\(\int \frac {1}{x^4 (a+b x^3)^{2/3}} \, dx\) [566]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^3}}{3 a x^3}+\frac {2 b \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 44, 59, 631, 210, 31} \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {2 b \arctan \left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a x^3} \]

[In]

Int[1/(x^4*(a + b*x^3)^(2/3)),x]

[Out]

-1/3*(a + b*x^3)^(1/3)/(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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 59

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 272

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 631

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*} \text {integral}& = \frac {1}{3} \text {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) \text {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 \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{3 a^{5/3}}+\frac {b \text {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) \text {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*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {-3 a^{2/3} \sqrt [3]{a+b x^3}+2 \sqrt {3} b x^3 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 b x^3 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )+b x^3 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{9 a^{5/3} x^3} \]

[In]

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

[Out]

(-3*a^(2/3)*(a + b*x^3)^(1/3) + 2*Sqrt[3]*b*x^3*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] - 2*b*x^3*
Log[-a^(1/3) + (a + b*x^3)^(1/3)] + b*x^3*Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(9*a^(
5/3)*x^3)

Maple [A] (verified)

Time = 3.96 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, b \,x^{3}-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) b \,x^{3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) b \,x^{3}-3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}}{9 a^{\frac {5}{3}} x^{3}}\)	\(111\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (81) = 162\).

Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.65 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {2 \, \sqrt {3} a b x^{3} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}}{9 \, a^{3} x^{3}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=- \frac {\Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{5} \Gamma \left (\frac {8}{3}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {2 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{3 \, {\left ({\left (b x^{3} + a\right )} a - a^{2}\right )}} + \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {5}{3}}} - \frac {2 \, b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {5}{3}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {b^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, b^{2} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{a x^{3}}}{9 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {\ln \left (\frac {b-\sqrt {3}\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}+\frac {\ln \left (\frac {b+\sqrt {3}\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}-\frac {2\,b\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-a^{1/3}\right )}{9\,a^{5/3}}-\frac {{\left (b\,x^3+a\right )}^{1/3}}{3\,a\,x^3} \]

[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)