3.515 \(\int \frac {\sqrt [3]{a+b x^3}}{x^4} \, dx\)

Optimal. Leaf size=107 \[ \frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{2/3}}-\frac {b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b \log (x)}{6 a^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{3 x^3} \]

[Out]

-1/3*(b*x^3+a)^(1/3)/x^3-1/6*b*ln(x)/a^(2/3)+1/6*b*ln(a^(1/3)-(b*x^3+a)^(1/3))/a^(2/3)-1/9*b*arctan(1/3*(a^(1/
3)+2*(b*x^3+a)^(1/3))/a^(1/3)*3^(1/2))/a^(2/3)*3^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 107, 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, 47, 57, 617, 204, 31} \[ \frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{2/3}}-\frac {b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b \log (x)}{6 a^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/x^4,x]

[Out]

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

Rule 31

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {\sqrt [3]{a+b x^3}}{x^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{x^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 x^3}+\frac {1}{9} b \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 x^3}-\frac {b \log (x)}{6 a^{2/3}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 a^{2/3}}-\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 )}{6 \sqrt [3]{a}}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 x^3}-\frac {b \log (x)}{6 a^{2/3}}+\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{2/3}}+\frac {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^{2/3}}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 x^3}-\frac {b \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b \log (x)}{6 a^{2/3}}+\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.35 \[ \frac {b \left (a+b x^3\right )^{4/3} \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};\frac {b x^3}{a}+1\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/x^4,x]

[Out]

(b*(a + b*x^3)^(4/3)*Hypergeometric2F1[4/3, 2, 7/3, 1 + (b*x^3)/a])/(4*a^2)

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fricas [A]  time = 0.81, size = 155, normalized size = 1.45 \[ -\frac {2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a b x^{3} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + {\left (a^{2}\right )}^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}}{18 \, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*sqrt(3)*(a^2)^(1/6)*a*b*x^3*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(b*x^3 + a)^(1/
3)*(a^2)^(2/3))/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)) + 6*(b*x^3 + a)^(1/3)*a^2)/(a^2*x^3)

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giac [A]  time = 0.69, size = 115, normalized size = 1.07 \[ -\frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {b^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {2 \, b^{2} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {2}{3}}} + \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x^{3}}}{18 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(2/3) + b^2*log((b*x^3 + a)
^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) - 2*b^2*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/a^(2/3) +
6*(b*x^3 + a)^(1/3)*b/x^3)/b

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 103, normalized size = 0.96 \[ -\frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {2}{3}}} + \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(2/3) - 1/18*b*log((b*x^3 + a)^(2
/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) + 1/9*b*log((b*x^3 + a)^(1/3) - a^(1/3))/a^(2/3) - 1/3*(b*x
^3 + a)^(1/3)/x^3

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mupad [B]  time = 1.21, size = 122, normalized size = 1.14 \[ \frac {b\,\ln \left (b\,{\left (b\,x^3+a\right )}^{1/3}-a^{1/3}\,b\right )}{9\,a^{2/3}}-\frac {{\left (b\,x^3+a\right )}^{1/3}}{3\,x^3}-\frac {\ln \left (\frac {a^{1/3}\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2}+b\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{18\,a^{2/3}}-\frac {\ln \left (\frac {a^{1/3}\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2}+b\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{18\,a^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/x^4,x)

[Out]

(b*log(b*(a + b*x^3)^(1/3) - a^(1/3)*b))/(9*a^(2/3)) - (a + b*x^3)^(1/3)/(3*x^3) - (log((a^(1/3)*(b - 3^(1/2)*
b*1i))/2 + b*(a + b*x^3)^(1/3))*(b - 3^(1/2)*b*1i))/(18*a^(2/3)) - (log((a^(1/3)*(b + 3^(1/2)*b*1i))/2 + b*(a
+ b*x^3)^(1/3))*(b + 3^(1/2)*b*1i))/(18*a^(2/3))

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sympy [C]  time = 2.62, size = 41, normalized size = 0.38 \[ - \frac {\sqrt [3]{b} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 x^{2} \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(1/3)*gamma(2/3)*hyper((-1/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**3))/(3*x**2*gamma(5/3))

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