\(\int \frac {1}{(1+x^2) (3+\frac {10 x}{1+x^2})} \, dx\) [377]

Optimal result

Integrand size = 22, antiderivative size = 19 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=-\frac {1}{8} \log (3+x)+\frac {1}{8} \log (1+3 x) \]

[Out]

-1/8*ln(3+x)+1/8*ln(1+3*x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6820, 630, 31} \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=\frac {1}{8} \log (3 x+1)-\frac {1}{8} \log (x+3) \]

[In]

Int[1/((1 + x^2)*(3 + (10*x)/(1 + x^2))),x]

[Out]

-1/8*Log[3 + x] + Log[1 + 3*x]/8

Rule 31

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

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{3+10 x+3 x^2} \, dx \\ & = \frac {3}{8} \int \frac {1}{1+3 x} \, dx-\frac {3}{8} \int \frac {1}{9+3 x} \, dx \\ & = -\frac {1}{8} \log (3+x)+\frac {1}{8} \log (1+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=-\frac {1}{8} \log (3+x)+\frac {1}{8} \log (1+3 x) \]

[In]

Integrate[1/((1 + x^2)*(3 + (10*x)/(1 + x^2))),x]

[Out]

-1/8*Log[3 + x] + Log[1 + 3*x]/8

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-\frac {\ln \left (3+x \right )}{8}+\frac {\ln \left (\frac {1}{3}+x \right )}{8}\)	\(14\)
default	\(-\frac {\ln \left (3+x \right )}{8}+\frac {\ln \left (1+3 x \right )}{8}\)	\(16\)
norman	\(-\frac {\ln \left (3+x \right )}{8}+\frac {\ln \left (1+3 x \right )}{8}\)	\(16\)
risch	\(-\frac {\ln \left (3+x \right )}{8}+\frac {\ln \left (1+3 x \right )}{8}\)	\(16\)

[In]

int(1/(x^2+1)/(3+10*x/(x^2+1)),x,method=_RETURNVERBOSE)

[Out]

-1/8*ln(3+x)+1/8*ln(1/3+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=\frac {1}{8} \, \log \left (3 \, x + 1\right ) - \frac {1}{8} \, \log \left (x + 3\right ) \]

[In]

integrate(1/(x^2+1)/(3+10*x/(x^2+1)),x, algorithm="fricas")

[Out]

1/8*log(3*x + 1) - 1/8*log(x + 3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=\frac {\log {\left (x + \frac {1}{3} \right )}}{8} - \frac {\log {\left (x + 3 \right )}}{8} \]

[In]

integrate(1/(x**2+1)/(3+10*x/(x**2+1)),x)

[Out]

log(x + 1/3)/8 - log(x + 3)/8

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=\frac {1}{8} \, \log \left (3 \, x + 1\right ) - \frac {1}{8} \, \log \left (x + 3\right ) \]

[In]

integrate(1/(x^2+1)/(3+10*x/(x^2+1)),x, algorithm="maxima")

[Out]

1/8*log(3*x + 1) - 1/8*log(x + 3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=\frac {1}{8} \, \log \left ({\left | 3 \, x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x + 3 \right |}\right ) \]

[In]

integrate(1/(x^2+1)/(3+10*x/(x^2+1)),x, algorithm="giac")

[Out]

1/8*log(abs(3*x + 1)) - 1/8*log(abs(x + 3))

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (1+x^2\right ) \left (3+\frac {10 x}{1+x^2}\right )} \, dx=-\frac {\mathrm {atanh}\left (\frac {3\,x}{4}+\frac {5}{4}\right )}{4} \]

[In]

int(1/((x^2 + 1)*((10*x)/(x^2 + 1) + 3)),x)

[Out]

-atanh((3*x)/4 + 5/4)/4