3.4.33 \(\int \frac {1}{x^4 (a+b x^3)^2} \, dx\) [333]

Optimal. Leaf size=52 \[ -\frac {1}{3 a^2 x^3}-\frac {b}{3 a^2 \left (a+b x^3\right )}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log \left (a+b x^3\right )}{3 a^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 52, 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, 46} \begin {gather*} \frac {2 b \log \left (a+b x^3\right )}{3 a^3}-\frac {2 b \log (x)}{a^3}-\frac {b}{3 a^2 \left (a+b x^3\right )}-\frac {1}{3 a^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)^2),x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 {1}{x^4 \left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {1}{3 a^2 x^3}-\frac {b}{3 a^2 \left (a+b x^3\right )}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log \left (a+b x^3\right )}{3 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 41, normalized size = 0.79 \begin {gather*} -\frac {a \left (\frac {1}{x^3}+\frac {b}{a+b x^3}\right )+6 b \log (x)-2 b \log \left (a+b x^3\right )}{3 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^3)^2),x]

[Out]

-1/3*(a*(x^(-3) + b/(a + b*x^3)) + 6*b*Log[x] - 2*b*Log[a + b*x^3])/a^3

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Maple [A]
time = 0.12, size = 55, normalized size = 1.06

method result size
norman \(\frac {-\frac {1}{3 a}+\frac {2 b^{2} x^{6}}{3 a^{3}}}{x^{3} \left (b \,x^{3}+a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (b \,x^{3}+a \right )}{3 a^{3}}\) \(54\)
default \(\frac {b^{2} \left (-\frac {a}{b \left (b \,x^{3}+a \right )}+\frac {2 \ln \left (b \,x^{3}+a \right )}{b}\right )}{3 a^{3}}-\frac {1}{3 a^{2} x^{3}}-\frac {2 b \ln \left (x \right )}{a^{3}}\) \(55\)
risch \(\frac {-\frac {2 b \,x^{3}}{3 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b \,x^{3}+a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (-b \,x^{3}-a \right )}{3 a^{3}}\) \(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 53, normalized size = 1.02 \begin {gather*} -\frac {2 \, b x^{3} + a}{3 \, {\left (a^{2} b x^{6} + a^{3} x^{3}\right )}} + \frac {2 \, b \log \left (b x^{3} + a\right )}{3 \, a^{3}} - \frac {2 \, b \log \left (x^{3}\right )}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 73, normalized size = 1.40 \begin {gather*} -\frac {2 \, a b x^{3} + a^{2} - 2 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \log \left (x\right )}{3 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.22, size = 54, normalized size = 1.04 \begin {gather*} \frac {- a - 2 b x^{3}}{3 a^{3} x^{3} + 3 a^{2} b x^{6}} - \frac {2 b \log {\left (x \right )}}{a^{3}} + \frac {2 b \log {\left (\frac {a}{b} + x^{3} \right )}}{3 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.35, size = 51, normalized size = 0.98 \begin {gather*} \frac {2 \, b \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, b x^{3} + a}{3 \, {\left (b x^{6} + a x^{3}\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 53, normalized size = 1.02 \begin {gather*} \frac {2\,b\,\ln \left (b\,x^3+a\right )}{3\,a^3}-\frac {\frac {1}{3\,a}+\frac {2\,b\,x^3}{3\,a^2}}{b\,x^6+a\,x^3}-\frac {2\,b\,\ln \left (x\right )}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^3)^2),x)

[Out]

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

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