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

Optimal. Leaf size=116 \[ -\frac {\left (a+b x^2\right )^{4/3}}{2 x^2}+2 b \sqrt [3]{a+b x^2}+\sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {2 \sqrt [3]{a} b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3}}-\frac {2}{3} \sqrt [3]{a} b \log (x) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 116, 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, 50, 57, 617, 204, 31} \[ -\frac {\left (a+b x^2\right )^{4/3}}{2 x^2}+2 b \sqrt [3]{a+b x^2}+\sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {2 \sqrt [3]{a} b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3}}-\frac {2}{3} \sqrt [3]{a} b \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^3,x]

[Out]

2*b*(a + b*x^2)^(1/3) - (a + b*x^2)^(4/3)/(2*x^2) - (2*a^(1/3)*b*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(Sqrt[
3]*a^(1/3))])/Sqrt[3] - (2*a^(1/3)*b*Log[x])/3 + a^(1/3)*b*Log[a^(1/3) - (a + b*x^2)^(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 50

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.32 \[ \frac {3 b \left (a+b x^2\right )^{7/3} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {b x^2}{a}+1\right )}{14 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^3,x]

[Out]

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

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fricas [A]  time = 0.98, size = 129, normalized size = 1.11 \[ -\frac {4 \, \sqrt {3} a^{\frac {1}{3}} b x^{2} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 2 \, a^{\frac {1}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, b x^{2} - a\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{6 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.42, size = 131, normalized size = 1.13 \[ -\frac {4 \, \sqrt {3} a^{\frac {1}{3}} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, a^{\frac {1}{3}} b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} b^{2} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) - 9 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b}{x^{2}}}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(4*sqrt(3)*a^(1/3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) + 2*a^(1/3)*b^2*log((b
*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) - 4*a^(1/3)*b^2*log(abs((b*x^2 + a)^(1/3) - a^(1/3))) -
9*(b*x^2 + a)^(1/3)*b^2 + 3*(b*x^2 + a)^(1/3)*a*b/x^2)/b

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.90, size = 116, normalized size = 1.00 \[ -\frac {2}{3} \, \sqrt {3} a^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{3} \, a^{\frac {1}{3}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {2}{3} \, a^{\frac {1}{3}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} b - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}} a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.42, size = 141, normalized size = 1.22 \[ \frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {a\,{\left (b\,x^2+a\right )}^{1/3}}{2\,x^2}+\frac {2\,a^{1/3}\,b\,\ln \left (6\,a^{4/3}\,b-6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}\right )}{3}+\frac {a^{1/3}\,b\,\ln \left (6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}-3\,a^{4/3}\,b\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{3}-\frac {a^{1/3}\,b\,\ln \left (3\,a^{4/3}\,b\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )+6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(4/3)/x^3,x)

[Out]

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

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sympy [C]  time = 1.41, size = 46, normalized size = 0.40 \[ - \frac {b^{\frac {4}{3}} x^{\frac {2}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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