3.20.78 \(\int \frac {x^5}{(a+\frac {b}{x^3})^2} \, dx\) [1978]

Optimal. Leaf size=56 \[ -\frac {2 b x^3}{3 a^3}+\frac {x^6}{6 a^2}+\frac {b^3}{3 a^4 \left (b+a x^3\right )}+\frac {b^2 \log \left (b+a x^3\right )}{a^4} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 272, 45} \begin {gather*} \frac {b^3}{3 a^4 \left (a x^3+b\right )}+\frac {b^2 \log \left (a x^3+b\right )}{a^4}-\frac {2 b x^3}{3 a^3}+\frac {x^6}{6 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b/x^3)^2,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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]]

Rubi steps

\begin {align*} \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^2} \, dx &=\int \frac {x^{11}}{\left (b+a x^3\right )^2} \, dx\\ &=\frac {1}{3} \text {Subst}\left (\int \frac {x^3}{(b+a x)^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {2 b}{a^3}+\frac {x}{a^2}-\frac {b^3}{a^3 (b+a x)^2}+\frac {3 b^2}{a^3 (b+a x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 b x^3}{3 a^3}+\frac {x^6}{6 a^2}+\frac {b^3}{3 a^4 \left (b+a x^3\right )}+\frac {b^2 \log \left (b+a x^3\right )}{a^4}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 49, normalized size = 0.88 \begin {gather*} \frac {-4 a b x^3+a^2 x^6+\frac {2 b^3}{b+a x^3}+6 b^2 \log \left (b+a x^3\right )}{6 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b/x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.09, size = 54, normalized size = 0.96

method	result	size
norman	\(\frac {\frac {b^{3}}{a^{4}}+\frac {x^{9}}{6 a}-\frac {b \,x^{6}}{2 a^{2}}}{a \,x^{3}+b}+\frac {b^{2} \ln \left (a \,x^{3}+b \right )}{a^{4}}\)	\(52\)
default	\(\frac {b^{2} \left (\frac {b}{a \left (a \,x^{3}+b \right )}+\frac {3 \ln \left (a \,x^{3}+b \right )}{a}\right )}{3 a^{3}}+\frac {\left (a \,x^{3}-2 b \right )^{2}}{6 a^{4}}\)	\(54\)
risch	\(\frac {x^{6}}{6 a^{2}}-\frac {2 b \,x^{3}}{3 a^{3}}+\frac {2 b^{2}}{3 a^{4}}+\frac {b^{3}}{3 a^{4} \left (a \,x^{3}+b \right )}+\frac {b^{2} \ln \left (a \,x^{3}+b \right )}{a^{4}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a+b/x^3)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 53, normalized size = 0.95 \begin {gather*} \frac {b^{3}}{3 \, {\left (a^{5} x^{3} + a^{4} b\right )}} + \frac {b^{2} \log \left (a x^{3} + b\right )}{a^{4}} + \frac {a x^{6} - 4 \, b x^{3}}{6 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^3/(a^5*x^3 + a^4*b) + b^2*log(a*x^3 + b)/a^4 + 1/6*(a*x^6 - 4*b*x^3)/a^3

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Fricas [A]
time = 0.36, size = 70, normalized size = 1.25 \begin {gather*} \frac {a^{3} x^{9} - 3 \, a^{2} b x^{6} - 4 \, a b^{2} x^{3} + 2 \, b^{3} + 6 \, {\left (a b^{2} x^{3} + b^{3}\right )} \log \left (a x^{3} + b\right )}{6 \, {\left (a^{5} x^{3} + a^{4} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^3*x^9 - 3*a^2*b*x^6 - 4*a*b^2*x^3 + 2*b^3 + 6*(a*b^2*x^3 + b^3)*log(a*x^3 + b))/(a^5*x^3 + a^4*b)

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Sympy [A]
time = 0.14, size = 53, normalized size = 0.95 \begin {gather*} \frac {b^{3}}{3 a^{5} x^{3} + 3 a^{4} b} + \frac {x^{6}}{6 a^{2}} - \frac {2 b x^{3}}{3 a^{3}} + \frac {b^{2} \log {\left (a x^{3} + b \right )}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**3/(3*a**5*x**3 + 3*a**4*b) + x**6/(6*a**2) - 2*b*x**3/(3*a**3) + b**2*log(a*x**3 + b)/a**4

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Giac [A]
time = 1.37, size = 54, normalized size = 0.96 \begin {gather*} \frac {b^{2} \log \left ({\left | a x^{3} + b \right |}\right )}{a^{4}} + \frac {b^{3}}{3 \, {\left (a x^{3} + b\right )} a^{4}} + \frac {a^{2} x^{6} - 4 \, a b x^{3}}{6 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.13, size = 56, normalized size = 1.00 \begin {gather*} \frac {x^6}{6\,a^2}+\frac {b^3}{3\,a\,\left (a^4\,x^3+b\,a^3\right )}-\frac {2\,b\,x^3}{3\,a^3}+\frac {b^2\,\ln \left (a\,x^3+b\right )}{a^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b/x^3)^2,x)

[Out]

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

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