3.751 \(\int \frac{x^2}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=167 \[ -\frac{1}{\sqrt [3]{a+b x^3} (b c-a d)}-\frac{\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} (b c-a d)^{4/3}} \]

[Out]

-(1/((b*c - a*d)*(a + b*x^3)^(1/3))) + (d^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/S
qrt[3]])/(Sqrt[3]*(b*c - a*d)^(4/3)) - (d^(1/3)*Log[c + d*x^3])/(6*(b*c - a*d)^(4/3)) + (d^(1/3)*Log[(b*c - a*
d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*(b*c - a*d)^(4/3))

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Rubi [A]  time = 0.165478, antiderivative size = 167, 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, 51, 56, 617, 204, 31} \[ -\frac{1}{\sqrt [3]{a+b x^3} (b c-a d)}-\frac{\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} (b c-a d)^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-(1/((b*c - a*d)*(a + b*x^3)^(1/3))) + (d^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/S
qrt[3]])/(Sqrt[3]*(b*c - a*d)^(4/3)) - (d^(1/3)*Log[c + d*x^3])/(6*(b*c - a*d)^(4/3)) + (d^(1/3)*Log[(b*c - a*
d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*(b*c - a*d)^(4/3))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 )^{4/3} \left (c+d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{4/3} (c+d x)} \, dx,x,x^3\right )\\ &=-\frac{1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac{d \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 (b c-a d)}\\ &=-\frac{1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac{\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \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 (b c-a d)^{4/3}}-\frac{\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 (b c-a d)}\\ &=-\frac{1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac{\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}-\frac{\sqrt [3]{d} \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 )}{(b c-a d)^{4/3}}\\ &=-\frac{1}{(b c-a d) \sqrt [3]{a+b x^3}}+\frac{\sqrt [3]{d} \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} (b c-a d)^{4/3}}-\frac{\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac{\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}\\ \end{align*}

Mathematica [C]  time = 0.0145839, size = 50, normalized size = 0.3 \[ -\frac{\, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )}{\sqrt [3]{a+b x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

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

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^2/(b*x^3+a)^(4/3)/(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)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33494, size = 599, normalized size = 3.59 \begin{align*} -\frac{2 \, \sqrt{3}{\left (b x^{3} + a\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{1}{3}} \arctan \left (\frac{2}{3} \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-\frac{d}{b c - a d}\right )^{\frac{1}{3}} + \frac{1}{3} \, \sqrt{3}\right ) -{\left (b x^{3} + a\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{1}{3}} \log \left (-{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{2}{3}} d -{\left (b c - a d\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{1}{3}}\right ) + 2 \,{\left (b x^{3} + a\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{1}{3}} \log \left ({\left (b c - a d\right )} \left (-\frac{d}{b c - a d}\right )^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} d\right ) + 6 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{6 \,{\left ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/6*(2*sqrt(3)*(b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*arctan(2/3*sqrt(3)*(b*x^3 + a)^(1/3)*(-d/(b*c - a*d))^(1/3)
+ 1/3*sqrt(3)) - (b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*log(-(b*x^3 + a)^(1/3)*(b*c - a*d)*(-d/(b*c - a*d))^(2/3)
+ (b*x^3 + a)^(2/3)*d - (b*c - a*d)*(-d/(b*c - a*d))^(1/3)) + 2*(b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*log((b*c -
a*d)*(-d/(b*c - a*d))^(2/3) + (b*x^3 + a)^(1/3)*d) + 6*(b*x^3 + a)^(2/3))/((b^2*c - a*b*d)*x^3 + a*b*c - a^2*
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{4}{3}} \left (c + d x^{3}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**2/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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Giac [B]  time = 1.17892, size = 385, normalized size = 2.31 \begin{align*} \frac{d \left (-\frac{b c - a d}{d}\right )^{\frac{2}{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^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac{{\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 )}{\sqrt{3} b^{2} c^{2} d - 2 \, \sqrt{3} a b c d^{2} + \sqrt{3} a^{2} 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 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

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