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

Optimal. Leaf size=123 \[ -\frac{2 b}{a^2 \sqrt [3]{a+b x^2}}-\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{a^{7/3}}-\frac{2 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{7/3}}+\frac{2 b \log (x)}{3 a^{7/3}}-\frac{1}{2 a x^2 \sqrt [3]{a+b x^2}} \]

[Out]

(-2*b)/(a^2*(a + b*x^2)^(1/3)) - 1/(2*a*x^2*(a + b*x^2)^(1/3)) - (2*b*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(
Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(7/3)) + (2*b*Log[x])/(3*a^(7/3)) - (b*Log[a^(1/3) - (a + b*x^2)^(1/3)])/a^(7/3)

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Rubi [A]  time = 0.0801575, antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 51, 55, 617, 204, 31} \[ -\frac{2 \left (a+b x^2\right )^{2/3}}{a^2 x^2}-\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{a^{7/3}}-\frac{2 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{7/3}}+\frac{2 b \log (x)}{3 a^{7/3}}+\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)^(4/3)),x]

[Out]

3/(2*a*x^2*(a + b*x^2)^(1/3)) - (2*(a + b*x^2)^(2/3))/(a^2*x^2) - (2*b*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/
(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(7/3)) + (2*b*Log[x])/(3*a^(7/3)) - (b*Log[a^(1/3) - (a + b*x^2)^(1/3)])/a^(7/3
)

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 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 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]

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 31

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

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a+b x^2\right )^{4/3}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{4/3}} \, dx,x,x^2\right )\\ &=\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{a}\\ &=\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{2/3}}{a^2 x^2}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{3 a^2}\\ &=\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{2/3}}{a^2 x^2}+\frac{2 b \log (x)}{3 a^{7/3}}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{a^{7/3}}-\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 )}{a^2}\\ &=\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{2/3}}{a^2 x^2}+\frac{2 b \log (x)}{3 a^{7/3}}-\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{a^{7/3}}+\frac{(2 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 )}{a^{7/3}}\\ &=\frac{3}{2 a x^2 \sqrt [3]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{2/3}}{a^2 x^2}-\frac{2 b \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{7/3}}+\frac{2 b \log (x)}{3 a^{7/3}}-\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{a^{7/3}}\\ \end{align*}

Mathematica [C]  time = 0.0070998, size = 37, normalized size = 0.3 \[ -\frac{3 b \, _2F_1\left (-\frac{1}{3},2;\frac{2}{3};\frac{b x^2}{a}+1\right )}{2 a^2 \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89612, size = 1127, normalized size = 9.16 \begin{align*} \left [\frac{6 \, \sqrt{\frac{1}{3}}{\left (a b^{2} x^{4} + a^{2} b x^{2}\right )} \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 ) + 2 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \left (-a\right )^{\frac{2}{3}} \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 ) - 4 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \left (-a\right )^{\frac{2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac{1}{3}} + \left (-a\right )^{\frac{1}{3}}\right ) - 3 \,{\left (4 \, a b x^{2} + a^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{2}{3}}}{6 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, -\frac{12 \, \sqrt{\frac{1}{3}}{\left (a b^{2} x^{4} + a^{2} b x^{2}\right )} \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 ) - 2 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \left (-a\right )^{\frac{2}{3}} \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 ) + 4 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \left (-a\right )^{\frac{2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac{1}{3}} + \left (-a\right )^{\frac{1}{3}}\right ) + 3 \,{\left (4 \, a b x^{2} + a^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{2}{3}}}{6 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.58406, size = 41, normalized size = 0.33 \begin{align*} - \frac{\Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{4}{3}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac{4}{3}} x^{\frac{14}{3}} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(7/3)*hyper((4/3, 7/3), (10/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(4/3)*x**(14/3)*gamma(10/3))

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Giac [A]  time = 4.92595, size = 171, normalized size = 1.39 \begin{align*} -\frac{1}{6} \, b{\left (\frac{4 \, \sqrt{3} \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{7}{3}}} - \frac{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{7}{3}}} + \frac{4 \, \log \left ({\left |{\left (b x^{2} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{7}{3}}} + \frac{3 \,{\left (4 \, b x^{2} + a\right )}}{{\left ({\left (b x^{2} + a\right )}^{\frac{4}{3}} -{\left (b x^{2} + a\right )}^{\frac{1}{3}} a\right )} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*(4*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(7/3) - 2*log((b*x^2 + a)^(2/3
) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(7/3) + 4*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(7/3) + 3*(4*b*x^
2 + a)/(((b*x^2 + a)^(4/3) - (b*x^2 + a)^(1/3)*a)*a^2))