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

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

[Out]

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

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Rubi [A]  time = 0.07, 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, 47, 55, 617, 204, 31} \[ -\frac {\left (a+b x^2\right )^{2/3}}{2 x^2}+\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 \sqrt [3]{a}}+\frac {b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a}}-\frac {b \log (x)}{3 \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(2/3)/x^3,x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.36 \[ \frac {3 b \left (a+b x^2\right )^{5/3} \, _2F_1\left (\frac {5}{3},2;\frac {8}{3};\frac {b x^2}{a}+1\right )}{10 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(2/3)/x^3,x]

[Out]

(3*b*(a + b*x^2)^(5/3)*Hypergeometric2F1[5/3, 2, 8/3, 1 + (b*x^2)/a])/(10*a^2)

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fricas [A]  time = 1.01, size = 290, normalized size = 2.79 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} a b x^{2} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) - a^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{6 \, a x^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} a^{\frac {2}{3}} b x^{2} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right ) - a^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a}{6 \, a x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.37, size = 116, normalized size = 1.12 \[ \frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} + \frac {2 \, b^{2} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {1}{3}}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} b}{x^{2}}}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 103, normalized size = 0.99 \[ \frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {1}{3}}} - \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {1}{3}}} + \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {1}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {2}{3}}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.89, size = 136, normalized size = 1.31 \[ \frac {b\,\ln \left (a^{1/3}\,b^2-b^2\,{\left (b\,x^2+a\right )}^{1/3}\right )}{3\,a^{1/3}}-\frac {{\left (b\,x^2+a\right )}^{2/3}}{2\,x^2}-\frac {\ln \left (\frac {a^{1/3}\,{\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{4}-b^2\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{1/3}}-\frac {\ln \left (\frac {a^{1/3}\,{\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{4}-b^2\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(2/3)/x^3,x)

[Out]

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

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sympy [C]  time = 1.22, size = 42, normalized size = 0.40 \[ - \frac {b^{\frac {2}{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(2/3)*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), a*exp_polar(I*pi)/(b*x**2))/(2*x**(2/3)*gamma(4/3))

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