3.141 \(\int x (b \sqrt [3]{x}+a x)^{3/2} \, dx\)

Optimal. Leaf size=408 \[ -\frac{44 b^{21/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{a x+b \sqrt [3]{x}}}-\frac{88 b^5 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{1105 a^{7/2} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{a x+b \sqrt [3]{x}}}+\frac{88 b^{21/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{88 b^4 \sqrt [3]{x} \sqrt{a x+b \sqrt [3]{x}}}{3315 a^3}-\frac{88 b^3 x \sqrt{a x+b \sqrt [3]{x}}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{a x+b \sqrt [3]{x}}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{a x+b \sqrt [3]{x}}+\frac{2}{7} x^2 \left (a x+b \sqrt [3]{x}\right )^{3/2} \]

[Out]

(-88*b^5*(b + a*x^(2/3))*x^(1/3))/(1105*a^(7/2)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) + (88*b^4*x
^(1/3)*Sqrt[b*x^(1/3) + a*x])/(3315*a^3) - (88*b^3*x*Sqrt[b*x^(1/3) + a*x])/(4641*a^2) + (24*b^2*x^(5/3)*Sqrt[
b*x^(1/3) + a*x])/(1547*a) + (12*b*x^(7/3)*Sqrt[b*x^(1/3) + a*x])/119 + (2*x^2*(b*x^(1/3) + a*x)^(3/2))/7 + (8
8*b^(21/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticE[2
*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(1105*a^(15/4)*Sqrt[b*x^(1/3) + a*x]) - (44*b^(21/4)*(Sqrt[b] + Sqrt
[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/
b^(1/4)], 1/2])/(1105*a^(15/4)*Sqrt[b*x^(1/3) + a*x])

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Rubi [A]  time = 0.554065, antiderivative size = 408, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {2018, 2021, 2024, 2032, 329, 305, 220, 1196} \[ -\frac{88 b^5 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{1105 a^{7/2} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{a x+b \sqrt [3]{x}}}-\frac{44 b^{21/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{88 b^{21/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{88 b^4 \sqrt [3]{x} \sqrt{a x+b \sqrt [3]{x}}}{3315 a^3}-\frac{88 b^3 x \sqrt{a x+b \sqrt [3]{x}}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{a x+b \sqrt [3]{x}}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{a x+b \sqrt [3]{x}}+\frac{2}{7} x^2 \left (a x+b \sqrt [3]{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-88*b^5*(b + a*x^(2/3))*x^(1/3))/(1105*a^(7/2)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) + (88*b^4*x
^(1/3)*Sqrt[b*x^(1/3) + a*x])/(3315*a^3) - (88*b^3*x*Sqrt[b*x^(1/3) + a*x])/(4641*a^2) + (24*b^2*x^(5/3)*Sqrt[
b*x^(1/3) + a*x])/(1547*a) + (12*b*x^(7/3)*Sqrt[b*x^(1/3) + a*x])/119 + (2*x^2*(b*x^(1/3) + a*x)^(3/2))/7 + (8
8*b^(21/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticE[2
*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(1105*a^(15/4)*Sqrt[b*x^(1/3) + a*x]) - (44*b^(21/4)*(Sqrt[b] + Sqrt
[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/
b^(1/4)], 1/2])/(1105*a^(15/4)*Sqrt[b*x^(1/3) + a*x])

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2024

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

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int x \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx &=3 \operatorname{Subst}\left (\int x^5 \left (b x+a x^3\right )^{3/2} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac{1}{7} (6 b) \operatorname{Subst}\left (\int x^6 \sqrt{b x+a x^3} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac{1}{119} \left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{x^7}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac{\left (132 b^3\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{1547 a}\\ &=-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac{\left (44 b^4\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{663 a^2}\\ &=\frac{88 b^4 \sqrt [3]{x} \sqrt{b \sqrt [3]{x}+a x}}{3315 a^3}-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac{\left (44 b^5\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{1105 a^3}\\ &=\frac{88 b^4 \sqrt [3]{x} \sqrt{b \sqrt [3]{x}+a x}}{3315 a^3}-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac{\left (44 b^5 \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\sqrt{b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{1105 a^3 \sqrt{b \sqrt [3]{x}+a x}}\\ &=\frac{88 b^4 \sqrt [3]{x} \sqrt{b \sqrt [3]{x}+a x}}{3315 a^3}-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac{\left (88 b^5 \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{1105 a^3 \sqrt{b \sqrt [3]{x}+a x}}\\ &=\frac{88 b^4 \sqrt [3]{x} \sqrt{b \sqrt [3]{x}+a x}}{3315 a^3}-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac{\left (88 b^{11/2} \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{1105 a^{7/2} \sqrt{b \sqrt [3]{x}+a x}}+\frac{\left (88 b^{11/2} \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{a} x^2}{\sqrt{b}}}{\sqrt{b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{1105 a^{7/2} \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{88 b^5 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{1105 a^{7/2} \left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right ) \sqrt{b \sqrt [3]{x}+a x}}+\frac{88 b^4 \sqrt [3]{x} \sqrt{b \sqrt [3]{x}+a x}}{3315 a^3}-\frac{88 b^3 x \sqrt{b \sqrt [3]{x}+a x}}{4641 a^2}+\frac{24 b^2 x^{5/3} \sqrt{b \sqrt [3]{x}+a x}}{1547 a}+\frac{12}{119} b x^{7/3} \sqrt{b \sqrt [3]{x}+a x}+\frac{2}{7} x^2 \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac{88 b^{21/4} \left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right ) \sqrt{\frac{b+a x^{2/3}}{\left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{b \sqrt [3]{x}+a x}}-\frac{44 b^{21/4} \left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right ) \sqrt{\frac{b+a x^{2/3}}{\left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{1105 a^{15/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.10018, size = 123, normalized size = 0.3 \[ \frac{2 \sqrt [3]{x} \sqrt{a x+b \sqrt [3]{x}} \left (\left (a x^{2/3}+b\right )^2 \sqrt{\frac{a x^{2/3}}{b}+1} \left (221 a^2 x^{4/3}-143 a b x^{2/3}+77 b^2\right )-77 b^4 \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{a x^{2/3}}{b}\right )\right )}{1547 a^3 \sqrt{\frac{a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(2*x^(1/3)*Sqrt[b*x^(1/3) + a*x]*((b + a*x^(2/3))^2*Sqrt[1 + (a*x^(2/3))/b]*(77*b^2 - 143*a*b*x^(2/3) + 221*a^
2*x^(4/3)) - 77*b^4*Hypergeometric2F1[-3/2, 3/4, 7/4, -((a*x^(2/3))/b)]))/(1547*a^3*Sqrt[1 + (a*x^(2/3))/b])

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Maple [A]  time = 0.018, size = 263, normalized size = 0.6 \begin{align*} -{\frac{2}{23205\,{a}^{4}} \left ( -4665\,{x}^{8/3}{a}^{4}{b}^{2}-7800\,{x}^{10/3}{a}^{5}b+40\,{x}^{2}{a}^{3}{b}^{3}+924\,{b}^{6}\sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{a\sqrt [3]{x}}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) -462\,{b}^{6}\sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{a\sqrt [3]{x}}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) -3315\,{x}^{4}{a}^{6}-308\,{x}^{2/3}a{b}^{5}-88\,{x}^{4/3}{a}^{2}{b}^{4} \right ){\frac{1}{\sqrt{\sqrt [3]{x} \left ( b+a{x}^{{\frac{2}{3}}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^(1/3)+a*x)^(3/2),x)

[Out]

-2/23205/a^4*(-4665*x^(8/3)*a^4*b^2-7800*x^(10/3)*a^5*b+40*x^2*a^3*b^3+924*b^6*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b
)^(1/2))^(1/2)*2^(1/2)*((-a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*Elliptic
E(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-462*b^6*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(
1/2)*2^(1/2)*((-a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1
/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-3315*x^4*a^6-308*x^(2/3)*a*b^5-88*x^(4/3)*a^2*b^4)/(x^(1/3)
*(b+a*x^(2/3)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a x + b x^{\frac{1}{3}}\right )}^{\frac{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(1/3)+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*x + b*x^(1/3))^(3/2)*x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a x^{2} + b x^{\frac{4}{3}}\right )} \sqrt{a x + b x^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(1/3)+a*x)^(3/2),x, algorithm="fricas")

[Out]

integral((a*x^2 + b*x^(4/3))*sqrt(a*x + b*x^(1/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**(1/3)+a*x)**(3/2),x)

[Out]

Integral(x*(a*x + b*x**(1/3))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a x + b x^{\frac{1}{3}}\right )}^{\frac{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(1/3)+a*x)^(3/2),x, algorithm="giac")

[Out]

integrate((a*x + b*x^(1/3))^(3/2)*x, x)