3.566 \(\int \frac {1}{x^4 (a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a x^3} \]

[Out]

-1/3*(b*x^3+a)^(1/3)/a/x^3+1/3*b*ln(x)/a^(5/3)-1/3*b*ln(a^(1/3)-(b*x^3+a)^(1/3))/a^(5/3)+2/9*b*arctan(1/3*(a^(
1/3)+2*(b*x^3+a)^(1/3))/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 51, 57, 617, 204, 31} \[ -\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)^(2/3)),x]

[Out]

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

Rule 31

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

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 57

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

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 266

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

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]

Rubi steps

\begin {align*} \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{2/3}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a x^3}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )}{9 a}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a x^3}+\frac {b \log (x)}{3 a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{3 a^{4/3}}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a x^3}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{3 a^{5/3}}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a x^3}+\frac {2 b \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.31 \[ \frac {b \sqrt [3]{a+b x^3} \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};\frac {b x^3}{a}+1\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^3)^(2/3)),x]

[Out]

(b*(a + b*x^3)^(1/3)*Hypergeometric2F1[1/3, 2, 4/3, 1 + (b*x^3)/a])/a^2

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fricas [B]  time = 0.61, size = 182, normalized size = 1.65 \[ \frac {2 \, \sqrt {3} a b x^{3} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}}{9 \, a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/9*(2*sqrt(3)*a*b*x^3*sqrt(-(-a^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-a^2)^(1/3)*a - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(
-a^2)^(2/3))*sqrt(-(-a^2)^(1/3))/a^2) + (-a^2)^(2/3)*b*x^3*log((b*x^3 + a)^(2/3)*a - (-a^2)^(1/3)*a + (b*x^3 +
a)^(1/3)*(-a^2)^(2/3)) - 2*(-a^2)^(2/3)*b*x^3*log((b*x^3 + a)^(1/3)*a - (-a^2)^(2/3)) - 3*(b*x^3 + a)^(1/3)*a
^2)/(a^3*x^3)

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giac [A]  time = 0.70, size = 118, normalized size = 1.07 \[ \frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {b^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, b^{2} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{a x^{3}}}{9 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

1/9*(2*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) + b^2*log((b*x^3 + a)^(
2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2*b^2*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*
(b*x^3 + a)^(1/3)*b/(a*x^3))/b

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 118, normalized size = 1.07 \[ \frac {2 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{3 \, {\left ({\left (b x^{3} + a\right )} a - a^{2}\right )}} + \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {5}{3}}} - \frac {2 \, b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {5}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

2/9*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) - 1/3*(b*x^3 + a)^(1/3)*b/((
b*x^3 + a)*a - a^2) + 1/9*b*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2/9*b*log((
b*x^3 + a)^(1/3) - a^(1/3))/a^(5/3)

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mupad [B]  time = 1.32, size = 128, normalized size = 1.16 \[ \frac {\ln \left (\frac {b-\sqrt {3}\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}+\frac {\ln \left (\frac {b+\sqrt {3}\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}-\frac {2\,b\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-a^{1/3}\right )}{9\,a^{5/3}}-\frac {{\left (b\,x^3+a\right )}^{1/3}}{3\,a\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^3)^(2/3)),x)

[Out]

(log((b - 3^(1/2)*b*1i)/a^(2/3) + (2*b*(a + b*x^3)^(1/3))/a)*(b - 3^(1/2)*b*1i))/(9*a^(5/3)) + (log((b + 3^(1/
2)*b*1i)/a^(2/3) + (2*b*(a + b*x^3)^(1/3))/a)*(b + 3^(1/2)*b*1i))/(9*a^(5/3)) - (2*b*log((a + b*x^3)^(1/3) - a
^(1/3)))/(9*a^(5/3)) - (a + b*x^3)^(1/3)/(3*a*x^3)

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sympy [C]  time = 2.26, size = 39, normalized size = 0.35 \[ - \frac {\Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{5} \Gamma \left (\frac {8}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**3+a)**(2/3),x)

[Out]

-gamma(5/3)*hyper((2/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**3))/(3*b**(2/3)*x**5*gamma(8/3))

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