3.110 \(\int \frac{x^3}{(a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\)

Optimal. Leaf size=360 \[ \frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{10 \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{5 \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{10 \left (a+b x^3\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{243 \sqrt{3} a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

[Out]

(5*x)/(243*a^3*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) - x/(12*b*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + x
/(108*a*b*(a + b*x^3)^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + x/(81*a^2*b*(a + b*x^3)*Sqrt[a^2 + 2*a*b*x^3 + b^2*
x^6]) - (10*(a + b*x^3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(243*Sqrt[3]*a^(11/3)*b^(4/3)*Sqrt[
a^2 + 2*a*b*x^3 + b^2*x^6]) + (10*(a + b*x^3)*Log[a^(1/3) + b^(1/3)*x])/(729*a^(11/3)*b^(4/3)*Sqrt[a^2 + 2*a*b
*x^3 + b^2*x^6]) - (5*(a + b*x^3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(729*a^(11/3)*b^(4/3)*Sqrt[a
^2 + 2*a*b*x^3 + b^2*x^6])

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Rubi [A]  time = 0.179909, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {1355, 288, 199, 200, 31, 634, 617, 204, 628} \[ \frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{10 \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{5 \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{10 \left (a+b x^3\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{243 \sqrt{3} a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(5*x)/(243*a^3*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) - x/(12*b*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + x
/(108*a*b*(a + b*x^3)^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + x/(81*a^2*b*(a + b*x^3)*Sqrt[a^2 + 2*a*b*x^3 + b^2*
x^6]) - (10*(a + b*x^3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(243*Sqrt[3]*a^(11/3)*b^(4/3)*Sqrt[
a^2 + 2*a*b*x^3 + b^2*x^6]) + (10*(a + b*x^3)*Log[a^(1/3) + b^(1/3)*x])/(729*a^(11/3)*b^(4/3)*Sqrt[a^2 + 2*a*b
*x^3 + b^2*x^6]) - (5*(a + b*x^3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(729*a^(11/3)*b^(4/3)*Sqrt[a
^2 + 2*a*b*x^3 + b^2*x^6])

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^3}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^3\right )\right ) \int \frac{x^3}{\left (a b+b^2 x^3\right )^5} \, dx}{\sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (b^2 \left (a b+b^2 x^3\right )\right ) \int \frac{1}{\left (a b+b^2 x^3\right )^4} \, dx}{12 \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (2 b \left (a b+b^2 x^3\right )\right ) \int \frac{1}{\left (a b+b^2 x^3\right )^3} \, dx}{27 a \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (5 \left (a b+b^2 x^3\right )\right ) \int \frac{1}{\left (a b+b^2 x^3\right )^2} \, dx}{81 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (10 \left (a b+b^2 x^3\right )\right ) \int \frac{1}{a b+b^2 x^3} \, dx}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (10 \left (a b+b^2 x^3\right )\right ) \int \frac{1}{\sqrt [3]{a} \sqrt [3]{b}+b^{2/3} x} \, dx}{729 a^{11/3} b^{5/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (10 \left (a b+b^2 x^3\right )\right ) \int \frac{2 \sqrt [3]{a} \sqrt [3]{b}-b^{2/3} x}{a^{2/3} b^{2/3}-\sqrt [3]{a} b x+b^{4/3} x^2} \, dx}{729 a^{11/3} b^{5/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{10 \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{\left (5 \left (a b+b^2 x^3\right )\right ) \int \frac{-\sqrt [3]{a} b+2 b^{4/3} x}{a^{2/3} b^{2/3}-\sqrt [3]{a} b x+b^{4/3} x^2} \, dx}{729 a^{11/3} b^{7/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (5 \left (a b+b^2 x^3\right )\right ) \int \frac{1}{a^{2/3} b^{2/3}-\sqrt [3]{a} b x+b^{4/3} x^2} \, dx}{243 a^{10/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{10 \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{5 \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\left (10 \left (a b+b^2 x^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{243 a^{11/3} b^{7/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{5 x}{243 a^3 b \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{x}{12 b \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{108 a b \left (a+b x^3\right )^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{x}{81 a^2 b \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{10 \left (a+b x^3\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{243 \sqrt{3} a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{10 \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{5 \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{729 a^{11/3} b^{4/3} \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ \end{align*}

Mathematica [A]  time = 0.140106, size = 221, normalized size = 0.61 \[ \frac{\left (a+b x^3\right ) \left (-20 \left (a+b x^3\right )^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+60 a^{2/3} \sqrt [3]{b} x \left (a+b x^3\right )^3+36 a^{5/3} \sqrt [3]{b} x \left (a+b x^3\right )^2+27 a^{8/3} \sqrt [3]{b} x \left (a+b x^3\right )-243 a^{11/3} \sqrt [3]{b} x+40 \left (a+b x^3\right )^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+40 \sqrt{3} \left (a+b x^3\right )^4 \tan ^{-1}\left (\frac{2 \sqrt [3]{b} x-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )\right )}{2916 a^{11/3} b^{4/3} \left (\left (a+b x^3\right )^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

((a + b*x^3)*(-243*a^(11/3)*b^(1/3)*x + 27*a^(8/3)*b^(1/3)*x*(a + b*x^3) + 36*a^(5/3)*b^(1/3)*x*(a + b*x^3)^2
+ 60*a^(2/3)*b^(1/3)*x*(a + b*x^3)^3 + 40*Sqrt[3]*(a + b*x^3)^4*ArcTan[(-a^(1/3) + 2*b^(1/3)*x)/(Sqrt[3]*a^(1/
3))] + 40*(a + b*x^3)^4*Log[a^(1/3) + b^(1/3)*x] - 20*(a + b*x^3)^4*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*
x^2]))/(2916*a^(11/3)*b^(4/3)*((a + b*x^3)^2)^(5/2))

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Maple [B]  time = 0.016, size = 519, normalized size = 1.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)

[Out]

1/2916*(-40*3^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(a/b)^(1/3))/(a/b)^(1/3))*x^12*b^4+40*ln(x+(a/b)^(1/3))*x^12*b^4-
20*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*x^12*b^4+60*(a/b)^(2/3)*x^10*b^4-160*3^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(a/
b)^(1/3))/(a/b)^(1/3))*x^9*a*b^3+160*ln(x+(a/b)^(1/3))*x^9*a*b^3-80*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*x^9*a*b^
3+216*(a/b)^(2/3)*x^7*a*b^3-240*3^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(a/b)^(1/3))/(a/b)^(1/3))*x^6*a^2*b^2+240*ln(
x+(a/b)^(1/3))*x^6*a^2*b^2-120*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*x^6*a^2*b^2+279*(a/b)^(2/3)*x^4*a^2*b^2-160*3
^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(a/b)^(1/3))/(a/b)^(1/3))*x^3*a^3*b+160*ln(x+(a/b)^(1/3))*x^3*a^3*b-80*ln(x^2-
(a/b)^(1/3)*x+(a/b)^(2/3))*x^3*a^3*b-120*(a/b)^(2/3)*x*a^3*b-40*3^(1/2)*arctan(1/3*3^(1/2)*(-2*x+(a/b)^(1/3))/
(a/b)^(1/3))*a^4+40*ln(x+(a/b)^(1/3))*a^4-20*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*a^4)*(b*x^3+a)/(a/b)^(2/3)/b^2/
a^3/((b*x^3+a)^2)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85608, size = 1636, normalized size = 4.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

[1/2916*(60*a^2*b^4*x^10 + 216*a^3*b^3*x^7 + 279*a^4*b^2*x^4 - 120*a^5*b*x + 60*sqrt(1/3)*(a*b^5*x^12 + 4*a^2*
b^4*x^9 + 6*a^3*b^3*x^6 + 4*a^4*b^2*x^3 + a^5*b)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x -
a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - 20*(
b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2
*b)^(1/3)*a) + 40*(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4)*(a^2*b)^(2/3)*log(a*b*x + (a^2*
b)^(2/3)))/(a^5*b^6*x^12 + 4*a^6*b^5*x^9 + 6*a^7*b^4*x^6 + 4*a^8*b^3*x^3 + a^9*b^2), 1/2916*(60*a^2*b^4*x^10 +
216*a^3*b^3*x^7 + 279*a^4*b^2*x^4 - 120*a^5*b*x + 120*sqrt(1/3)*(a*b^5*x^12 + 4*a^2*b^4*x^9 + 6*a^3*b^3*x^6 +
4*a^4*b^2*x^3 + a^5*b)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2
*b)^(1/3)/b)/a^2) - 20*(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4)*(a^2*b)^(2/3)*log(a*b*x^2
- (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 40*(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4)*(a^2*b)
^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^5*b^6*x^12 + 4*a^6*b^5*x^9 + 6*a^7*b^4*x^6 + 4*a^8*b^3*x^3 + a^9*b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**3/((a + b*x**3)**2)**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x