3.708 \(\int \frac {1}{x^5 \sqrt [3]{a+b x^2}} \, dx\)

Optimal. Leaf size=138 \[ \frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{7/3}}+\frac {b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}-\frac {b^2 \log (x)}{9 a^{7/3}}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4} \]

[Out]

-1/4*(b*x^2+a)^(2/3)/a/x^4+1/3*b*(b*x^2+a)^(2/3)/a^2/x^2-1/9*b^2*ln(x)/a^(7/3)+1/6*b^2*ln(a^(1/3)-(b*x^2+a)^(1
/3))/a^(7/3)+1/9*b^2*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(7/3)*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, 55, 617, 204, 31} \[ \frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{7/3}}+\frac {b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}-\frac {b^2 \log (x)}{9 a^{7/3}}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2)^(1/3)),x]

[Out]

-(a + b*x^2)^(2/3)/(4*a*x^4) + (b*(a + b*x^2)^(2/3))/(3*a^2*x^2) + (b^2*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))
/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)) - (b^2*Log[x])/(9*a^(7/3)) + (b^2*Log[a^(1/3) - (a + b*x^2)^(1/3)])/(
6*a^(7/3))

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^5 \sqrt [3]{a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [3]{a+b x}} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4}-\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{3 a}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{9 a^2}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}-\frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{6 a^{7/3}}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )}{6 a^2}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}-\frac {b^2 \log (x)}{9 a^{7/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{7/3}}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )}{3 a^{7/3}}\\ &=-\frac {\left (a+b x^2\right )^{2/3}}{4 a x^4}+\frac {b \left (a+b x^2\right )^{2/3}}{3 a^2 x^2}+\frac {b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{7/3}}-\frac {b^2 \log (x)}{9 a^{7/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.28 \[ -\frac {3 b^2 \left (a+b x^2\right )^{2/3} \, _2F_1\left (\frac {2}{3},3;\frac {5}{3};\frac {b x^2}{a}+1\right )}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 326, normalized size = 2.36 \[ \left [\frac {6 \, \sqrt {\frac {1}{3}} a b^{2} x^{4} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) - 2 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 4 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, a^{3} x^{4}}, \frac {12 \, \sqrt {\frac {1}{3}} a^{\frac {2}{3}} b^{2} x^{4} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right ) - 2 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 4 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, a^{3} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/36*(6*sqrt(1/3)*a*b^2*x^4*sqrt(-1/a^(2/3))*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^(4/3))*sqrt(-1/a^(2/3)) - 3*(b*x^2 + a)^(1/3)*a^(2/3) + 3*a)/x^2) - 2*a^(2/3)*b^2*x^4*log((b
*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 4*a^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3) - a^(1/3)) +
3*(4*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(2/3))/(a^3*x^4), 1/36*(12*sqrt(1/3)*a^(2/3)*b^2*x^4*arctan(sqrt(1/3)*(2*(b*
x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) - 2*a^(2/3)*b^2*x^4*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(
2/3)) + 4*a^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3*(4*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(2/3))/(a^3*x^4
)]

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giac [A]  time = 1.13, size = 142, normalized size = 1.03 \[ \frac {\frac {4 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {7}{3}}} - \frac {2 \, b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {7}{3}}} + \frac {4 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {7}{3}}} + \frac {3 \, {\left (4 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} b^{3} - 7 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{4}}}{36 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(4*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(7/3) - 2*b^3*log((b*x^2 + a
)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(7/3) + 4*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(7/3) +
3*(4*(b*x^2 + a)^(5/3)*b^3 - 7*(b*x^2 + a)^(2/3)*a*b^3)/(a^2*b^2*x^4))/b

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{3}} x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 158, normalized size = 1.14 \[ \frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {7}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {7}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {7}{3}}} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} b^{2} - 7 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{2} + a\right )} a^{3} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(7/3) - 1/18*b^2*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(7/3) + 1/9*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(7/3) + 1/1
2*(4*(b*x^2 + a)^(5/3)*b^2 - 7*(b*x^2 + a)^(2/3)*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 = 5.06, size = 201, normalized size = 1.46 \[ \frac {b^2\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{9\,a^{7/3}}-\frac {\ln \left (\frac {b^4\,{\left (b\,x^2+a\right )}^{1/3}}{9\,a^4}-\frac {{\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}^2}{36\,a^{11/3}}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{18\,a^{7/3}}-\frac {\frac {7\,b^2\,{\left (b\,x^2+a\right )}^{2/3}}{6\,a}-\frac {2\,b^2\,{\left (b\,x^2+a\right )}^{5/3}}{3\,a^2}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {b^2\,\ln \left (\frac {b^4\,{\left (b\,x^2+a\right )}^{1/3}}{9\,a^4}-\frac {b^4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{9\,a^{11/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{7/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2)^(1/3)),x)

[Out]

(b^2*log((a + b*x^2)^(1/3) - a^(1/3)))/(9*a^(7/3)) - (log((b^4*(a + b*x^2)^(1/3))/(9*a^4) - (3^(1/2)*b^2*1i +
b^2)^2/(36*a^(11/3)))*(3^(1/2)*b^2*1i + b^2))/(18*a^(7/3)) - ((7*b^2*(a + b*x^2)^(2/3))/(6*a) - (2*b^2*(a + b*
x^2)^(5/3))/(3*a^2))/(2*(a + b*x^2)^2 - 4*a*(a + b*x^2) + 2*a^2) + (b^2*log((b^4*(a + b*x^2)^(1/3))/(9*a^4) -
(b^4*((3^(1/2)*1i)/2 - 1/2)^2)/(9*a^(11/3)))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(7/3))

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sympy [C]  time = 1.37, size = 41, normalized size = 0.30 \[ - \frac {\Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \sqrt [3]{b} x^{\frac {14}{3}} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(7/3)*hyper((1/3, 7/3), (10/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(1/3)*x**(14/3)*gamma(10/3))

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