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

Optimal. Leaf size=135 \[ -\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}}+\frac {b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{5/3}}+\frac {b^2 \log (x)}{18 a^{5/3}}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}-\frac {\sqrt [3]{a+b x^2}}{4 x^4} \]

[Out]

-1/4*(b*x^2+a)^(1/3)/x^4-1/12*b*(b*x^2+a)^(1/3)/a/x^2+1/18*b^2*ln(x)/a^(5/3)-1/12*b^2*ln(a^(1/3)-(b*x^2+a)^(1/
3))/a^(5/3)+1/18*b^2*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 = 0.09, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {266, 47, 51, 57, 617, 204, 31} \[ -\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}}+\frac {b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{5/3}}+\frac {b^2 \log (x)}{18 a^{5/3}}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}-\frac {\sqrt [3]{a+b x^2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/x^5,x]

[Out]

-(a + b*x^2)^(1/3)/(4*x^4) - (b*(a + b*x^2)^(1/3))/(12*a*x^2) + (b^2*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(S
qrt[3]*a^(1/3))])/(6*Sqrt[3]*a^(5/3)) + (b^2*Log[x])/(18*a^(5/3)) - (b^2*Log[a^(1/3) - (a + b*x^2)^(1/3)])/(12
*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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 {\sqrt [3]{a+b x^2}}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}+\frac {1}{12} b \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{2/3}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^2\right )}{18 a}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}+\frac {b^2 \log (x)}{18 a^{5/3}}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}}+\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 )}{12 a^{4/3}}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}+\frac {b^2 \log (x)}{18 a^{5/3}}-\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}}-\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 )}{6 a^{5/3}}\\ &=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}+\frac {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^{5/3}}+\frac {b^2 \log (x)}{18 a^{5/3}}-\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.29 \[ -\frac {3 b^2 \left (a+b x^2\right )^{4/3} \, _2F_1\left (\frac {4}{3},3;\frac {7}{3};\frac {b x^2}{a}+1\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/x^5,x]

[Out]

(-3*b^2*(a + b*x^2)^(4/3)*Hypergeometric2F1[4/3, 3, 7/3, 1 + (b*x^2)/a])/(8*a^3)

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fricas [A]  time = 0.69, size = 199, normalized size = 1.47 \[ \frac {2 \, \sqrt {3} a b^{2} x^{4} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (a^{2} b x^{2} + 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{36 \, a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.37, size = 140, normalized size = 1.04 \[ \frac {\frac {2 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{3} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a b^{2} x^{4}}}{36 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.91, size = 155, normalized size = 1.15 \[ \frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{18 \, a^{\frac {5}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{36 \, a^{\frac {5}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{18 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a - 2 \, {\left (b x^{2} + a\right )} a^{2} + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.11, size = 217, normalized size = 1.61 \[ \frac {b^2\,\ln \left (\frac {b^2}{2\,{\left (-a\right )}^{2/3}}-\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}\right )}{18\,{\left (-a\right )}^{5/3}}-\frac {\ln \left (\frac {b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{2/3}}+\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{36\,{\left (-a\right )}^{5/3}}-\frac {\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3}+\frac {b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6\,a}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {b^2\,\ln \left (\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}-\frac {b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,{\left (-a\right )}^{2/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,{\left (-a\right )}^{5/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/x^5,x)

[Out]

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

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sympy [C]  time = 1.39, size = 42, normalized size = 0.31 \[ - \frac {\sqrt [3]{b} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac {10}{3}} \Gamma \left (\frac {8}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(1/3)*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**2))/(2*x**(10/3)*gamma(8/3))

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