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

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

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Rubi [A]  time = 0.20, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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IntegrateAlgebraic [A]  time = 0.03, size = 46, normalized size = 1.00 \[ -\frac {7}{100} \log \left (x^2+1\right )+\frac {2}{5 (2 x+1)}-\frac {1}{2} \log (x+1)+\frac {16}{25} \log (2 x+1)+\frac {1}{50} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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

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fricas [A]  time = 0.85, size = 57, normalized size = 1.24 \[ \frac {2 \, {\left (2 \, x + 1\right )} \arctan \relax (x) - 7 \, {\left (2 \, x + 1\right )} \log \left (x^{2} + 1\right ) + 64 \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) - 50 \, {\left (2 \, x + 1\right )} \log \left (x + 1\right ) + 40}{100 \, {\left (2 \, x + 1\right )}} \]

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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giac [A]  time = 1.04, size = 62, normalized size = 1.35 \[ \frac {2}{5 \, {\left (2 \, x + 1\right )}} + \frac {1}{50} \, \arctan \left (-\frac {5}{2 \, {\left (2 \, x + 1\right )}} + \frac {1}{2}\right ) - \frac {7}{100} \, \log \left (-\frac {2}{2 \, x + 1} + \frac {5}{{\left (2 \, x + 1\right )}^{2}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | -\frac {1}{2 \, x + 1} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.32, size = 35, normalized size = 0.76

method	result	size
risch	\(\frac {1}{\frac {5}{2}+5 x}-\frac {\ln \left (1+x \right )}{2}+\frac {16 \ln \left (1+2 x \right )}{25}-\frac {7 \ln \left (x^{2}+1\right )}{100}+\frac {\arctan \relax (x )}{50}\)	\(35\)
default	\(\frac {2}{5 \left (1+2 x \right )}+\frac {\arctan \relax (x )}{50}-\frac {\ln \left (1+x \right )}{2}+\frac {16 \ln \left (1+2 x \right )}{25}-\frac {7 \ln \left (x^{2}+1\right )}{100}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(1+x)/(1+2*x)^2/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.98, size = 36, normalized size = 0.78 \[ \frac {2}{5 \, {\left (2 \, x + 1\right )}} + \frac {1}{50} \, \arctan \relax (x) - \frac {7}{100} \, \log \left (x^{2} + 1\right ) + \frac {16}{25} \, \log \left (2 \, x + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]

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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mupad [B]  time = 0.19, size = 38, normalized size = 0.83 \[ \frac {16\,\ln \left (x+\frac {1}{2}\right )}{25}-\frac {\ln \left (x+1\right )}{2}+\frac {1}{5\,\left (x+\frac {1}{2}\right )}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {7}{100}-\frac {1}{100}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {7}{100}+\frac {1}{100}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*log(x + 1/2))/25 - log(x + 1)/2 - log(x - 1i)*(7/100 + 1i/100) - log(x + 1i)*(7/100 - 1i/100) + 1/(5*(x +
1/2))

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sympy [A]  time = 0.22, size = 37, normalized size = 0.80 \[ \frac {16 \log {\left (x + \frac {1}{2} \right )}}{25} - \frac {\log {\left (x + 1 \right )}}{2} - \frac {7 \log {\left (x^{2} + 1 \right )}}{100} + \frac {\operatorname {atan}{\relax (x )}}{50} + \frac {2}{10 x + 5} \]

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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