3.695 \(\int \frac{(a+b x^2)^{4/3}}{x^5} \, dx\)

Optimal. Leaf size=132 \[ \frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{2/3}}-\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3}}-\frac{b^2 \log (x)}{9 a^{2/3}}-\frac{b \sqrt [3]{a+b x^2}}{3 x^2}-\frac{\left (a+b x^2\right )^{4/3}}{4 x^4} \]

[Out]

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

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Rubi [A]  time = 0.087383, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 47, 57, 617, 204, 31} \[ \frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{2/3}}-\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3}}-\frac{b^2 \log (x)}{9 a^{2/3}}-\frac{b \sqrt [3]{a+b x^2}}{3 x^2}-\frac{\left (a+b x^2\right )^{4/3}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^5,x]

[Out]

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

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 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 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 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]

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 0.009028, size = 39, normalized size = 0.3 \[ -\frac{3 b^2 \left (a+b x^2\right )^{7/3} \, _2F_1\left (\frac{7}{3},3;\frac{10}{3};\frac{b x^2}{a}+1\right )}{14 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^5,x]

[Out]

(-3*b^2*(a + b*x^2)^(7/3)*Hypergeometric2F1[7/3, 3, 10/3, 1 + (b*x^2)/a])/(14*a^3)

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/x^5,x)

[Out]

int((b*x^2+a)^(4/3)/x^5,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74868, size = 471, normalized size = 3.57 \begin{align*} -\frac{4 \, \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 ) + 2 \,{\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 ) - 4 \,{\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 (7 \, a^{2} b x^{2} + 3 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{3}}}{36 \, a^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 1.84654, size = 42, normalized size = 0.32 \begin{align*} - \frac{b^{\frac{4}{3}} \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac{4}{3}} \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/x**5,x)

[Out]

-b**(4/3)*gamma(2/3)*hyper((-4/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**2))/(2*x**(4/3)*gamma(5/3))

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Giac [A]  time = 4.57951, size = 167, normalized size = 1.27 \begin{align*} -\frac{1}{36} \, b^{2}{\left (\frac{4 \, \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 )}{a^{\frac{2}{3}}} + \frac{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{2}{3}}} - \frac{4 \, \log \left ({\left |{\left (b x^{2} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{2}{3}}} + \frac{3 \,{\left (7 \,{\left (b x^{2} + a\right )}^{\frac{4}{3}} - 4 \,{\left (b x^{2} + a\right )}^{\frac{1}{3}} a\right )}}{b^{2} x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="giac")

[Out]

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