\(\int \frac {-x+4 x^3}{(5+x^2)^2} \, dx\) [110]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=\frac {21}{2 \left (5+x^2\right )}+2 \log \left (5+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1607, 455, 45} \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=\frac {21}{2 \left (x^2+5\right )}+2 \log \left (x^2+5\right ) \]

[In]

Int[(-x + 4*x^3)/(5 + x^2)^2,x]

[Out]

21/(2*(5 + x^2)) + 2*Log[5 + x^2]

Rule 45

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])

Rule 455

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 1607

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (-1+4 x^2\right )}{\left (5+x^2\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {-1+4 x}{(5+x)^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=\frac {21}{2 \left (5+x^2\right )}+2 \log \left (5+x^2\right ) \]

[In]

Integrate[(-x + 4*x^3)/(5 + x^2)^2,x]

[Out]

21/(2*(5 + x^2)) + 2*Log[5 + x^2]

Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {21}{2 \left (x^{2}+5\right )}+2 \ln \left (x^{2}+5\right )\)	\(19\)
norman	\(\frac {21}{2 \left (x^{2}+5\right )}+2 \ln \left (x^{2}+5\right )\)	\(19\)
risch	\(\frac {21}{2 \left (x^{2}+5\right )}+2 \ln \left (x^{2}+5\right )\)	\(19\)
meijerg	\(-\frac {21 x^{2}}{50 \left (1+\frac {x^{2}}{5}\right )}+2 \ln \left (1+\frac {x^{2}}{5}\right )\)	\(26\)
parallelrisch	\(\frac {4 \ln \left (x^{2}+5\right ) x^{2}+21+20 \ln \left (x^{2}+5\right )}{2 x^{2}+10}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=\frac {4 \, {\left (x^{2} + 5\right )} \log \left (x^{2} + 5\right ) + 21}{2 \, {\left (x^{2} + 5\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=2 \log {\left (x^{2} + 5 \right )} + \frac {21}{2 x^{2} + 10} \]

[In]

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

[Out]

2*log(x**2 + 5) + 21/(2*x**2 + 10)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=\frac {21}{2 \, {\left (x^{2} + 5\right )}} + 2 \, \log \left (x^{2} + 5\right ) \]

[In]

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

[Out]

21/2/(x^2 + 5) + 2*log(x^2 + 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=-\frac {4 \, x^{2} - 1}{2 \, {\left (x^{2} + 5\right )}} + 2 \, \log \left (x^{2} + 5\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-x+4 x^3}{\left (5+x^2\right )^2} \, dx=2\,\ln \left (x^2+5\right )+\frac {21}{2\,\left (x^2+5\right )} \]

[In]

int(-(x - 4*x^3)/(x^2 + 5)^2,x)

[Out]

2*log(x^2 + 5) + 21/(2*(x^2 + 5))