3.140 \(\int \frac{x^2}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=435 \[ \frac{2 i x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{2 i x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}+\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}-\frac{2 x}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}-\frac{x^2 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f \sqrt{a \sin (e+f x)+a}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

[Out]

(-2*x)/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (x^2*Cot[e/2 + Pi/4 + (f*x)/2])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - (
x^2*ArcTanh[E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - (4*ArcTa
nh[Cos[e/2 + Pi/4 + (f*x)/2]]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]]) + ((2*I)*x*PolyLog[2
, -E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - ((2*I)*x*PolyLo
g[2, E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (4*PolyLog[3,
-E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]]) + (4*PolyLog[3, E^
((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.235191, antiderivative size = 435, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3319, 4186, 3770, 4183, 2531, 2282, 6589} \[ \frac{2 i x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{2 i x \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}+\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}-\frac{2 x}{a f^2 \sqrt{a \sin (e+f x)+a}}-\frac{4 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )\right )}{a f^3 \sqrt{a \sin (e+f x)+a}}-\frac{x^2 \sin \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+2 f x+\pi )}\right )}{a f \sqrt{a \sin (e+f x)+a}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-2*x)/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (x^2*Cot[e/2 + Pi/4 + (f*x)/2])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - (
x^2*ArcTanh[E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - (4*ArcTa
nh[Cos[e/2 + Pi/4 + (f*x)/2]]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]]) + ((2*I)*x*PolyLog[2
, -E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - ((2*I)*x*PolyLo
g[2, E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^2*Sqrt[a + a*Sin[e + f*x]]) - (4*PolyLog[3,
-E^((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]]) + (4*PolyLog[3, E^
((I/4)*(2*e + Pi + 2*f*x))]*Sin[e/2 + Pi/4 + (f*x)/2])/(a*f^3*Sqrt[a + a*Sin[e + f*x]])

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4183

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

Rubi steps

\begin{align*} \int \frac{x^2}{(a+a \sin (e+f x))^{3/2}} \, dx &=\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x^2 \csc ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{2 a \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 x}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}+\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x^2 \csc \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{4 a \sqrt{a+a \sin (e+f x)}}+\frac{\left (2 \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \int \csc \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{a f^2 \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 x}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}-\frac{4 \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}-\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x \log \left (1-e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{a f \sqrt{a+a \sin (e+f x)}}+\frac{\sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \int x \log \left (1+e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{a f \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 x}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}-\frac{4 \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}+\frac{2 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{2 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{\left (2 i \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \int \text{Li}_2\left (-e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2 \sqrt{a+a \sin (e+f x)}}+\frac{\left (2 i \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \int \text{Li}_2\left (e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2 \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 x}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}-\frac{4 \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}+\frac{2 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{2 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{\left (4 \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}+\frac{\left (4 \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 x}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{2 a f \sqrt{a+a \sin (e+f x)}}-\frac{x^2 \tanh ^{-1}\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f \sqrt{a+a \sin (e+f x)}}-\frac{4 \tanh ^{-1}\left (\cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}+\frac{2 i x \text{Li}_2\left (-e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{2 i x \text{Li}_2\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^2 \sqrt{a+a \sin (e+f x)}}-\frac{4 \text{Li}_3\left (-e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}+\frac{4 \text{Li}_3\left (e^{\frac{1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f^3 \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.03136, size = 352, normalized size = 0.81 \[ -\frac{x \sqrt{a (\sin (e+f x)+1)} \left ((4-f x) \sin \left (\frac{1}{2} (e+f x)\right )+(f x+4) \cos \left (\frac{1}{2} (e+f x)\right )\right )}{2 a^2 f^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}+\frac{\sqrt [4]{-1} e^{-\frac{3}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )^3 \left (-4 i f x \text{PolyLog}\left (2,-\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )+4 i f x \text{PolyLog}\left (2,\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )+8 \text{PolyLog}\left (3,-\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )-8 \text{PolyLog}\left (3,\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )-f^2 x^2 \log \left (1-\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )+f^2 x^2 \log \left (1+\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )+16 \tanh ^{-1}\left (\sqrt [4]{-1} e^{\frac{1}{2} i (e+f x)}\right )\right )}{2 \sqrt{2} f^3 \left (-i a e^{-i (e+f x)} \left (e^{i (e+f x)}+i\right )^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

((-1)^(1/4)*(I + E^(I*(e + f*x)))^3*(16*ArcTanh[(-1)^(1/4)*E^((I/2)*(e + f*x))] - f^2*x^2*Log[1 - (-1)^(1/4)*E
^((I/2)*(e + f*x))] + f^2*x^2*Log[1 + (-1)^(1/4)*E^((I/2)*(e + f*x))] - (4*I)*f*x*PolyLog[2, -((-1)^(1/4)*E^((
I/2)*(e + f*x)))] + (4*I)*f*x*PolyLog[2, (-1)^(1/4)*E^((I/2)*(e + f*x))] + 8*PolyLog[3, -((-1)^(1/4)*E^((I/2)*
(e + f*x)))] - 8*PolyLog[3, (-1)^(1/4)*E^((I/2)*(e + f*x))]))/(2*Sqrt[2]*E^(((3*I)/2)*(e + f*x))*(((-I)*a*(I +
E^(I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*f^3) - (x*((4 + f*x)*Cos[(e + f*x)/2] + (4 - f*x)*Sin[(e + f*x)/2]
)*Sqrt[a*(1 + Sin[e + f*x])])/(2*a^2*f^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+a*sin(f*x+e))^(3/2),x)

[Out]

int(x^2/(a+a*sin(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^2/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sin \left (f x + e\right ) + a} x^{2}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*x^2/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral(x**2/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(x^2/(a*sin(f*x + e) + a)^(3/2), x)