3.680 \(\int \frac{x^2 (a+b x^3)^{2/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=162 \[ -\frac{(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}}+\frac{(b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}+\frac{(b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{5/3}}+\frac{\left (a+b x^3\right )^{2/3}}{2 d} \]

[Out]

(a + b*x^3)^(2/3)/(2*d) + ((b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt
[3]])/(Sqrt[3]*d^(5/3)) - ((b*c - a*d)^(2/3)*Log[c + d*x^3])/(6*d^(5/3)) + ((b*c - a*d)^(2/3)*Log[(b*c - a*d)^
(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(5/3))

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Rubi [A]  time = 0.170246, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {444, 50, 56, 617, 204, 31} \[ -\frac{(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}}+\frac{(b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}+\frac{(b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{5/3}}+\frac{\left (a+b x^3\right )^{2/3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^3)^(2/3))/(c + d*x^3),x]

[Out]

(a + b*x^3)^(2/3)/(2*d) + ((b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt
[3]])/(Sqrt[3]*d^(5/3)) - ((b*c - a*d)^(2/3)*Log[c + d*x^3])/(6*d^(5/3)) + ((b*c - a*d)^(2/3)*Log[(b*c - a*d)^
(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(5/3))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

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 31

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

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{c+d x} \, dx,x,x^3\right )\\ &=\frac{\left (a+b x^3\right )^{2/3}}{2 d}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d}\\ &=\frac{\left (a+b x^3\right )^{2/3}}{2 d}-\frac{(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}}+\frac{(b c-a d)^{2/3} \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{\frac{(b c-a d)^{2/3}}{d^{2/3}}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^2}\\ &=\frac{\left (a+b x^3\right )^{2/3}}{2 d}-\frac{(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}}+\frac{(b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}-\frac{(b c-a d)^{2/3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{d^{5/3}}\\ &=\frac{\left (a+b x^3\right )^{2/3}}{2 d}+\frac{(b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{5/3}}-\frac{(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{5/3}}+\frac{(b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3}}\\ \end{align*}

Mathematica [C]  time = 0.0155015, size = 47, normalized size = 0.29 \[ -\frac{\left (a+b x^3\right )^{2/3} \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )-1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x^3)^(2/3)*(-1 + Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c) + a*d)]))/(2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53618, size = 736, normalized size = 4.54 \begin{align*} \frac{2 \, \sqrt{3} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \arctan \left (-\frac{2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} d \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} - \sqrt{3}{\left (b c - a d\right )}}{3 \,{\left (b c - a d\right )}}\right ) - \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} d \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{2}{3}}{\left (b c - a d\right )} -{\left (b c - a d\right )} \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}}\right ) + 2 \, \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{1}{3}} \log \left (-d \left (\frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )}\right ) + 3 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*((b^2*c^
2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) - ((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2
)^(1/3)*log((b*x^3 + a)^(1/3)*d*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(2/3)*(b*c - a*d) -
(b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) + 2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*log(-
d*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d)) + 3*(b*x^3 + a)^(2/3))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.23453, size = 350, normalized size = 2.16 \begin{align*} \frac{{\left (b c d \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} - a d^{2} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \right |}\right )}{3 \,{\left (b c d^{2} - a d^{3}\right )}} + \frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{2 \, d} + \frac{\sqrt{3}{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}}\right )}{3 \, d^{3}} - \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}}\right )}{6 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*c*d*(-(b*c - a*d)/d)^(1/3) - a*d^2*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(
1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c*d^2 - a*d^3) + 1/2*(b*x^3 + a)^(2/3)/d + 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(
2/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/d^3 - 1/6*(-b*c
*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))
/d^3