3.720 \(\int \frac {1}{x^5 (a+b x^2)^{2/3}} \, dx\)

Optimal. Leaf size=138 \[ \frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}}-\frac {5 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {\sqrt [3]{a+b x^2}}{4 a x^4} \]

[Out]

-1/4*(b*x^2+a)^(1/3)/a/x^4+5/12*b*(b*x^2+a)^(1/3)/a^2/x^2-5/18*b^2*ln(x)/a^(8/3)+5/12*b^2*ln(a^(1/3)-(b*x^2+a)
^(1/3))/a^(8/3)-5/18*b^2*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(8/3)*3^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 51, 57, 617, 204, 31} \[ \frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}}-\frac {5 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {\sqrt [3]{a+b x^2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2)^(2/3)),x]

[Out]

-(a + b*x^2)^(1/3)/(4*a*x^4) + (5*b*(a + b*x^2)^(1/3))/(12*a^2*x^2) - (5*b^2*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(
1/3))/(Sqrt[3]*a^(1/3))])/(6*Sqrt[3]*a^(8/3)) - (5*b^2*Log[x])/(18*a^(8/3)) + (5*b^2*Log[a^(1/3) - (a + b*x^2)
^(1/3)])/(12*a^(8/3))

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 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 {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x)^{2/3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{2/3}} \, dx,x,x^2\right )}{12 a}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^2\right )}{18 a^2}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {5 b^2 \log (x)}{18 a^{8/3}}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )}{12 a^{7/3}}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )}{6 a^{8/3}}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {5 b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{6 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.28 \[ -\frac {3 b^2 \sqrt [3]{a+b x^2} \, _2F_1\left (\frac {1}{3},3;\frac {4}{3};\frac {b x^2}{a}+1\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 174, normalized size = 1.26 \[ -\frac {10 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{4} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + 5 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 3 \, {\left (5 \, a^{2} b x^{2} - 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{36 \, a^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

-1/36*(10*sqrt(3)*(a^2)^(1/6)*a*b^2*x^4*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(b*x^2 + a)^
(1/3)*(a^2)^(2/3))/a^2) + 5*(a^2)^(2/3)*b^2*x^4*log((b*x^2 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^2 + a)^(1/3)*(a
^2)^(2/3)) - 10*(a^2)^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3)*a - (a^2)^(2/3)) - 3*(5*a^2*b*x^2 - 3*a^3)*(b*x^2 +
a)^(1/3))/(a^4*x^4)

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giac [A]  time = 1.13, size = 142, normalized size = 1.03 \[ -\frac {\frac {10 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {8}{3}}} + \frac {5 \, b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {8}{3}}} - \frac {10 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{3} - 8 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{4}}}{36 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

-1/36*(10*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) + 5*b^3*log((b*x^2 +
a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) - 10*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(8/3
) - 3*(5*(b*x^2 + a)^(4/3)*b^3 - 8*(b*x^2 + a)^(1/3)*a*b^3)/(a^2*b^2*x^4))/b

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^2+a)^(2/3),x)

[Out]

int(1/x^5/(b*x^2+a)^(2/3),x)

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maxima [A]  time = 2.98, size = 158, normalized size = 1.14 \[ -\frac {5 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{18 \, a^{\frac {8}{3}}} - \frac {5 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{36 \, a^{\frac {8}{3}}} + \frac {5 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{18 \, a^{\frac {8}{3}}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} - 8 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{2} + a\right )} a^{3} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

-5/18*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) - 5/36*b^2*log((b*x^2 +
a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) + 5/18*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(8/3) +
1/12*(5*(b*x^2 + a)^(4/3)*b^2 - 8*(b*x^2 + a)^(1/3)*a*b^2)/((b*x^2 + a)^2*a^2 - 2*(b*x^2 + a)*a^3 + a^4)

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mupad [B]  time = 5.14, size = 193, normalized size = 1.40 \[ \frac {5\,b^2\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{18\,a^{8/3}}-\frac {\frac {4\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3\,a}-\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6\,a^2}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a^2}-\frac {5\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{5/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{8/3}}-\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a^2}+\frac {5\,b^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{5/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{8/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2)^(2/3)),x)

[Out]

(5*b^2*log((a + b*x^2)^(1/3) - a^(1/3)))/(18*a^(8/3)) - ((4*b^2*(a + b*x^2)^(1/3))/(3*a) - (5*b^2*(a + b*x^2)^
(4/3))/(6*a^2))/(2*(a + b*x^2)^2 - 4*a*(a + b*x^2) + 2*a^2) + (5*b^2*log((5*b^2*(a + b*x^2)^(1/3))/(2*a^2) - (
5*b^2*((3^(1/2)*1i)/2 - 1/2))/(2*a^(5/3)))*((3^(1/2)*1i)/2 - 1/2))/(18*a^(8/3)) - (5*b^2*log((5*b^2*(a + b*x^2
)^(1/3))/(2*a^2) + (5*b^2*((3^(1/2)*1i)/2 + 1/2))/(2*a^(5/3)))*((3^(1/2)*1i)/2 + 1/2))/(18*a^(8/3))

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sympy [C]  time = 1.45, size = 41, normalized size = 0.30 \[ - \frac {\Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {2}{3}} x^{\frac {16}{3}} \Gamma \left (\frac {11}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**2+a)**(2/3),x)

[Out]

-gamma(8/3)*hyper((2/3, 8/3), (11/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(2/3)*x**(16/3)*gamma(11/3))

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