3.707 \(\int \frac {1}{x^3 \sqrt [3]{a+b x^2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}-\frac {b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt {3} a^{4/3}}+\frac {b \log (x)}{6 a^{4/3}}-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2} \]

[Out]

-1/2*(b*x^2+a)^(2/3)/a/x^2+1/6*b*ln(x)/a^(4/3)-1/4*b*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(4/3)-1/6*b*arctan(1/3*(a^(
1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(4/3)*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, 55, 617, 204, 31} \[ -\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}-\frac {b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt {3} a^{4/3}}+\frac {b \log (x)}{6 a^{4/3}}-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)^(1/3)),x]

[Out]

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

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 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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] && 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^3 \sqrt [3]{a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{a+b x}} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2}-\frac {b \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2}+\frac {b \log (x)}{6 a^{4/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )}{4 a}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2}+\frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )}{2 a^{4/3}}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{2 a x^2}-\frac {b \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{2 \sqrt {3} a^{4/3}}+\frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.34 \[ \frac {3 b \left (a+b x^2\right )^{2/3} \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {b x^2}{a}+1\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(a + b*x^2)^(2/3)*Hypergeometric2F1[2/3, 2, 5/3, 1 + (b*x^2)/a])/(4*a^2)

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fricas [A]  time = 0.91, size = 344, normalized size = 3.13 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} a b x^{2} \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a + \left (-a\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) + \left (-a\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right ) - 6 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{12 \, a^{2} x^{2}}, -\frac {6 \, \sqrt {\frac {1}{3}} a b x^{2} \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \arctan \left (\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}}\right ) - \left (-a\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) + 2 \, \left (-a\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right ) + 6 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{12 \, a^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*sqrt(1/3)*a*b*x^2*sqrt((-a)^(1/3)/a)*log((2*b*x^2 - 3*sqrt(1/3)*(2*(b*x^2 + a)^(2/3)*(-a)^(2/3) - (b*
x^2 + a)^(1/3)*a + (-a)^(1/3)*a)*sqrt((-a)^(1/3)/a) - 3*(b*x^2 + a)^(1/3)*(-a)^(2/3) + 3*a)/x^2) + (-a)^(2/3)*
b*x^2*log((b*x^2 + a)^(2/3) - (b*x^2 + a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) - 2*(-a)^(2/3)*b*x^2*log((b*x^2 + a)^
(1/3) + (-a)^(1/3)) - 6*(b*x^2 + a)^(2/3)*a)/(a^2*x^2), -1/12*(6*sqrt(1/3)*a*b*x^2*sqrt(-(-a)^(1/3)/a)*arctan(
sqrt(1/3)*(2*(b*x^2 + a)^(1/3) - (-a)^(1/3))*sqrt(-(-a)^(1/3)/a)) - (-a)^(2/3)*b*x^2*log((b*x^2 + a)^(2/3) - (
b*x^2 + a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) + 2*(-a)^(2/3)*b*x^2*log((b*x^2 + a)^(1/3) + (-a)^(1/3)) + 6*(b*x^2
+ a)^(2/3)*a)/(a^2*x^2)]

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giac [A]  time = 1.14, size = 119, normalized size = 1.08 \[ -\frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {4}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {4}{3}}} + \frac {2 \, b^{2} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {4}{3}}} + \frac {6 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} b}{a x^{2}}}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(4/3) - b^2*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(4/3) + 2*b^2*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(4/3) +
6*(b*x^2 + a)^(2/3)*b/(a*x^2))/b

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{3}} x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.95, size = 118, normalized size = 1.07 \[ -\frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{6 \, a^{\frac {4}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {2}{3}} b}{2 \, {\left ({\left (b x^{2} + a\right )} a - a^{2}\right )}} + \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{12 \, a^{\frac {4}{3}}} - \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{6 \, a^{\frac {4}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(4/3) - 1/2*(b*x^2 + a)^(2/3)*b/(
(b*x^2 + a)*a - a^2) + 1/12*b*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(4/3) - 1/6*b*log
((b*x^2 + a)^(1/3) - a^(1/3))/a^(4/3)

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mupad [B]  time = 5.01, size = 138, normalized size = 1.25 \[ -\frac {b\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{6\,a^{4/3}}-\frac {{\left (b\,x^2+a\right )}^{2/3}}{2\,a\,x^2}+\frac {\ln \left (\frac {{\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{16\,a^{5/3}}-\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{4\,a^2}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{12\,a^{4/3}}+\frac {\ln \left (\frac {{\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{16\,a^{5/3}}-\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{4\,a^2}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{12\,a^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^2)^(1/3)),x)

[Out]

(log((b - 3^(1/2)*b*1i)^2/(16*a^(5/3)) - (b^2*(a + b*x^2)^(1/3))/(4*a^2))*(b - 3^(1/2)*b*1i))/(12*a^(4/3)) - (
a + b*x^2)^(2/3)/(2*a*x^2) - (b*log((a + b*x^2)^(1/3) - a^(1/3)))/(6*a^(4/3)) + (log((b + 3^(1/2)*b*1i)^2/(16*
a^(5/3)) - (b^2*(a + b*x^2)^(1/3))/(4*a^2))*(b + 3^(1/2)*b*1i))/(12*a^(4/3))

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sympy [C]  time = 1.19, size = 41, normalized size = 0.37 \[ - \frac {\Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \sqrt [3]{b} x^{\frac {8}{3}} \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(4/3)*hyper((1/3, 4/3), (7/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(1/3)*x**(8/3)*gamma(7/3))

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