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

Optimal. Leaf size=86 \[ \frac{3 c^2}{2 b^4 \left (b+c x^2\right )}+\frac{c^2}{4 b^3 \left (b+c x^2\right )^2}-\frac{3 c^2 \log \left (b+c x^2\right )}{b^5}+\frac{6 c^2 \log (x)}{b^5}+\frac{3 c}{2 b^4 x^2}-\frac{1}{4 b^3 x^4} \]

[Out]

-1/(4*b^3*x^4) + (3*c)/(2*b^4*x^2) + c^2/(4*b^3*(b + c*x^2)^2) + (3*c^2)/(2*b^4*(b + c*x^2)) + (6*c^2*Log[x])/
b^5 - (3*c^2*Log[b + c*x^2])/b^5

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Rubi [A]  time = 0.0718563, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1584, 266, 44} \[ \frac{3 c^2}{2 b^4 \left (b+c x^2\right )}+\frac{c^2}{4 b^3 \left (b+c x^2\right )^2}-\frac{3 c^2 \log \left (b+c x^2\right )}{b^5}+\frac{6 c^2 \log (x)}{b^5}+\frac{3 c}{2 b^4 x^2}-\frac{1}{4 b^3 x^4} \]

Antiderivative was successfully verified.

[In]

Int[x/(b*x^2 + c*x^4)^3,x]

[Out]

-1/(4*b^3*x^4) + (3*c)/(2*b^4*x^2) + c^2/(4*b^3*(b + c*x^2)^2) + (3*c^2)/(2*b^4*(b + c*x^2)) + (6*c^2*Log[x])/
b^5 - (3*c^2*Log[b + c*x^2])/b^5

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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0456188, size = 74, normalized size = 0.86 \[ \frac{\frac{b \left (4 b^2 c x^2-b^3+18 b c^2 x^4+12 c^3 x^6\right )}{x^4 \left (b+c x^2\right )^2}-12 c^2 \log \left (b+c x^2\right )+24 c^2 \log (x)}{4 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b*x^2 + c*x^4)^3,x]

[Out]

((b*(-b^3 + 4*b^2*c*x^2 + 18*b*c^2*x^4 + 12*c^3*x^6))/(x^4*(b + c*x^2)^2) + 24*c^2*Log[x] - 12*c^2*Log[b + c*x
^2])/(4*b^5)

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Maple [A]  time = 0.056, size = 79, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,{b}^{3}{x}^{4}}}+{\frac{3\,c}{2\,{b}^{4}{x}^{2}}}+{\frac{{c}^{2}}{4\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{3\,{c}^{2}}{2\,{b}^{4} \left ( c{x}^{2}+b \right ) }}+6\,{\frac{{c}^{2}\ln \left ( x \right ) }{{b}^{5}}}-3\,{\frac{{c}^{2}\ln \left ( c{x}^{2}+b \right ) }{{b}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b^3/x^4+3/2*c/b^4/x^2+1/4*c^2/b^3/(c*x^2+b)^2+3/2*c^2/b^4/(c*x^2+b)+6*c^2*ln(x)/b^5-3*c^2*ln(c*x^2+b)/b^5

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Maxima [A]  time = 0.978365, size = 124, normalized size = 1.44 \begin{align*} \frac{12 \, c^{3} x^{6} + 18 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} - b^{3}}{4 \,{\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} - \frac{3 \, c^{2} \log \left (c x^{2} + b\right )}{b^{5}} + \frac{3 \, c^{2} \log \left (x^{2}\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*c^3*x^6 + 18*b*c^2*x^4 + 4*b^2*c*x^2 - b^3)/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4) - 3*c^2*log(c*x^2 +
b)/b^5 + 3*c^2*log(x^2)/b^5

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Fricas [A]  time = 1.50158, size = 274, normalized size = 3.19 \begin{align*} \frac{12 \, b c^{3} x^{6} + 18 \, b^{2} c^{2} x^{4} + 4 \, b^{3} c x^{2} - b^{4} - 12 \,{\left (c^{4} x^{8} + 2 \, b c^{3} x^{6} + b^{2} c^{2} x^{4}\right )} \log \left (c x^{2} + b\right ) + 24 \,{\left (c^{4} x^{8} + 2 \, b c^{3} x^{6} + b^{2} c^{2} x^{4}\right )} \log \left (x\right )}{4 \,{\left (b^{5} c^{2} x^{8} + 2 \, b^{6} c x^{6} + b^{7} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*b*c^3*x^6 + 18*b^2*c^2*x^4 + 4*b^3*c*x^2 - b^4 - 12*(c^4*x^8 + 2*b*c^3*x^6 + b^2*c^2*x^4)*log(c*x^2 +
b) + 24*(c^4*x^8 + 2*b*c^3*x^6 + b^2*c^2*x^4)*log(x))/(b^5*c^2*x^8 + 2*b^6*c*x^6 + b^7*x^4)

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Sympy [A]  time = 1.2709, size = 90, normalized size = 1.05 \begin{align*} \frac{- b^{3} + 4 b^{2} c x^{2} + 18 b c^{2} x^{4} + 12 c^{3} x^{6}}{4 b^{6} x^{4} + 8 b^{5} c x^{6} + 4 b^{4} c^{2} x^{8}} + \frac{6 c^{2} \log{\left (x \right )}}{b^{5}} - \frac{3 c^{2} \log{\left (\frac{b}{c} + x^{2} \right )}}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b**3 + 4*b**2*c*x**2 + 18*b*c**2*x**4 + 12*c**3*x**6)/(4*b**6*x**4 + 8*b**5*c*x**6 + 4*b**4*c**2*x**8) + 6*c
**2*log(x)/b**5 - 3*c**2*log(b/c + x**2)/b**5

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Giac [A]  time = 1.28543, size = 107, normalized size = 1.24 \begin{align*} -\frac{3 \, c^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{b^{5}} + \frac{6 \, c^{2} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac{12 \, c^{3} x^{6} + 18 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} - b^{3}}{4 \,{\left (c x^{4} + b x^{2}\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*c^2*log(abs(c*x^2 + b))/b^5 + 6*c^2*log(abs(x))/b^5 + 1/4*(12*c^3*x^6 + 18*b*c^2*x^4 + 4*b^2*c*x^2 - b^3)/(
(c*x^4 + b*x^2)^2*b^4)