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

Optimal. Leaf size=104 \[ \frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{4/3}}-\frac {\log (x)}{2 a^{4/3}}+\frac {3}{2 a \sqrt [3]{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 104, 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 = {266, 51, 55, 617, 204, 31} \[ \frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{4/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{4/3}}-\frac {\log (x)}{2 a^{4/3}}+\frac {3}{2 a \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)^(4/3)),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.35 \[ \frac {3 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {b x^2}{a}+1\right )}{2 a \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)^(4/3)),x]

[Out]

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

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fricas [A]  time = 1.01, size = 327, normalized size = 3.14 \[ \left [\frac {\sqrt {3} {\left (a b x^{2} + a^{2}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} + \sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) - {\left (b x^{2} + a\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, {\left (b x^{2} + a\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 6 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{4 \, {\left (a^{2} b x^{2} + a^{3}\right )}}, -\frac {{\left (b x^{2} + a\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 2 \, {\left (b x^{2} + a\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - \frac {2 \, \sqrt {3} {\left (a b x^{2} + a^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - 6 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{4 \, {\left (a^{2} b x^{2} + a^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.10, size = 101, normalized size = 0.97 \[ \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {4}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {4}{3}}} + \frac {\log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{2 \, a^{\frac {4}{3}}} + \frac {3}{2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {4}{3}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.02, size = 100, normalized size = 0.96 \[ \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {4}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {4}{3}}} + \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{2 \, a^{\frac {4}{3}}} + \frac {3}{2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.59, size = 123, normalized size = 1.18 \[ \frac {\ln \left (18\,a\,{\left (b\,x^2+a\right )}^{1/3}-18\,a^{4/3}\right )}{2\,a^{4/3}}+\frac {3}{2\,a\,{\left (b\,x^2+a\right )}^{1/3}}+\frac {\ln \left (18\,a\,{\left (b\,x^2+a\right )}^{1/3}-\frac {9\,a^{4/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{4/3}}-\frac {\ln \left (18\,a\,{\left (b\,x^2+a\right )}^{1/3}-\frac {9\,a^{4/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^2)^(4/3)),x)

[Out]

log(18*a*(a + b*x^2)^(1/3) - 18*a^(4/3))/(2*a^(4/3)) + 3/(2*a*(a + b*x^2)^(1/3)) + (log(18*a*(a + b*x^2)^(1/3)
- (9*a^(4/3)*(3^(1/2)*1i - 1)^2)/2)*(3^(1/2)*1i - 1))/(4*a^(4/3)) - (log(18*a*(a + b*x^2)^(1/3) - (9*a^(4/3)*
(3^(1/2)*1i + 1)^2)/2)*(3^(1/2)*1i + 1))/(4*a^(4/3))

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sympy [C]  time = 1.13, size = 41, normalized size = 0.39 \[ - \frac {\Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {4}{3}} x^{\frac {8}{3}} \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(4/3)*hyper((4/3, 4/3), (7/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(4/3)*x**(8/3)*gamma(7/3))

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