3.111 \(\int \frac{x}{(1+x) (1+2 x)^2 (1+x^2)} \, dx\)

Optimal. Leaf size=46 \[ -\frac{7}{100} \log \left (x^2+1\right )+\frac{2}{5 (2 x+1)}-\frac{1}{2} \log (x+1)+\frac{16}{25} \log (2 x+1)+\frac{1}{50} \tan ^{-1}(x) \]

[Out]

2/(5*(1 + 2*x)) + ArcTan[x]/50 - Log[1 + x]/2 + (16*Log[1 + 2*x])/25 - (7*Log[1 + x^2])/100

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Rubi [A]  time = 0.204399, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {6725, 635, 203, 260} \[ -\frac{7}{100} \log \left (x^2+1\right )+\frac{2}{5 (2 x+1)}-\frac{1}{2} \log (x+1)+\frac{16}{25} \log (2 x+1)+\frac{1}{50} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x)*(1 + 2*x)^2*(1 + x^2)),x]

[Out]

2/(5*(1 + 2*x)) + ArcTan[x]/50 - Log[1 + x]/2 + (16*Log[1 + 2*x])/25 - (7*Log[1 + x^2])/100

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx &=\int \left (-\frac{1}{2 (1+x)}-\frac{4}{5 (1+2 x)^2}+\frac{32}{25 (1+2 x)}+\frac{1-7 x}{50 \left (1+x^2\right )}\right ) \, dx\\ &=\frac{2}{5 (1+2 x)}-\frac{1}{2} \log (1+x)+\frac{16}{25} \log (1+2 x)+\frac{1}{50} \int \frac{1-7 x}{1+x^2} \, dx\\ &=\frac{2}{5 (1+2 x)}-\frac{1}{2} \log (1+x)+\frac{16}{25} \log (1+2 x)+\frac{1}{50} \int \frac{1}{1+x^2} \, dx-\frac{7}{50} \int \frac{x}{1+x^2} \, dx\\ &=\frac{2}{5 (1+2 x)}+\frac{1}{50} \tan ^{-1}(x)-\frac{1}{2} \log (1+x)+\frac{16}{25} \log (1+2 x)-\frac{7}{100} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0187063, size = 40, normalized size = 0.87 \[ \frac{1}{100} \left (-7 \log \left (x^2+1\right )+\frac{40}{2 x+1}-50 \log (x+1)+64 \log (2 x+1)+2 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 + x)*(1 + 2*x)^2*(1 + x^2)),x]

[Out]

(40/(1 + 2*x) + 2*ArcTan[x] - 50*Log[1 + x] + 64*Log[1 + 2*x] - 7*Log[1 + x^2])/100

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Maple [A]  time = 0.008, size = 37, normalized size = 0.8 \begin{align*}{\frac{2}{5+10\,x}}+{\frac{\arctan \left ( x \right ) }{50}}-{\frac{\ln \left ( 1+x \right ) }{2}}+{\frac{16\,\ln \left ( 1+2\,x \right ) }{25}}-{\frac{7\,\ln \left ({x}^{2}+1 \right ) }{100}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(1+x)/(1+2*x)^2/(x^2+1),x)

[Out]

2/5/(1+2*x)+1/50*arctan(x)-1/2*ln(1+x)+16/25*ln(1+2*x)-7/100*ln(x^2+1)

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Maxima [A]  time = 1.42559, size = 49, normalized size = 1.07 \begin{align*} \frac{2}{5 \,{\left (2 \, x + 1\right )}} + \frac{1}{50} \, \arctan \left (x\right ) - \frac{7}{100} \, \log \left (x^{2} + 1\right ) + \frac{16}{25} \, \log \left (2 \, x + 1\right ) - \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/(1+2*x)^2/(x^2+1),x, algorithm="maxima")

[Out]

2/5/(2*x + 1) + 1/50*arctan(x) - 7/100*log(x^2 + 1) + 16/25*log(2*x + 1) - 1/2*log(x + 1)

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Fricas [A]  time = 2.11362, size = 171, normalized size = 3.72 \begin{align*} \frac{2 \,{\left (2 \, x + 1\right )} \arctan \left (x\right ) - 7 \,{\left (2 \, x + 1\right )} \log \left (x^{2} + 1\right ) + 64 \,{\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) - 50 \,{\left (2 \, x + 1\right )} \log \left (x + 1\right ) + 40}{100 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/(1+2*x)^2/(x^2+1),x, algorithm="fricas")

[Out]

1/100*(2*(2*x + 1)*arctan(x) - 7*(2*x + 1)*log(x^2 + 1) + 64*(2*x + 1)*log(2*x + 1) - 50*(2*x + 1)*log(x + 1)
+ 40)/(2*x + 1)

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Sympy [A]  time = 0.186602, size = 37, normalized size = 0.8 \begin{align*} \frac{16 \log{\left (x + \frac{1}{2} \right )}}{25} - \frac{\log{\left (x + 1 \right )}}{2} - \frac{7 \log{\left (x^{2} + 1 \right )}}{100} + \frac{\operatorname{atan}{\left (x \right )}}{50} + \frac{2}{10 x + 5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/(1+2*x)**2/(x**2+1),x)

[Out]

16*log(x + 1/2)/25 - log(x + 1)/2 - 7*log(x**2 + 1)/100 + atan(x)/50 + 2/(10*x + 5)

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Giac [A]  time = 1.06118, size = 84, normalized size = 1.83 \begin{align*} \frac{2}{5 \,{\left (2 \, x + 1\right )}} + \frac{1}{50} \, \arctan \left (-\frac{5}{2 \,{\left (2 \, x + 1\right )}} + \frac{1}{2}\right ) - \frac{7}{100} \, \log \left (-\frac{2}{2 \, x + 1} + \frac{5}{{\left (2 \, x + 1\right )}^{2}} + 1\right ) - \frac{1}{2} \, \log \left ({\left | -\frac{1}{2 \, x + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/(1+2*x)^2/(x^2+1),x, algorithm="giac")

[Out]

2/5/(2*x + 1) + 1/50*arctan(-5/2/(2*x + 1) + 1/2) - 7/100*log(-2/(2*x + 1) + 5/(2*x + 1)^2 + 1) - 1/2*log(abs(
-1/(2*x + 1) - 1))