3.187 \(\int \frac {1}{x^3 (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=49 \[ -\frac {c^2 \log \left (b+c x^2\right )}{2 b^3}+\frac {c^2 \log (x)}{b^3}+\frac {c}{2 b^2 x^2}-\frac {1}{4 b x^4} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, 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} \[ -\frac {c^2 \log \left (b+c x^2\right )}{2 b^3}+\frac {c^2 \log (x)}{b^3}+\frac {c}{2 b^2 x^2}-\frac {1}{4 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(b*x^2 + c*x^4)),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.00 \[ -\frac {c^2 \log \left (b+c x^2\right )}{2 b^3}+\frac {c^2 \log (x)}{b^3}+\frac {c}{2 b^2 x^2}-\frac {1}{4 b x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(b*x^2 + c*x^4)),x]

[Out]

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

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fricas [A]  time = 0.84, size = 45, normalized size = 0.92 \[ -\frac {2 \, c^{2} x^{4} \log \left (c x^{2} + b\right ) - 4 \, c^{2} x^{4} \log \relax (x) - 2 \, b c x^{2} + b^{2}}{4 \, b^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 57, normalized size = 1.16 \[ \frac {c^{2} \log \left (x^{2}\right )}{2 \, b^{3}} - \frac {c^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3}} - \frac {3 \, c^{2} x^{4} - 2 \, b c x^{2} + b^{2}}{4 \, b^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 44, normalized size = 0.90 \[ \frac {c^{2} \ln \relax (x )}{b^{3}}-\frac {c^{2} \ln \left (c \,x^{2}+b \right )}{2 b^{3}}+\frac {c}{2 b^{2} x^{2}}-\frac {1}{4 b \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 1.32, size = 47, normalized size = 0.96 \[ -\frac {c^{2} \log \left (c x^{2} + b\right )}{2 \, b^{3}} + \frac {c^{2} \log \left (x^{2}\right )}{2 \, b^{3}} + \frac {2 \, c x^{2} - b}{4 \, b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 46, normalized size = 0.94 \[ \frac {c^2\,\ln \relax (x)}{b^3}-\frac {c^2\,\ln \left (c\,x^2+b\right )}{2\,b^3}-\frac {\frac {1}{4\,b}-\frac {c\,x^2}{2\,b^2}}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x^2 + c*x^4)),x)

[Out]

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

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sympy [A]  time = 0.36, size = 42, normalized size = 0.86 \[ \frac {- b + 2 c x^{2}}{4 b^{2} x^{4}} + \frac {c^{2} \log {\relax (x )}}{b^{3}} - \frac {c^{2} \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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