3.2.40 \(\int \frac {x^2}{(a+a \sin (e+f x))^{3/2}} \, dx\) [140]

Optimal. Leaf size=435 \[ -\frac {2 x}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {x^2 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f \sqrt {a+a \sin (e+f x)}}-\frac {4 \tanh ^{-1}\left (\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}}+\frac {2 i x \text {Li}_2\left (-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {2 i x \text {Li}_2\left (e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {4 \text {Li}_3\left (-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}}+\frac {4 \text {Li}_3\left (e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}} \]

[Out]

-2*x/a/f^2/(a+a*sin(f*x+e))^(1/2)-1/2*x^2*cot(1/2*e+1/4*Pi+1/2*f*x)/a/f/(a+a*sin(f*x+e))^(1/2)-x^2*arctanh(exp
(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f/(a+a*sin(f*x+e))^(1/2)-4*arctanh(cos(1/2*e+1/4*Pi+1/2*f*
x))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a+a*sin(f*x+e))^(1/2)+2*I*x*polylog(2,-exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2
*e+1/4*Pi+1/2*f*x)/a/f^2/(a+a*sin(f*x+e))^(1/2)-2*I*x*polylog(2,exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/
2*f*x)/a/f^2/(a+a*sin(f*x+e))^(1/2)-4*polylog(3,-exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a
+a*sin(f*x+e))^(1/2)+4*polylog(3,exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a+a*sin(f*x+e))^(
1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 435, normalized size of antiderivative = 1.00, 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 = {3400, 4271, 3855, 4268, 2611, 2320, 6724} \begin {gather*} -\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 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 ) \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 {2 x}{a f^2 \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}} \end {gather*}

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 2320

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 2611

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 3400

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 3855

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

Rule 4268

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 4271

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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 6724

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 ) \text {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 ) \text {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 = 1.21, size = 352, normalized size = 0.81 \begin {gather*} \frac {\sqrt [4]{-1} e^{-\frac {3}{2} i (e+f x)} \left (i+e^{i (e+f x)}\right )^3 \left (16 \tanh ^{-1}\left (\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 )-4 i f x \text {Li}_2\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )+4 i f x \text {Li}_2\left (\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )+8 \text {Li}_3\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )-8 \text {Li}_3\left (\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )\right )}{2 \sqrt {2} \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^{3/2} f^3}-\frac {x \left ((4+f x) \cos \left (\frac {1}{2} (e+f x)\right )+(4-f x) \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))}}{2 a^2 f^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \end {gather*}

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)

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + a*sin(e + f*x))^(3/2),x)

[Out]

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

________________________________________________________________________________________