3.732 \(\int \frac {1}{x^5 (a+b x^2)^{4/3}} \, dx\)

Optimal. Leaf size=159 \[ \frac {7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{10/3}}+\frac {7 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{10/3}}-\frac {7 b^2 \log (x)}{9 a^{10/3}}+\frac {7 b^2}{3 a^3 \sqrt [3]{a+b x^2}}+\frac {7 b}{12 a^2 x^2 \sqrt [3]{a+b x^2}}-\frac {1}{4 a x^4 \sqrt [3]{a+b x^2}} \]

[Out]

7/3*b^2/a^3/(b*x^2+a)^(1/3)-1/4/a/x^4/(b*x^2+a)^(1/3)+7/12*b/a^2/x^2/(b*x^2+a)^(1/3)-7/9*b^2*ln(x)/a^(10/3)+7/
6*b^2*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(10/3)+7/9*b^2*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(
10/3)*3^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{10/3}}+\frac {7 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{10/3}}-\frac {7 b^2 \log (x)}{9 a^{10/3}}+\frac {7 b \left (a+b x^2\right )^{2/3}}{3 a^3 x^2}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}+\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2)^(4/3)),x]

[Out]

3/(2*a*x^4*(a + b*x^2)^(1/3)) - (7*(a + b*x^2)^(2/3))/(4*a^2*x^4) + (7*b*(a + b*x^2)^(2/3))/(3*a^3*x^2) + (7*b
^2*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(10/3)) - (7*b^2*Log[x])/(9*a^(10/3
)) + (7*b^2*Log[a^(1/3) - (a + b*x^2)^(1/3)])/(6*a^(10/3))

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^5 \left (a+b x^2\right )^{4/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x)^{4/3}} \, dx,x,x^2\right )\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{2 a}\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{3 a^2}\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}+\frac {7 b \left (a+b x^2\right )^{2/3}}{3 a^3 x^2}+\frac {\left (7 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{a+b x}} \, dx,x,x^2\right )}{9 a^3}\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}+\frac {7 b \left (a+b x^2\right )^{2/3}}{3 a^3 x^2}-\frac {7 b^2 \log (x)}{9 a^{10/3}}-\frac {\left (7 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{6 a^{10/3}}+\frac {\left (7 b^2\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 )}{6 a^3}\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}+\frac {7 b \left (a+b x^2\right )^{2/3}}{3 a^3 x^2}-\frac {7 b^2 \log (x)}{9 a^{10/3}}+\frac {7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{10/3}}-\frac {\left (7 b^2\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 )}{3 a^{10/3}}\\ &=\frac {3}{2 a x^4 \sqrt [3]{a+b x^2}}-\frac {7 \left (a+b x^2\right )^{2/3}}{4 a^2 x^4}+\frac {7 b \left (a+b x^2\right )^{2/3}}{3 a^3 x^2}+\frac {7 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^{10/3}}-\frac {7 b^2 \log (x)}{9 a^{10/3}}+\frac {7 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{10/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.25 \[ \frac {3 b^2 \, _2F_1\left (-\frac {1}{3},3;\frac {2}{3};\frac {b x^2}{a}+1\right )}{2 a^3 \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^2)^(4/3)),x]

[Out]

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

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fricas [A]  time = 0.81, size = 437, normalized size = 2.75 \[ \left [\frac {42 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{6} + a^{2} b^{2} x^{4}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) - 14 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 28 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (28 \, a b^{2} x^{4} + 7 \, a^{2} b x^{2} - 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}, -\frac {14 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 28 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - \frac {84 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{6} + a^{2} b^{2} x^{4}\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - 3 \, {\left (28 \, a b^{2} x^{4} + 7 \, a^{2} b x^{2} - 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(4/3),x, algorithm="fricas")

[Out]

[1/36*(42*sqrt(1/3)*(a*b^3*x^6 + a^2*b^2*x^4)*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) - 14*(b^3
*x^6 + a*b^2*x^4)*a^(2/3)*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 28*(b^3*x^6 + a*b^2*x
^4)*a^(2/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3*(28*a*b^2*x^4 + 7*a^2*b*x^2 - 3*a^3)*(b*x^2 + a)^(2/3))/(a^4*
b*x^6 + a^5*x^4), -1/36*(14*(b^3*x^6 + a*b^2*x^4)*a^(2/3)*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) +
a^(2/3)) - 28*(b^3*x^6 + a*b^2*x^4)*a^(2/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) - 84*sqrt(1/3)*(a*b^3*x^6 + a^2*b
^2*x^4)*arctan(sqrt(1/3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(1/3) - 3*(28*a*b^2*x^4 + 7*a^2*b*x^2 - 3*
a^3)*(b*x^2 + a)^(2/3))/(a^4*b*x^6 + a^5*x^4)]

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giac [A]  time = 1.10, size = 154, normalized size = 0.97 \[ \frac {7 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {10}{3}}} - \frac {7 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {10}{3}}} + \frac {7 \, b^{2} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{9 \, a^{\frac {10}{3}}} + \frac {3 \, b^{2}}{2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{3}} + \frac {10 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} b^{2} - 13 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a b^{2}}{12 \, a^{3} b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(4/3),x, algorithm="giac")

[Out]

7/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(10/3) - 7/18*b^2*log((b*x^2 + a
)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(10/3) + 7/9*b^2*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(10/
3) + 3/2*b^2/((b*x^2 + a)^(1/3)*a^3) + 1/12*(10*(b*x^2 + a)^(5/3)*b^2 - 13*(b*x^2 + a)^(2/3)*a*b^2)/(a^3*b^2*x
^4)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {4}{3}} x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^2+a)^(4/3),x)

[Out]

int(1/x^5/(b*x^2+a)^(4/3),x)

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maxima [A]  time = 2.97, size = 176, normalized size = 1.11 \[ \frac {7 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {10}{3}}} + \frac {28 \, {\left (b x^{2} + a\right )}^{2} b^{2} - 49 \, {\left (b x^{2} + a\right )} a b^{2} + 18 \, a^{2} b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{\frac {7}{3}} a^{3} - 2 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} a^{4} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{5}\right )}} - \frac {7 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {10}{3}}} + \frac {7 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {10}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(4/3),x, algorithm="maxima")

[Out]

7/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(10/3) + 1/12*(28*(b*x^2 + a)^2*
b^2 - 49*(b*x^2 + a)*a*b^2 + 18*a^2*b^2)/((b*x^2 + a)^(7/3)*a^3 - 2*(b*x^2 + a)^(4/3)*a^4 + (b*x^2 + a)^(1/3)*
a^5) - 7/18*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(10/3) + 7/9*b^2*log((b*x^2 + a
)^(1/3) - a^(1/3))/a^(10/3)

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mupad [B]  time = 5.67, size = 224, normalized size = 1.41 \[ \frac {\frac {3\,b^2}{a}-\frac {49\,b^2\,\left (b\,x^2+a\right )}{6\,a^2}+\frac {14\,b^2\,{\left (b\,x^2+a\right )}^2}{3\,a^3}}{2\,{\left (b\,x^2+a\right )}^{7/3}-4\,a\,{\left (b\,x^2+a\right )}^{4/3}+2\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}+\frac {7\,b^2\,\ln \left (147\,a^3\,b^4\,{\left (b\,x^2+a\right )}^{1/3}-147\,a^{10/3}\,b^4\right )}{9\,a^{10/3}}+\frac {7\,b^2\,\ln \left (147\,a^3\,b^4\,{\left (b\,x^2+a\right )}^{1/3}-147\,a^{10/3}\,b^4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}}-\frac {7\,b^2\,\ln \left (147\,a^3\,b^4\,{\left (b\,x^2+a\right )}^{1/3}-147\,a^{10/3}\,b^4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2)^(4/3)),x)

[Out]

((3*b^2)/a - (49*b^2*(a + b*x^2))/(6*a^2) + (14*b^2*(a + b*x^2)^2)/(3*a^3))/(2*(a + b*x^2)^(7/3) - 4*a*(a + b*
x^2)^(4/3) + 2*a^2*(a + b*x^2)^(1/3)) + (7*b^2*log(147*a^3*b^4*(a + b*x^2)^(1/3) - 147*a^(10/3)*b^4))/(9*a^(10
/3)) + (7*b^2*log(147*a^3*b^4*(a + b*x^2)^(1/3) - 147*a^(10/3)*b^4*((3^(1/2)*1i)/2 - 1/2)^2)*((3^(1/2)*1i)/2 -
1/2))/(9*a^(10/3)) - (7*b^2*log(147*a^3*b^4*(a + b*x^2)^(1/3) - 147*a^(10/3)*b^4*((3^(1/2)*1i)/2 + 1/2)^2)*((
3^(1/2)*1i)/2 + 1/2))/(9*a^(10/3))

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sympy [C]  time = 1.62, size = 41, normalized size = 0.26 \[ - \frac {\Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {4}{3}} x^{\frac {20}{3}} \Gamma \left (\frac {13}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**2+a)**(4/3),x)

[Out]

-gamma(10/3)*hyper((4/3, 10/3), (13/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(4/3)*x**(20/3)*gamma(13/3))

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