3.8.19 \(\int \frac {1}{x^3 (a+b x^2)^{2/3}} \, dx\) [719]

Optimal. Leaf size=107 \[ -\frac {\sqrt [3]{a+b x^2}}{2 a x^2}+\frac {b \tan ^{-1}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{5/3}} \]

[Out]

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

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Rubi [A]
time = 0.05, 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 = {272, 44, 59, 631, 210, 31} \begin {gather*} \frac {b \text {ArcTan}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {\sqrt [3]{a+b x^2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)^(2/3)),x]

[Out]

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

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Mathematica [A]
time = 0.10, size = 135, normalized size = 1.26 \begin {gather*} \frac {-3 a^{2/3} \sqrt [3]{a+b x^2}+2 \sqrt {3} b x^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 b x^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )+b x^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{6 a^{5/3} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^(2/3)*(a + b*x^2)^(1/3) + 2*Sqrt[3]*b*x^2*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] - 2*b*x^2*
Log[-a^(1/3) + (a + b*x^2)^(1/3)] + b*x^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/(6*a^(
5/3)*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 118, normalized size = 1.10 \begin {gather*} \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 )}{3 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{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 )}{6 \, a^{\frac {5}{3}}} - \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {5}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) - 1/2*(b*x^2 + a)^(1/3)*b/((
b*x^2 + a)*a - a^2) + 1/6*b*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 1/3*b*log((
b*x^2 + a)^(1/3) - a^(1/3))/a^(5/3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (81) = 162\).
time = 0.66, size = 182, normalized size = 1.70 \begin {gather*} \frac {2 \, \sqrt {3} a b x^{2} \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^{2} + 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^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{2}}{6 \, a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.68, size = 41, normalized size = 0.38 \begin {gather*} - \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^{2}}} \right )}}{2 b^{\frac {2}{3}} x^{\frac {10}{3}} \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(5/3)*hyper((2/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(2/3)*x**(10/3)*gamma(8/3))

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Giac [A]
time = 1.19, size = 118, normalized size = 1.10 \begin {gather*} \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 {5}{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 {5}{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 {5}{3}}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} b}{a x^{2}}}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) + b^2*log((b*x^2 + a)^(
2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2*b^2*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*
(b*x^2 + a)^(1/3)*b/(a*x^2))/b

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Mupad [B]
time = 5.05, size = 130, normalized size = 1.21 \begin {gather*} \frac {\ln \left (\frac {3\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2\,a^{2/3}}+\frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{a}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{5/3}}+\frac {\ln \left (\frac {3\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2\,a^{2/3}}+\frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{a}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{5/3}}-\frac {b\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{3\,a^{5/3}}-\frac {{\left (b\,x^2+a\right )}^{1/3}}{2\,a\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^2)^(2/3)),x)

[Out]

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

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