3.1.71 \(\int \frac {1}{x^5 (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=63 \[ \frac {c^3 \log \left (b+c x^2\right )}{2 b^4}-\frac {c^3 \log (x)}{b^4}-\frac {c^2}{2 b^3 x^2}+\frac {c}{4 b^2 x^4}-\frac {1}{6 b x^6} \]

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1584, 266, 44} \begin {gather*} -\frac {c^2}{2 b^3 x^2}+\frac {c^3 \log \left (b+c x^2\right )}{2 b^4}-\frac {c^3 \log (x)}{b^4}+\frac {c}{4 b^2 x^4}-\frac {1}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(b*x^2 + c*x^4)),x]

[Out]

-1/(6*b*x^6) + c/(4*b^2*x^4) - c^2/(2*b^3*x^2) - (c^3*Log[x])/b^4 + (c^3*Log[b + c*x^2])/(2*b^4)

Rule 44

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] && L
tQ[m + n + 2, 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 1584

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

Rubi steps

\begin {align*} \int \frac {1}{x^5 \left (b x^2+c x^4\right )} \, dx &=\int \frac {1}{x^7 \left (b+c x^2\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 (b+c x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{b x^4}-\frac {c}{b^2 x^3}+\frac {c^2}{b^3 x^2}-\frac {c^3}{b^4 x}+\frac {c^4}{b^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{6 b x^6}+\frac {c}{4 b^2 x^4}-\frac {c^2}{2 b^3 x^2}-\frac {c^3 \log (x)}{b^4}+\frac {c^3 \log \left (b+c x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 63, normalized size = 1.00 \begin {gather*} \frac {c^3 \log \left (b+c x^2\right )}{2 b^4}-\frac {c^3 \log (x)}{b^4}-\frac {c^2}{2 b^3 x^2}+\frac {c}{4 b^2 x^4}-\frac {1}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(b*x^2 + c*x^4)),x]

[Out]

-1/6*1/(b*x^6) + c/(4*b^2*x^4) - c^2/(2*b^3*x^2) - (c^3*Log[x])/b^4 + (c^3*Log[b + c*x^2])/(2*b^4)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^5 \left (b x^2+c x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^5*(b*x^2 + c*x^4)),x]

[Out]

IntegrateAlgebraic[1/(x^5*(b*x^2 + c*x^4)), x]

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fricas [A]  time = 0.74, size = 58, normalized size = 0.92 \begin {gather*} \frac {6 \, c^{3} x^{6} \log \left (c x^{2} + b\right ) - 12 \, c^{3} x^{6} \log \relax (x) - 6 \, b c^{2} x^{4} + 3 \, b^{2} c x^{2} - 2 \, b^{3}}{12 \, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*c^3*x^6*log(c*x^2 + b) - 12*c^3*x^6*log(x) - 6*b*c^2*x^4 + 3*b^2*c*x^2 - 2*b^3)/(b^4*x^6)

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giac [A]  time = 0.15, size = 70, normalized size = 1.11 \begin {gather*} -\frac {c^{3} \log \left (x^{2}\right )}{2 \, b^{4}} + \frac {c^{3} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4}} + \frac {11 \, c^{3} x^{6} - 6 \, b c^{2} x^{4} + 3 \, b^{2} c x^{2} - 2 \, b^{3}}{12 \, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^3*log(x^2)/b^4 + 1/2*c^3*log(abs(c*x^2 + b))/b^4 + 1/12*(11*c^3*x^6 - 6*b*c^2*x^4 + 3*b^2*c*x^2 - 2*b^3
)/(b^4*x^6)

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maple [A]  time = 0.01, size = 56, normalized size = 0.89 \begin {gather*} -\frac {c^{3} \ln \relax (x )}{b^{4}}+\frac {c^{3} \ln \left (c \,x^{2}+b \right )}{2 b^{4}}-\frac {c^{2}}{2 b^{3} x^{2}}+\frac {c}{4 b^{2} x^{4}}-\frac {1}{6 b \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 58, normalized size = 0.92 \begin {gather*} \frac {c^{3} \log \left (c x^{2} + b\right )}{2 \, b^{4}} - \frac {c^{3} \log \left (x^{2}\right )}{2 \, b^{4}} - \frac {6 \, c^{2} x^{4} - 3 \, b c x^{2} + 2 \, b^{2}}{12 \, b^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^3*log(c*x^2 + b)/b^4 - 1/2*c^3*log(x^2)/b^4 - 1/12*(6*c^2*x^4 - 3*b*c*x^2 + 2*b^2)/(b^3*x^6)

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mupad [B]  time = 0.07, size = 58, normalized size = 0.92 \begin {gather*} \frac {c^3\,\ln \left (c\,x^2+b\right )}{2\,b^4}-\frac {\frac {1}{6\,b}-\frac {c\,x^2}{4\,b^2}+\frac {c^2\,x^4}{2\,b^3}}{x^6}-\frac {c^3\,\ln \relax (x)}{b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(b*x^2 + c*x^4)),x)

[Out]

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

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sympy [A]  time = 0.40, size = 56, normalized size = 0.89 \begin {gather*} \frac {- 2 b^{2} + 3 b c x^{2} - 6 c^{2} x^{4}}{12 b^{3} x^{6}} - \frac {c^{3} \log {\relax (x )}}{b^{4}} + \frac {c^{3} \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*b**2 + 3*b*c*x**2 - 6*c**2*x**4)/(12*b**3*x**6) - c**3*log(x)/b**4 + c**3*log(b/c + x**2)/(2*b**4)

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