\(\int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+(6 x^2+6 x^3) \log (1+x)+(x^2+x^3) \log ^2(1+x)} \, dx\) [3999]

Optimal result

Integrand size = 57, antiderivative size = 14 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=\frac {1}{x (-3-\log (1+x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 6819} \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=-\frac {1}{x (\log (x+1)+3)} \]

[In]

Int[(3 + 4*x + (1 + x)*Log[1 + x])/(9*x^2 + 9*x^3 + (6*x^2 + 6*x^3)*Log[1 + x] + (x^2 + x^3)*Log[1 + x]^2),x]

[Out]

-(1/(x*(3 + Log[1 + x])))

Rule 6819

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[q*y^(m +
1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

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 {3+4 x+(1+x) \log (1+x)}{x^2 (1+x) (3+\log (1+x))^2} \, dx \\ & = -\frac {1}{x (3+\log (1+x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=-\frac {1}{x (3+\log (1+x))} \]

[In]

Integrate[(3 + 4*x + (1 + x)*Log[1 + x])/(9*x^2 + 9*x^3 + (6*x^2 + 6*x^3)*Log[1 + x] + (x^2 + x^3)*Log[1 + x]^
2),x]

[Out]

-(1/(x*(3 + Log[1 + x])))

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

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

[In]

int(((1+x)*ln(1+x)+3+4*x)/((x^3+x^2)*ln(1+x)^2+(6*x^3+6*x^2)*ln(1+x)+9*x^3+9*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=-\frac {1}{x \log \left (x + 1\right ) + 3 \, x} \]

[In]

integrate(((1+x)*log(1+x)+3+4*x)/((x^3+x^2)*log(1+x)^2+(6*x^3+6*x^2)*log(1+x)+9*x^3+9*x^2),x, algorithm="frica
s")

[Out]

-1/(x*log(x + 1) + 3*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=- \frac {1}{x \log {\left (x + 1 \right )} + 3 x} \]

[In]

integrate(((1+x)*ln(1+x)+3+4*x)/((x**3+x**2)*ln(1+x)**2+(6*x**3+6*x**2)*ln(1+x)+9*x**3+9*x**2),x)

[Out]

-1/(x*log(x + 1) + 3*x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=-\frac {1}{x \log \left (x + 1\right ) + 3 \, x} \]

[In]

integrate(((1+x)*log(1+x)+3+4*x)/((x^3+x^2)*log(1+x)^2+(6*x^3+6*x^2)*log(1+x)+9*x^3+9*x^2),x, algorithm="maxim
a")

[Out]

-1/(x*log(x + 1) + 3*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=-\frac {1}{x \log \left (x + 1\right ) + 3 \, x} \]

[In]

integrate(((1+x)*log(1+x)+3+4*x)/((x^3+x^2)*log(1+x)^2+(6*x^3+6*x^2)*log(1+x)+9*x^3+9*x^2),x, algorithm="giac"
)

[Out]

-1/(x*log(x + 1) + 3*x)

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 77, normalized size of antiderivative = 5.50 \[ \int \frac {3+4 x+(1+x) \log (1+x)}{9 x^2+9 x^3+\left (6 x^2+6 x^3\right ) \log (1+x)+\left (x^2+x^3\right ) \log ^2(1+x)} \, dx=\frac {\frac {2\,x\,\ln \left (x+1\right )}{3}+\frac {4\,x^2\,\ln \left (x+1\right )}{3}+\frac {2\,x^3\,\ln \left (x+1\right )}{3}+3\,x^2+2\,x^3-1}{3\,x+x\,\ln \left (x+1\right )+2\,x^2\,\ln \left (x+1\right )+x^3\,\ln \left (x+1\right )+6\,x^2+3\,x^3} \]

[In]

int((4*x + log(x + 1)*(x + 1) + 3)/(log(x + 1)^2*(x^2 + x^3) + log(x + 1)*(6*x^2 + 6*x^3) + 9*x^2 + 9*x^3),x)

[Out]

((2*x*log(x + 1))/3 + (4*x^2*log(x + 1))/3 + (2*x^3*log(x + 1))/3 + 3*x^2 + 2*x^3 - 1)/(3*x + x*log(x + 1) + 2
*x^2*log(x + 1) + x^3*log(x + 1) + 6*x^2 + 3*x^3)