\(\int \frac {15-15 x+e^{x^2} (-3+3 x+6 x^2)+e^{x+x^2} (-3+6 x+6 x^2)}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx\) [5760]

Optimal result

Integrand size = 62, antiderivative size = 27 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 \left (x+\log \left (\frac {5-e^{x^2}-e^{x+x^2}}{x}\right )\right ) \]

[Out]

3*ln((5-exp(x^2+x)-exp(x^2))/x)+3*x

Rubi [F]

\[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=\int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx \]

[In]

Int[(15 - 15*x + E^x^2*(-3 + 3*x + 6*x^2) + E^(x + x^2)*(-3 + 6*x + 6*x^2))/(-5*x + E^x^2*x + E^(x + x^2)*x),x
]

[Out]

3*x + 3*x^2 + 3*Log[1 + E^x] - 3*Log[x] + 15*Defer[Int][E^x/((1 + E^x)*(-5 + E^x^2 + E^(x + x^2))), x] + 30*De
fer[Int][x/((1 + E^x)*(-5 + E^x^2 + E^(x + x^2))), x] + 30*Defer[Int][(E^x*x)/((1 + E^x)*(-5 + E^x^2 + E^(x +
x^2))), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {15 \left (e^x+2 x+2 e^x x\right )}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )}+\frac {3 \left (-1-e^x+x+2 e^x x+2 x^2+2 e^x x^2\right )}{\left (1+e^x\right ) x}\right ) \, dx \\ & = 3 \int \frac {-1-e^x+x+2 e^x x+2 x^2+2 e^x x^2}{\left (1+e^x\right ) x} \, dx+15 \int \frac {e^x+2 x+2 e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx \\ & = 3 \int \left (-\frac {1}{1+e^x}+\frac {-1+2 x+2 x^2}{x}\right ) \, dx+15 \int \left (\frac {e^x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )}+\frac {2 x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )}+\frac {2 e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )}\right ) \, dx \\ & = -\left (3 \int \frac {1}{1+e^x} \, dx\right )+3 \int \frac {-1+2 x+2 x^2}{x} \, dx+15 \int \frac {e^x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx \\ & = 3 \int \left (2-\frac {1}{x}+2 x\right ) \, dx-3 \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,e^x\right )+15 \int \frac {e^x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx \\ & = 6 x+3 x^2-3 \log (x)-3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+3 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^x\right )+15 \int \frac {e^x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx \\ & = 3 x+3 x^2+3 \log \left (1+e^x\right )-3 \log (x)+15 \int \frac {e^x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx+30 \int \frac {e^x x}{\left (1+e^x\right ) \left (-5+e^{x^2}+e^{x+x^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 \left (x+\log \left (5-e^{x^2}-e^{x+x^2}\right )-\log (x)\right ) \]

[In]

Integrate[(15 - 15*x + E^x^2*(-3 + 3*x + 6*x^2) + E^(x + x^2)*(-3 + 6*x + 6*x^2))/(-5*x + E^x^2*x + E^(x + x^2
)*x),x]

[Out]

3*(x + Log[5 - E^x^2 - E^(x + x^2)] - Log[x])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
norman	\(3 x -3 \ln \left (x \right )+3 \ln \left ({\mathrm e}^{x^{2}+x}+{\mathrm e}^{x^{2}}-5\right )\)	\(24\)
risch	\(3 x -3 \ln \left (x \right )+3 \ln \left ({\mathrm e}^{\left (1+x \right ) x}+{\mathrm e}^{x^{2}}-5\right )\)	\(24\)
parallelrisch	\(3 x -3 \ln \left (x \right )+3 \ln \left ({\mathrm e}^{x^{2}+x}+{\mathrm e}^{x^{2}}-5\right )\)	\(24\)

[In]

int(((6*x^2+6*x-3)*exp(x^2+x)+(6*x^2+3*x-3)*exp(x^2)-15*x+15)/(x*exp(x^2+x)+exp(x^2)*x-5*x),x,method=_RETURNVE
RBOSE)

[Out]

3*x-3*ln(x)+3*ln(exp(x^2+x)+exp(x^2)-5)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 \, x - 3 \, \log \left (x\right ) + 3 \, \log \left (e^{\left (x^{2} + x\right )} + e^{\left (x^{2}\right )} - 5\right ) \]

[In]

integrate(((6*x^2+6*x-3)*exp(x^2+x)+(6*x^2+3*x-3)*exp(x^2)-15*x+15)/(x*exp(x^2+x)+exp(x^2)*x-5*x),x, algorithm
="fricas")

[Out]

3*x - 3*log(x) + 3*log(e^(x^2 + x) + e^(x^2) - 5)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 x - 3 \log {\left (x \right )} + 3 \log {\left (e^{x^{2}} + e^{x^{2} + x} - 5 \right )} \]

[In]

integrate(((6*x**2+6*x-3)*exp(x**2+x)+(6*x**2+3*x-3)*exp(x**2)-15*x+15)/(x*exp(x**2+x)+exp(x**2)*x-5*x),x)

[Out]

3*x - 3*log(x) + 3*log(exp(x**2) + exp(x**2 + x) - 5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 \, x - 3 \, \log \left (x\right ) + 3 \, \log \left (\frac {{\left (e^{x} + 1\right )} e^{\left (x^{2}\right )} - 5}{e^{x} + 1}\right ) + 3 \, \log \left (e^{x} + 1\right ) \]

[In]

integrate(((6*x^2+6*x-3)*exp(x^2+x)+(6*x^2+3*x-3)*exp(x^2)-15*x+15)/(x*exp(x^2+x)+exp(x^2)*x-5*x),x, algorithm
="maxima")

[Out]

3*x - 3*log(x) + 3*log(((e^x + 1)*e^(x^2) - 5)/(e^x + 1)) + 3*log(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3 \, x - 3 \, \log \left (x\right ) + 3 \, \log \left (e^{\left (x^{2} + x\right )} + e^{\left (x^{2}\right )} - 5\right ) \]

[In]

integrate(((6*x^2+6*x-3)*exp(x^2+x)+(6*x^2+3*x-3)*exp(x^2)-15*x+15)/(x*exp(x^2+x)+exp(x^2)*x-5*x),x, algorithm
="giac")

[Out]

3*x - 3*log(x) + 3*log(e^(x^2 + x) + e^(x^2) - 5)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {15-15 x+e^{x^2} \left (-3+3 x+6 x^2\right )+e^{x+x^2} \left (-3+6 x+6 x^2\right )}{-5 x+e^{x^2} x+e^{x+x^2} x} \, dx=3\,x+3\,\ln \left ({\mathrm {e}}^{x^2}+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x-5\right )-3\,\ln \left (x\right ) \]

[In]

int((exp(x^2)*(3*x + 6*x^2 - 3) - 15*x + exp(x + x^2)*(6*x + 6*x^2 - 3) + 15)/(x*exp(x^2) - 5*x + x*exp(x + x^
2)),x)

[Out]

3*x + 3*log(exp(x^2) + exp(x^2)*exp(x) - 5) - 3*log(x)