3.110 \(\int \frac{-x+4 x^3}{(5+x^2)^2} \, dx\)

Optimal. Leaf size=20 \[ \frac{21}{2 \left (x^2+5\right )}+2 \log \left (x^2+5\right ) \]

[Out]

21/(2*(5 + x^2)) + 2*Log[5 + x^2]

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Rubi [A]  time = 0.0230683, antiderivative size = 20, 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 = {1593, 444, 43} \[ \frac{21}{2 \left (x^2+5\right )}+2 \log \left (x^2+5\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-x + 4*x^3)/(5 + x^2)^2,x]

[Out]

21/(2*(5 + x^2)) + 2*Log[5 + x^2]

Rule 1593

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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+4 x^3}{\left (5+x^2\right )^2} \, dx &=\int \frac{x \left (-1+4 x^2\right )}{\left (5+x^2\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{-1+4 x}{(5+x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{21}{(5+x)^2}+\frac{4}{5+x}\right ) \, dx,x,x^2\right )\\ &=\frac{21}{2 \left (5+x^2\right )}+2 \log \left (5+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0071059, size = 20, normalized size = 1. \[ \frac{21}{2 \left (x^2+5\right )}+2 \log \left (x^2+5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-x + 4*x^3)/(5 + x^2)^2,x]

[Out]

21/(2*(5 + x^2)) + 2*Log[5 + x^2]

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Maple [A]  time = 0.007, size = 19, normalized size = 1. \begin{align*}{\frac{21}{2\,{x}^{2}+10}}+2\,\ln \left ({x}^{2}+5 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3-x)/(x^2+5)^2,x)

[Out]

21/2/(x^2+5)+2*ln(x^2+5)

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Maxima [A]  time = 0.987198, size = 24, normalized size = 1.2 \begin{align*} \frac{21}{2 \,{\left (x^{2} + 5\right )}} + 2 \, \log \left (x^{2} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x)/(x^2+5)^2,x, algorithm="maxima")

[Out]

21/2/(x^2 + 5) + 2*log(x^2 + 5)

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Fricas [A]  time = 1.39457, size = 63, normalized size = 3.15 \begin{align*} \frac{4 \,{\left (x^{2} + 5\right )} \log \left (x^{2} + 5\right ) + 21}{2 \,{\left (x^{2} + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x)/(x^2+5)^2,x, algorithm="fricas")

[Out]

1/2*(4*(x^2 + 5)*log(x^2 + 5) + 21)/(x^2 + 5)

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Sympy [A]  time = 0.092506, size = 15, normalized size = 0.75 \begin{align*} 2 \log{\left (x^{2} + 5 \right )} + \frac{21}{2 x^{2} + 10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3-x)/(x**2+5)**2,x)

[Out]

2*log(x**2 + 5) + 21/(2*x**2 + 10)

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Giac [A]  time = 1.15171, size = 34, normalized size = 1.7 \begin{align*} -\frac{4 \, x^{2} - 1}{2 \,{\left (x^{2} + 5\right )}} + 2 \, \log \left (x^{2} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x)/(x^2+5)^2,x, algorithm="giac")

[Out]

-1/2*(4*x^2 - 1)/(x^2 + 5) + 2*log(x^2 + 5)