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

Optimal. Leaf size=27 \[ \frac{x^2}{2 c}-\frac{b \log \left (b+c x^2\right )}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0266021, antiderivative size = 27, normalized size of antiderivative = 1., 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, 43} \[ \frac{x^2}{2 c}-\frac{b \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(b*x^2 + c*x^4),x]

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0046211, size = 27, normalized size = 1. \[ \frac{x^2}{2 c}-\frac{b \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(b*x^2 + c*x^4),x]

[Out]

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

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Maple [A]  time = 0.045, size = 24, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,c}}-{\frac{b\ln \left ( c{x}^{2}+b \right ) }{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989689, size = 31, normalized size = 1.15 \begin{align*} \frac{x^{2}}{2 \, c} - \frac{b \log \left (c x^{2} + b\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/c - 1/2*b*log(c*x^2 + b)/c^2

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Fricas [A]  time = 1.44295, size = 49, normalized size = 1.81 \begin{align*} \frac{c x^{2} - b \log \left (c x^{2} + b\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*x^2 - b*log(c*x^2 + b))/c^2

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Sympy [A]  time = 0.31091, size = 20, normalized size = 0.74 \begin{align*} - \frac{b \log{\left (b + c x^{2} \right )}}{2 c^{2}} + \frac{x^{2}}{2 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29538, size = 32, normalized size = 1.19 \begin{align*} \frac{x^{2}}{2 \, c} - \frac{b \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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