\(\int \frac {10 x+x^2+e^x (-76-17 x-x^2)}{475+190 x+24 x^2+x^3+e^x (95+19 x+x^2)} \, dx\) [6526]

Optimal result

Integrand size = 49, antiderivative size = 24 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=-4-\log \left (5+e^x+x\right )+\log \left (4 \left (-5-x+(10+x)^2\right )\right ) \]

[Out]

ln(4*(x+10)^2-20-4*x)-4-ln(exp(x)+5+x)

Rubi [F]

\[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=\int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx \]

[In]

Int[(10*x + x^2 + E^x*(-76 - 17*x - x^2))/(475 + 190*x + 24*x^2 + x^3 + E^x*(95 + 19*x + x^2)),x]

[Out]

-x + Log[95 + 19*x + x^2] + 4*Defer[Int][(5 + E^x + x)^(-1), x] + Defer[Int][x/(5 + E^x + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{\left (5+e^x+x\right ) \left (95+19 x+x^2\right )} \, dx \\ & = \int \left (\frac {4+x}{5+e^x+x}+\frac {-76-17 x-x^2}{95+19 x+x^2}\right ) \, dx \\ & = \int \frac {4+x}{5+e^x+x} \, dx+\int \frac {-76-17 x-x^2}{95+19 x+x^2} \, dx \\ & = \int \left (\frac {4}{5+e^x+x}+\frac {x}{5+e^x+x}\right ) \, dx+\int \left (-1+\frac {19+2 x}{95+19 x+x^2}\right ) \, dx \\ & = -x+4 \int \frac {1}{5+e^x+x} \, dx+\int \frac {x}{5+e^x+x} \, dx+\int \frac {19+2 x}{95+19 x+x^2} \, dx \\ & = -x+\log \left (95+19 x+x^2\right )+4 \int \frac {1}{5+e^x+x} \, dx+\int \frac {x}{5+e^x+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=-\log \left (5+e^x+x\right )+\log \left (95+19 x+x^2\right ) \]

[In]

Integrate[(10*x + x^2 + E^x*(-76 - 17*x - x^2))/(475 + 190*x + 24*x^2 + x^3 + E^x*(95 + 19*x + x^2)),x]

[Out]

-Log[5 + E^x + x] + Log[95 + 19*x + x^2]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
norman	\(-\ln \left ({\mathrm e}^{x}+5+x \right )+\ln \left (x^{2}+19 x +95\right )\)	\(19\)
risch	\(-\ln \left ({\mathrm e}^{x}+5+x \right )+\ln \left (x^{2}+19 x +95\right )\)	\(19\)
parallelrisch	\(-\ln \left ({\mathrm e}^{x}+5+x \right )+\ln \left (x^{2}+19 x +95\right )\)	\(19\)

[In]

int(((-x^2-17*x-76)*exp(x)+x^2+10*x)/((x^2+19*x+95)*exp(x)+x^3+24*x^2+190*x+475),x,method=_RETURNVERBOSE)

[Out]

-ln(exp(x)+5+x)+ln(x^2+19*x+95)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=\log \left (x^{2} + 19 \, x + 95\right ) - \log \left (x + e^{x} + 5\right ) \]

[In]

integrate(((-x^2-17*x-76)*exp(x)+x^2+10*x)/((x^2+19*x+95)*exp(x)+x^3+24*x^2+190*x+475),x, algorithm="fricas")

[Out]

log(x^2 + 19*x + 95) - log(x + e^x + 5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=- \log {\left (x + e^{x} + 5 \right )} + \log {\left (x^{2} + 19 x + 95 \right )} \]

[In]

integrate(((-x**2-17*x-76)*exp(x)+x**2+10*x)/((x**2+19*x+95)*exp(x)+x**3+24*x**2+190*x+475),x)

[Out]

-log(x + exp(x) + 5) + log(x**2 + 19*x + 95)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=\log \left (x^{2} + 19 \, x + 95\right ) - \log \left (x + e^{x} + 5\right ) \]

[In]

integrate(((-x^2-17*x-76)*exp(x)+x^2+10*x)/((x^2+19*x+95)*exp(x)+x^3+24*x^2+190*x+475),x, algorithm="maxima")

[Out]

log(x^2 + 19*x + 95) - log(x + e^x + 5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=\log \left (x^{2} + 19 \, x + 95\right ) - \log \left (-x - e^{x} - 5\right ) \]

[In]

integrate(((-x^2-17*x-76)*exp(x)+x^2+10*x)/((x^2+19*x+95)*exp(x)+x^3+24*x^2+190*x+475),x, algorithm="giac")

[Out]

log(x^2 + 19*x + 95) - log(-x - e^x - 5)

Mupad [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {10 x+x^2+e^x \left (-76-17 x-x^2\right )}{475+190 x+24 x^2+x^3+e^x \left (95+19 x+x^2\right )} \, dx=\ln \left (x^2+19\,x+95\right )-\ln \left (x+{\mathrm {e}}^x+5\right ) \]

[In]

int((10*x - exp(x)*(17*x + x^2 + 76) + x^2)/(190*x + exp(x)*(19*x + x^2 + 95) + 24*x^2 + x^3 + 475),x)

[Out]

log(19*x + x^2 + 95) - log(x + exp(x) + 5)