\(\int \frac {20+4 x}{6 x^2+e^x x^2+x^3+(-12 x-2 e^x x-2 x^2) \log (30+5 e^x+5 x)+(6+e^x+x) \log ^2(30+5 e^x+5 x)} \, dx\) [252]

Optimal result

Integrand size = 71, antiderivative size = 17 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=\frac {4}{-x+\log \left (5 \left (6+e^x+x\right )\right )} \]

[Out]

4/(ln(5*exp(x)+5*x+30)-x)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6820, 12, 6818} \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x-\log \left (5 \left (x+e^x+6\right )\right )} \]

[In]

Int[(20 + 4*x)/(6*x^2 + E^x*x^2 + x^3 + (-12*x - 2*E^x*x - 2*x^2)*Log[30 + 5*E^x + 5*x] + (6 + E^x + x)*Log[30
+ 5*E^x + 5*x]^2),x]

[Out]

-4/(x - Log[5*(6 + E^x + x)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, 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 {4 (5+x)}{\left (6+e^x+x\right ) \left (x-\log \left (5 \left (6+e^x+x\right )\right )\right )^2} \, dx \\ & = 4 \int \frac {5+x}{\left (6+e^x+x\right ) \left (x-\log \left (5 \left (6+e^x+x\right )\right )\right )^2} \, dx \\ & = -\frac {4}{x-\log \left (5 \left (6+e^x+x\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x-\log \left (5 \left (6+e^x+x\right )\right )} \]

[In]

Integrate[(20 + 4*x)/(6*x^2 + E^x*x^2 + x^3 + (-12*x - 2*E^x*x - 2*x^2)*Log[30 + 5*E^x + 5*x] + (6 + E^x + x)*
Log[30 + 5*E^x + 5*x]^2),x]

[Out]

-4/(x - Log[5*(6 + E^x + x)])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {4}{x -\ln \left (5 \,{\mathrm e}^{x}+5 x +30\right )}\)	\(19\)
parallelrisch	\(-\frac {4}{x -\ln \left (5 \,{\mathrm e}^{x}+5 x +30\right )}\)	\(19\)

[In]

int((20+4*x)/((exp(x)+x+6)*ln(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*ln(5*exp(x)+5*x+30)+exp(x)*x^2+x^3+6
*x^2),x,method=_RETURNVERBOSE)

[Out]

-4/(x-ln(5*exp(x)+5*x+30))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x - \log \left (5 \, x + 5 \, e^{x} + 30\right )} \]

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="fricas")

[Out]

-4/(x - log(5*x + 5*e^x + 30))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=\frac {4}{- x + \log {\left (5 x + 5 e^{x} + 30 \right )}} \]

[In]

integrate((20+4*x)/((exp(x)+x+6)*ln(5*exp(x)+5*x+30)**2+(-2*exp(x)*x-2*x**2-12*x)*ln(5*exp(x)+5*x+30)+exp(x)*x
**2+x**3+6*x**2),x)

[Out]

4/(-x + log(5*x + 5*exp(x) + 30))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x - \log \left (5\right ) - \log \left (x + e^{x} + 6\right )} \]

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="maxima")

[Out]

-4/(x - log(5) - log(x + e^x + 6))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x - \log \left (5 \, x + 5 \, e^{x} + 30\right )} \]

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="giac")

[Out]

-4/(x - log(5*x + 5*e^x + 30))

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {20+4 x}{6 x^2+e^x x^2+x^3+\left (-12 x-2 e^x x-2 x^2\right ) \log \left (30+5 e^x+5 x\right )+\left (6+e^x+x\right ) \log ^2\left (30+5 e^x+5 x\right )} \, dx=-\frac {4}{x-\ln \left (5\,x+5\,{\mathrm {e}}^x+30\right )} \]

[In]

int((4*x + 20)/(x^2*exp(x) + log(5*x + 5*exp(x) + 30)^2*(x + exp(x) + 6) - log(5*x + 5*exp(x) + 30)*(12*x + 2*
x*exp(x) + 2*x^2) + 6*x^2 + x^3),x)

[Out]

-4/(x - log(5*x + 5*exp(x) + 30))