3.2.34 \(\int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx\) [134]

Optimal. Leaf size=417 \[ -\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}-\frac {96 i \text {Li}_4\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}+\frac {96 i \text {Li}_4\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}} \]

[Out]

-4*x^3*arctanh(exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d/(a+a*sin(d*x+c))^(1/2)+12*I*x^2*polylog(
2,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-12*I*x^2*polylog(2,exp(1/4*
I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-48*x*polylog(3,-exp(1/4*I*(2*d*x+Pi+2*
c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^3/(a+a*sin(d*x+c))^(1/2)+48*x*polylog(3,exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+
1/4*Pi+1/2*d*x)/d^3/(a+a*sin(d*x+c))^(1/2)-96*I*polylog(4,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x
)/d^4/(a+a*sin(d*x+c))^(1/2)+96*I*polylog(4,exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^4/(a+a*sin(
d*x+c))^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3400, 4268, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {96 i \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (4,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^4 \sqrt {a \sin (c+d x)+a}}+\frac {96 i \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (4,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^4 \sqrt {a \sin (c+d x)+a}}-\frac {48 x \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}+\frac {48 x \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}+\frac {12 i x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {12 i x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {4 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt {a \sin (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-4*x^3*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d*Sqrt[a + a*Sin[c + d*x]]) + ((12*I
)*x^2*PolyLog[2, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) - ((
12*I)*x^2*PolyLog[2, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) -
(48*x*PolyLog[3, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^3*Sqrt[a + a*Sin[c + d*x]]) + (
48*x*PolyLog[3, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^3*Sqrt[a + a*Sin[c + d*x]]) - ((96
*I)*PolyLog[4, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^4*Sqrt[a + a*Sin[c + d*x]]) + ((96
*I)*PolyLog[4, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^4*Sqrt[a + a*Sin[c + d*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 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 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]

Rule 6744

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

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \int x^3 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{\sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}-\frac {\left (6 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x^2 \log \left (1-e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \sin (c+d x)}}+\frac {\left (6 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x^2 \log \left (1+e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {\left (24 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \text {Li}_2\left (-e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \sin (c+d x)}}+\frac {\left (24 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \text {Li}_2\left (e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {\left (48 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_3\left (-e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^3 \sqrt {a+a \sin (c+d x)}}-\frac {\left (48 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_3\left (e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^3 \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}-\frac {\left (96 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}+\frac {\left (96 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}-\frac {96 i \text {Li}_4\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}+\frac {96 i \text {Li}_4\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^4 \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.52, size = 306, normalized size = 0.73 \begin {gather*} \frac {\sqrt [4]{-1} \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \left (i+e^{i (c+d x)}\right ) \left (-i d^3 x^3 \log \left (1-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+i d^3 x^3 \log \left (1+\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+6 d^2 x^2 \text {Li}_2\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-6 d^2 x^2 \text {Li}_2\left (\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+24 i d x \text {Li}_3\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-24 i d x \text {Li}_3\left (\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-48 \text {Li}_4\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+48 \text {Li}_4\left (\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^4 \sqrt {-i a e^{-i (c+d x)} \left (i+e^{i (c+d x)}\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((-1)^(1/4)*Sqrt[2]*(I + E^(I*(c + d*x)))*((-I)*d^3*x^3*Log[1 - (-1)^(1/4)*E^((I/2)*(c + d*x))] + I*d^3*x^3*Lo
g[1 + (-1)^(1/4)*E^((I/2)*(c + d*x))] + 6*d^2*x^2*PolyLog[2, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] - 6*d^2*x^2*Po
lyLog[2, (-1)^(1/4)*E^((I/2)*(c + d*x))] + (24*I)*d*x*PolyLog[3, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] - (24*I)*d
*x*PolyLog[3, (-1)^(1/4)*E^((I/2)*(c + d*x))] - 48*PolyLog[4, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] + 48*PolyLog[
4, (-1)^(1/4)*E^((I/2)*(c + d*x))]))/(d^4*E^((I/2)*(c + d*x))*Sqrt[((-I)*a*(I + E^(I*(c + d*x)))^2)/E^(I*(c +
d*x))])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{3}}{\sqrt {a +a \sin \left (d x +c \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+a*sin(d*x+c))^(1/2),x)

[Out]

int(x^3/(a+a*sin(d*x+c))^(1/2),x)

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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^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/sqrt(a*sin(d*x + c) + a), x)

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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^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(x^3/sqrt(a*sin(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(x**3/sqrt(a*(sin(c + d*x) + 1)), x)

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Giac [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^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/sqrt(a*sin(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(x^3/(a + a*sin(c + d*x))^(1/2), x)

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