3.7.95 \(\int \frac {(a+b x^2)^{4/3}}{x^5} \, dx\) [695]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 43, 59, 631, 210, 31} \begin {gather*} -\frac {b^2 \text {ArcTan}\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 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/x^5,x]

[Out]

-1/3*(b*(a + b*x^2)^(1/3))/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 31

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 59

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 272

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 631

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^5} \, dx &=\frac {1}{2} \text {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 \text {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 \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.18, size = 148, normalized size = 1.12 \begin {gather*} \frac {1}{36} \left (-\frac {3 \sqrt [3]{a+b x^2} \left (3 a+7 b x^2\right )}{x^4}-\frac {4 \sqrt {3} b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {4 b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )}{a^{2/3}}-\frac {2 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{a^{2/3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/x^5,x]

[Out]

((-3*(a + b*x^2)^(1/3)*(3*a + 7*b*x^2))/x^4 - (4*Sqrt[3]*b^2*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3
]])/a^(2/3) + (4*b^2*Log[-a^(1/3) + (a + b*x^2)^(1/3)])/a^(2/3) - (2*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/
3) + (a + b*x^2)^(2/3)])/a^(2/3))/36

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{x^{5}}\, dx\]

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 [A]
time = 0.53, size = 152, normalized size = 1.15 \begin {gather*} -\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 )}{9 \, a^{\frac {2}{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 )}{18 \, a^{\frac {2}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} - 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(2/3) - 1/18*b^2*log((b*x^2 + a
)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) + 1/9*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(2/3) - 1/
12*(7*(b*x^2 + a)^(4/3)*b^2 - 4*(b*x^2 + a)^(1/3)*a*b^2)/((b*x^2 + a)^2 - 2*(b*x^2 + a)*a + a^2)

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Fricas [A]
time = 0.99, size = 174, normalized size = 1.32 \begin {gather*} -\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 {gather*}

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] Result contains complex when optimal does not.
time = 0.96, size = 42, normalized size = 0.32 \begin {gather*} - \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 {gather*}

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 = 1.74, size = 139, normalized size = 1.05 \begin {gather*} -\frac {\frac {4 \, \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 {2}{3}}} + \frac {2 \, 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 {2}{3}}} - \frac {4 \, b^{3} \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}} b^{3} - 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{b^{2} x^{4}}}{36 \, b} \end {gather*}

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

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Mupad [B]
time = 5.49, size = 191, normalized size = 1.45 \begin {gather*} \frac {b^2\,\ln \left (b^2\,{\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\,b^2\right )}{9\,a^{2/3}}-\frac {\ln \left (\frac {a^{1/3}\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{2}+b^2\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{18\,a^{2/3}}-\frac {\frac {7\,b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6}-\frac {2\,a\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {b^2\,\ln \left (b^2\,{\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(4/3)/x^5,x)

[Out]

(b^2*log(b^2*(a + b*x^2)^(1/3) - a^(1/3)*b^2))/(9*a^(2/3)) - (log((a^(1/3)*(3^(1/2)*b^2*1i + b^2))/2 + b^2*(a
+ b*x^2)^(1/3))*(3^(1/2)*b^2*1i + b^2))/(18*a^(2/3)) - ((7*b^2*(a + b*x^2)^(4/3))/6 - (2*a*b^2*(a + b*x^2)^(1/
3))/3)/(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) - a^(1/3)*b^2*((3^(1/2)*1i
)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(2/3))

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