}}\\ &=\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+c \sin (e+f x)}}\\ &=\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+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int (g+h x)^2 \sec (e+f x) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int (g+h x)^2 \tan (e+f x) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\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 \tan ^{-1}\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{(2 i a \cos (e+f x)) \int \frac{e^{2 i (e+f x)} (g+h x)^2}{1+e^{2 i (e+f x)}} \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(2 a h \cos (e+f x)) \int (g+h x) \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(2 a h \cos (e+f x)) \int (g+h x) \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\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 \tan ^{-1}\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) \text{Li}_2\left (-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) \text{Li}_2\left (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 a h \cos (e+f x)) \int (g+h x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{\left (2 i a h^2 \cos (e+f x)\right ) \int \text{Li}_2\left (-i e^{i (e+f x)}\right ) \, dx}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{\left (2 i a h^2 \cos (e+f x)\right ) \int \text{Li}_2\left (i e^{i (e+f x)}\right ) \, dx}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\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 \tan ^{-1}\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) \text{Li}_2\left (-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) \text{Li}_2\left (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) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{\left (2 a h^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{\left (2 a h^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{\left (i a h^2 \cos (e+f x)\right ) \int \text{Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\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 \tan ^{-1}\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) \text{Li}_2\left (-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) \text{Li}_2\left (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) \text{Li}_2\left (-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) \text{Li}_3\left (-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) \text{Li}_3\left (i e^{i (e+f x)}\right )}{f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{\left (a h^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\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 \tan ^{-1}\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) \text{Li}_2\left (-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) \text{Li}_2\left (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) \text{Li}_2\left (-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) \text{Li}_3\left (-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) \text{Li}_3\left (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) \text{Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.92293, size = 194, normalized size = 0.35 \[ \frac{\sqrt{2} \left (e^{i (e+f x)}+i\right ) \sqrt{a-a \sin (e+f x)} \left (12 f h^2 (g+h x) \text{PolyLog}\left (2,-i e^{-i (e+f x)}\right )-12 i h^3 \text{PolyLog}\left (3,-i e^{-i (e+f x)}\right )+f^2 (g+h x)^2 \left (f (g+h x)-6 i h \log \left (1+i e^{-i (e+f x)}\right )\right )\right )}{3 f^3 h \left (e^{i (e+f x)}-i\right ) \sqrt{-i c e^{-i (e+f x)} \left (e^{i (e+f x)}+i\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*x)^2*Sqrt[a - a*Sin[e + f*x]])/Sqrt[c + c*Sin[e + f*x]],x]

[Out]

(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)

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int{ \left ( hx+g \right ) ^{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{c+c\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)^2*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x)

[Out]

int((h*x+g)^2*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^2*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((h*x + g)^2*sqrt(-a*sin(f*x + e) + a)/sqrt(c*sin(f*x + e) + c), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^2*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)**2*(a-a*sin(f*x+e))**(1/2)/(c+c*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1))*(g + h*x)**2/sqrt(c*(sin(e + f*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^2*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((h*x + g)^2*sqrt(-a*sin(f*x + e) + a)/sqrt(c*sin(f*x + e) + c), x)