\(\int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx\) [6441]

Optimal result

Integrand size = 64, antiderivative size = 29 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x-\log \left (\left (e^x+x\right )^2-\frac {1}{5} x (3-x+4 \log (2 x))\right ) \]

[Out]

x-ln((exp(x)+x)^2-1/5*(3+4*ln(2*x)-x)*x)

Rubi [F]

\[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=\int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx \]

[In]

Int[(-7 + 10*E^x + 5*E^(2*x) + 15*x - 6*x^2 + (-4 + 4*x)*Log[2*x])/(-5*E^(2*x) + 3*x - 10*E^x*x - 6*x^2 + 4*x*
Log[2*x]),x]

[Out]

-x + 7*Defer[Int][(5*E^(2*x) - 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2*x])^(-1), x] - 10*Defer[Int][E^x/(5*E^(2*x)
- 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2*x]), x] - 18*Defer[Int][x/(5*E^(2*x) - 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2
*x]), x] + 10*Defer[Int][(E^x*x)/(5*E^(2*x) - 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2*x]), x] + 12*Defer[Int][x^2/(
5*E^(2*x) - 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2*x]), x] - 8*Defer[Int][(x*Log[2*x])/(5*E^(2*x) - 3*x + 10*E^x*x
+ 6*x^2 - 4*x*Log[2*x]), x] - 4*Defer[Int][Log[2*x]/(-5*E^(2*x) + 3*x - 10*E^x*x - 6*x^2 + 4*x*Log[2*x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+\frac {7-10 e^x-18 x+10 e^x x+12 x^2+4 \log (2 x)-8 x \log (2 x)}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}\right ) \, dx \\ & = -x+\int \frac {7-10 e^x-18 x+10 e^x x+12 x^2+4 \log (2 x)-8 x \log (2 x)}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx \\ & = -x+\int \left (\frac {7}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}-\frac {10 e^x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}-\frac {18 x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}+\frac {10 e^x x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}+\frac {12 x^2}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}-\frac {8 x \log (2 x)}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)}-\frac {4 \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)}\right ) \, dx \\ & = -x-4 \int \frac {\log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx+7 \int \frac {1}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx-8 \int \frac {x \log (2 x)}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx-10 \int \frac {e^x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx+10 \int \frac {e^x x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx+12 \int \frac {x^2}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx-18 \int \frac {x}{5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x-\log \left (5 e^{2 x}-3 x+10 e^x x+6 x^2-4 x \log (2 x)\right ) \]

[In]

Integrate[(-7 + 10*E^x + 5*E^(2*x) + 15*x - 6*x^2 + (-4 + 4*x)*Log[2*x])/(-5*E^(2*x) + 3*x - 10*E^x*x - 6*x^2
+ 4*x*Log[2*x]),x]

[Out]

x - Log[5*E^(2*x) - 3*x + 10*E^x*x + 6*x^2 - 4*x*Log[2*x]]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(-\ln \left (x^{2}+\frac {5 \,{\mathrm e}^{x} x}{3}-\frac {2 x \ln \left (2 x \right )}{3}+\frac {5 \,{\mathrm e}^{2 x}}{6}-\frac {x}{2}\right )+x\)	\(31\)
norman	\(x -\ln \left (-4 x \ln \left (2 x \right )+10 \,{\mathrm e}^{x} x +6 x^{2}+5 \,{\mathrm e}^{2 x}-3 x \right )\)	\(33\)
risch	\(x -\ln \left (x \right )-\ln \left (\ln \left (2 x \right )-\frac {6 x^{2}+10 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{2 x}-3 x}{4 x}\right )\)	\(40\)

[In]

int(((-4+4*x)*ln(2*x)+5*exp(x)^2+10*exp(x)-6*x^2+15*x-7)/(4*x*ln(2*x)-5*exp(x)^2-10*exp(x)*x-6*x^2+3*x),x,meth
od=_RETURNVERBOSE)

[Out]

-ln(x^2+5/3*exp(x)*x-2/3*x*ln(2*x)+5/6*exp(x)^2-1/2*x)+x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x - \log \left (2 \, x\right ) - \log \left (-\frac {6 \, x^{2} + 10 \, x e^{x} - 4 \, x \log \left (2 \, x\right ) - 3 \, x + 5 \, e^{\left (2 \, x\right )}}{x}\right ) \]

[In]

integrate(((-4+4*x)*log(2*x)+5*exp(x)^2+10*exp(x)-6*x^2+15*x-7)/(4*x*log(2*x)-5*exp(x)^2-10*exp(x)*x-6*x^2+3*x
),x, algorithm="fricas")

[Out]

x - log(2*x) - log(-(6*x^2 + 10*x*e^x - 4*x*log(2*x) - 3*x + 5*e^(2*x))/x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x - \log {\left (\frac {6 x^{2}}{5} + 2 x e^{x} - \frac {4 x \log {\left (2 x \right )}}{5} - \frac {3 x}{5} + e^{2 x} \right )} \]

[In]

integrate(((-4+4*x)*ln(2*x)+5*exp(x)**2+10*exp(x)-6*x**2+15*x-7)/(4*x*ln(2*x)-5*exp(x)**2-10*exp(x)*x-6*x**2+3
*x),x)

[Out]

x - log(6*x**2/5 + 2*x*exp(x) - 4*x*log(2*x)/5 - 3*x/5 + exp(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x - \log \left (\frac {6}{5} \, x^{2} - \frac {1}{5} \, x {\left (4 \, \log \left (2\right ) + 3\right )} + 2 \, x e^{x} - \frac {4}{5} \, x \log \left (x\right ) + e^{\left (2 \, x\right )}\right ) \]

[In]

integrate(((-4+4*x)*log(2*x)+5*exp(x)^2+10*exp(x)-6*x^2+15*x-7)/(4*x*log(2*x)-5*exp(x)^2-10*exp(x)*x-6*x^2+3*x
),x, algorithm="maxima")

[Out]

x - log(6/5*x^2 - 1/5*x*(4*log(2) + 3) + 2*x*e^x - 4/5*x*log(x) + e^(2*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x - \log \left (6 \, x^{2} + 10 \, x e^{x} - 4 \, x \log \left (2 \, x\right ) - 3 \, x + 5 \, e^{\left (2 \, x\right )}\right ) \]

[In]

integrate(((-4+4*x)*log(2*x)+5*exp(x)^2+10*exp(x)-6*x^2+15*x-7)/(4*x*log(2*x)-5*exp(x)^2-10*exp(x)*x-6*x^2+3*x
),x, algorithm="giac")

[Out]

x - log(6*x^2 + 10*x*e^x - 4*x*log(2*x) - 3*x + 5*e^(2*x))

Mupad [B] (verification not implemented)

Time = 12.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-7+10 e^x+5 e^{2 x}+15 x-6 x^2+(-4+4 x) \log (2 x)}{-5 e^{2 x}+3 x-10 e^x x-6 x^2+4 x \log (2 x)} \, dx=x-\ln \left (5\,{\mathrm {e}}^{2\,x}-3\,x-4\,x\,\ln \left (2\,x\right )+10\,x\,{\mathrm {e}}^x+6\,x^2\right ) \]

[In]

int(-(15*x + 5*exp(2*x) + 10*exp(x) - 6*x^2 + log(2*x)*(4*x - 4) - 7)/(5*exp(2*x) - 3*x - 4*x*log(2*x) + 10*x*
exp(x) + 6*x^2),x)

[Out]

x - log(5*exp(2*x) - 3*x - 4*x*log(2*x) + 10*x*exp(x) + 6*x^2)