3.2.16 \(\int \frac {1}{x (a+b x^3) (c+d x^3)} \, dx\)

Optimal. Leaf size=62 \[ -\frac {b \log \left (a+b x^3\right )}{3 a (b c-a d)}+\frac {d \log \left (c+d x^3\right )}{3 c (b c-a d)}+\frac {\log (x)}{a c} \]

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {446, 72} \begin {gather*} -\frac {b \log \left (a+b x^3\right )}{3 a (b c-a d)}+\frac {d \log \left (c+d x^3\right )}{3 c (b c-a d)}+\frac {\log (x)}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)*(c + d*x^3)),x]

[Out]

Log[x]/(a*c) - (b*Log[a + b*x^3])/(3*a*(b*c - a*d)) + (d*Log[c + d*x^3])/(3*c*(b*c - a*d))

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.87 \begin {gather*} \frac {-b c \log \left (a+b x^3\right )+a d \log \left (c+d x^3\right )-3 a d \log (x)+3 b c \log (x)}{3 a b c^2-3 a^2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)*(c + d*x^3)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x*(a + b*x^3)*(c + d*x^3)),x]

[Out]

IntegrateAlgebraic[1/(x*(a + b*x^3)*(c + d*x^3)), x]

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fricas [A]  time = 1.56, size = 54, normalized size = 0.87 \begin {gather*} -\frac {b c \log \left (b x^{3} + a\right ) - a d \log \left (d x^{3} + c\right ) - 3 \, {\left (b c - a d\right )} \log \relax (x)}{3 \, {\left (a b c^{2} - a^{2} c d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 71, normalized size = 1.15 \begin {gather*} -\frac {b^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, {\left (a b^{2} c - a^{2} b d\right )}} + \frac {d^{2} \log \left ({\left | d x^{3} + c \right |}\right )}{3 \, {\left (b c^{2} d - a c d^{2}\right )}} + \frac {\log \left ({\left | x \right |}\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 59, normalized size = 0.95 \begin {gather*} \frac {b \ln \left (b \,x^{3}+a \right )}{3 \left (a d -b c \right ) a}-\frac {d \ln \left (d \,x^{3}+c \right )}{3 \left (a d -b c \right ) c}+\frac {\ln \relax (x )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^3+a)/(d*x^3+c),x)

[Out]

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

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maxima [A]  time = 0.54, size = 61, normalized size = 0.98 \begin {gather*} -\frac {b \log \left (b x^{3} + a\right )}{3 \, {\left (a b c - a^{2} d\right )}} + \frac {d \log \left (d x^{3} + c\right )}{3 \, {\left (b c^{2} - a c d\right )}} + \frac {\log \left (x^{3}\right )}{3 \, a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.84, size = 58, normalized size = 0.94 \begin {gather*} \frac {b\,\ln \left (b\,x^3+a\right )}{3\,a^2\,d-3\,a\,b\,c}+\frac {d\,\ln \left (d\,x^3+c\right )}{3\,b\,c^2-3\,a\,c\,d}+\frac {\ln \relax (x)}{a\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3)*(c + d*x^3)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**3+a)/(d*x**3+c),x)

[Out]

Timed out

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