3.4.30 \(\int \frac {x^5}{(a+b x^3)^2} \, dx\) [330]

Optimal. Leaf size=33 \[ \frac {a}{3 b^2 \left (a+b x^3\right )}+\frac {\log \left (a+b x^3\right )}{3 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \begin {gather*} \frac {a}{3 b^2 \left (a+b x^3\right )}+\frac {\log \left (a+b x^3\right )}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^3)^2,x]

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.82 \begin {gather*} \frac {\frac {a}{a+b x^3}+\log \left (a+b x^3\right )}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.12, size = 30, normalized size = 0.91

method	result	size
default	\(\frac {a}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {\ln \left (b \,x^{3}+a \right )}{3 b^{2}}\)	\(30\)
norman	\(\frac {a}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {\ln \left (b \,x^{3}+a \right )}{3 b^{2}}\)	\(30\)
risch	\(\frac {a}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {\ln \left (b \,x^{3}+a \right )}{3 b^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 32, normalized size = 0.97 \begin {gather*} \frac {a}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {\log \left (b x^{3} + a\right )}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 35, normalized size = 1.06 \begin {gather*} \frac {{\left (b x^{3} + a\right )} \log \left (b x^{3} + a\right ) + a}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.11, size = 29, normalized size = 0.88 \begin {gather*} \frac {a}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {\log {\left (a + b x^{3} \right )}}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(3*a*b**2 + 3*b**3*x**3) + log(a + b*x**3)/(3*b**2)

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Giac [A]
time = 1.06, size = 48, normalized size = 1.45 \begin {gather*} -\frac {\frac {\log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{3} + a\right )} b}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 29, normalized size = 0.88 \begin {gather*} \frac {\ln \left (b\,x^3+a\right )}{3\,b^2}+\frac {a}{3\,b^2\,\left (b\,x^3+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^3)^2,x)

[Out]

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

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