\(\int \frac {78+26 x+e^x (-6-5 x-2 x^2)+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x (-26 x^3+4 x^3 \log (3))} \, dx\) [4880]

Optimal result

Integrand size = 78, antiderivative size = 28 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {3+2 x}{x^2 \left (-5+e^x+2 x-2 (4+x-\log (3))\right )} \]

[Out]

(3+2*x)/(exp(x)-13+2*ln(3))/x^2

Rubi [F]

\[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx \]

[In]

Int[(78 + 26*x + E^x*(-6 - 5*x - 2*x^2) + (-12 - 4*x)*Log[3])/(169*x^3 + E^(2*x)*x^3 - 52*x^3*Log[3] + 4*x^3*L
og[3]^2 + E^x*(-26*x^3 + 4*x^3*Log[3])),x]

[Out]

-((39 - Log[729])*Defer[Int][1/(x^2*(E^x - 13*(1 - (2*Log[3])/13))^2), x]) - (26 - Log[81])*Defer[Int][1/(x*(E
^x - 13*(1 - (2*Log[3])/13))^2), x] + 6*Defer[Int][1/(x^3*(-E^x + 13*(1 - (2*Log[3])/13))), x] + 5*Defer[Int][
1/(x^2*(-E^x + 13*(1 - (2*Log[3])/13))), x] + 2*Defer[Int][1/(x*(-E^x + 13*(1 - (2*Log[3])/13))), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{e^{2 x} x^3+x^3 (169-52 \log (3))+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx \\ & = \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{e^{2 x} x^3+e^x \left (-26 x^3+4 x^3 \log (3)\right )+x^3 \left (169-52 \log (3)+4 \log ^2(3)\right )} \, dx \\ & = \int \frac {-e^x \left (6+5 x+2 x^2\right )-2 (3+x) (-13+\log (9))}{x^3 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2} \, dx \\ & = \int \left (\frac {-6-5 x-2 x^2}{x^3 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )}+\frac {-39-x (26-\log (81))+\log (729)}{x^2 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2}\right ) \, dx \\ & = \int \frac {-6-5 x-2 x^2}{x^3 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )} \, dx+\int \frac {-39-x (26-\log (81))+\log (729)}{x^2 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2} \, dx \\ & = \int \left (\frac {6}{x^3 \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )}+\frac {5}{x^2 \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )}+\frac {2}{x \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )}\right ) \, dx+\int \left (\frac {-26+\log (81)}{x \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2}+\frac {-39+\log (729)}{x^2 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2}\right ) \, dx \\ & = 2 \int \frac {1}{x \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )} \, dx+5 \int \frac {1}{x^2 \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )} \, dx+6 \int \frac {1}{x^3 \left (-e^x+13 \left (1-\frac {2 \log (3)}{13}\right )\right )} \, dx+(-26+\log (81)) \int \frac {1}{x \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2} \, dx+(-39+\log (729)) \int \frac {1}{x^2 \left (e^x-13 \left (1-\frac {2 \log (3)}{13}\right )\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=-\frac {-3-2 x}{x^2 \left (-13+e^x+\log (9)\right )} \]

[In]

Integrate[(78 + 26*x + E^x*(-6 - 5*x - 2*x^2) + (-12 - 4*x)*Log[3])/(169*x^3 + E^(2*x)*x^3 - 52*x^3*Log[3] + 4
*x^3*Log[3]^2 + E^x*(-26*x^3 + 4*x^3*Log[3])),x]

[Out]

-((-3 - 2*x)/(x^2*(-13 + E^x + Log[9])))

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71

method	result	size
norman	\(\frac {3+2 x}{\left ({\mathrm e}^{x}-13+2 \ln \left (3\right )\right ) x^{2}}\)	\(20\)
risch	\(\frac {3+2 x}{\left ({\mathrm e}^{x}-13+2 \ln \left (3\right )\right ) x^{2}}\)	\(20\)
parallelrisch	\(\frac {3+2 x}{\left ({\mathrm e}^{x}-13+2 \ln \left (3\right )\right ) x^{2}}\)	\(20\)

[In]

int(((-2*x^2-5*x-6)*exp(x)+(-4*x-12)*ln(3)+26*x+78)/(exp(x)^2*x^3+(4*x^3*ln(3)-26*x^3)*exp(x)+4*x^3*ln(3)^2-52
*x^3*ln(3)+169*x^3),x,method=_RETURNVERBOSE)

[Out]

(3+2*x)/(exp(x)-13+2*ln(3))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {2 \, x + 3}{x^{2} e^{x} + 2 \, x^{2} \log \left (3\right ) - 13 \, x^{2}} \]

[In]

integrate(((-2*x^2-5*x-6)*exp(x)+(-4*x-12)*log(3)+26*x+78)/(exp(x)^2*x^3+(4*x^3*log(3)-26*x^3)*exp(x)+4*x^3*lo
g(3)^2-52*x^3*log(3)+169*x^3),x, algorithm="fricas")

[Out]

(2*x + 3)/(x^2*e^x + 2*x^2*log(3) - 13*x^2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {2 x + 3}{x^{2} e^{x} - 13 x^{2} + 2 x^{2} \log {\left (3 \right )}} \]

[In]

integrate(((-2*x**2-5*x-6)*exp(x)+(-4*x-12)*ln(3)+26*x+78)/(exp(x)**2*x**3+(4*x**3*ln(3)-26*x**3)*exp(x)+4*x**
3*ln(3)**2-52*x**3*ln(3)+169*x**3),x)

[Out]

(2*x + 3)/(x**2*exp(x) - 13*x**2 + 2*x**2*log(3))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {2 \, x + 3}{x^{2} {\left (2 \, \log \left (3\right ) - 13\right )} + x^{2} e^{x}} \]

[In]

integrate(((-2*x^2-5*x-6)*exp(x)+(-4*x-12)*log(3)+26*x+78)/(exp(x)^2*x^3+(4*x^3*log(3)-26*x^3)*exp(x)+4*x^3*lo
g(3)^2-52*x^3*log(3)+169*x^3),x, algorithm="maxima")

[Out]

(2*x + 3)/(x^2*(2*log(3) - 13) + x^2*e^x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {2 \, x + 3}{x^{2} e^{x} + 2 \, x^{2} \log \left (3\right ) - 13 \, x^{2}} \]

[In]

integrate(((-2*x^2-5*x-6)*exp(x)+(-4*x-12)*log(3)+26*x+78)/(exp(x)^2*x^3+(4*x^3*log(3)-26*x^3)*exp(x)+4*x^3*lo
g(3)^2-52*x^3*log(3)+169*x^3),x, algorithm="giac")

[Out]

(2*x + 3)/(x^2*e^x + 2*x^2*log(3) - 13*x^2)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {78+26 x+e^x \left (-6-5 x-2 x^2\right )+(-12-4 x) \log (3)}{169 x^3+e^{2 x} x^3-52 x^3 \log (3)+4 x^3 \log ^2(3)+e^x \left (-26 x^3+4 x^3 \log (3)\right )} \, dx=\frac {2\,x+3}{x^2\,{\mathrm {e}}^x+2\,x^2\,\ln \left (3\right )-13\,x^2} \]

[In]

int((26*x - log(3)*(4*x + 12) - exp(x)*(5*x + 2*x^2 + 6) + 78)/(4*x^3*log(3)^2 + x^3*exp(2*x) + exp(x)*(4*x^3*
log(3) - 26*x^3) - 52*x^3*log(3) + 169*x^3),x)

[Out]

(2*x + 3)/(x^2*exp(x) + 2*x^2*log(3) - 13*x^2)