3.1.77 \(\int \frac {-2 x+e^x (-10 x+10 x^2)+e^{2 x} (25-25 x+50 x^2)}{x-10 e^x x^2+e^{2 x} (50 x-25 x^2+25 x^3+25 x \log (4))+25 e^{2 x} x \log (2 x)} \, dx\)

Optimal. Leaf size=29 \[ \log \left (-2+x-\left (-\frac {e^{-x}}{5}+x\right )^2-\log (4)-\log (2 x)\right ) \]

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Rubi [F]  time = 14.42, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + E^x*(-10*x + 10*x^2) + E^(2*x)*(25 - 25*x + 50*x^2))/(x - 10*E^x*x^2 + E^(2*x)*(50*x - 25*x^2 + 25
*x^3 + 25*x*Log[4]) + 25*E^(2*x)*x*Log[2*x]),x]

[Out]

Log[2 - x + x^2 + Log[8*x]] + Defer[Int][1/(x*(-1 + 10*E^x*x + 25*E^(2*x)*x - 25*E^(2*x)*x^2 - 50*E^(2*x)*(1 +
Log[2]) - 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x] + 2*Defer[Int][x^2/((-1 + 10*E^x*x + 25*E^(2*x)*
x - 25*E^(2*x)*x^2 - 50*E^(2*x)*(1 + Log[2]) - 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x] + 2*Defer[In
t][Log[2*x]/((-1 + 10*E^x*x + 25*E^(2*x)*x - 25*E^(2*x)*x^2 - 50*E^(2*x)*(1 + Log[2]) - 25*E^(2*x)*Log[2*x])*(
2 - x + x^2 + Log[8*x])), x] + 10*Defer[Int][(E^x*Log[2*x])/((-1 + 10*E^x*x + 25*E^(2*x)*x - 25*E^(2*x)*x^2 -
50*E^(2*x)*(1 + Log[2]) - 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x] - (3 + Log[16])*Defer[Int][1/((1
- 10*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 50*E^(2*x)*(1 + Log[2]) + 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log
[8*x])), x] - 10*(1 + Log[4])*Defer[Int][E^x/((1 - 10*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 50*E^(2*x)*(1 +
Log[2]) + 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x] + 20*(1 + Log[2])*Defer[Int][(E^x*x)/((1 - 10*E^x
*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 50*E^(2*x)*(1 + Log[2]) + 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])),
x] + 10*Defer[Int][(E^x*x^3)/((1 - 10*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 50*E^(2*x)*(1 + Log[2]) + 25*E^
(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x] + 10*Defer[Int][(E^x*x*Log[2*x])/((1 - 10*E^x*x - 25*E^(2*x)*x +
25*E^(2*x)*x^2 + 50*E^(2*x)*(1 + Log[2]) + 25*E^(2*x)*Log[2*x])*(2 - x + x^2 + Log[8*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1-x+2 x^2}{x \left (2-x+x^2+\log (8 x)\right )}+\frac {-1-2 x^3+10 e^x x^4+20 e^x x^2 (1+\log (2))-10 e^x x (1+\log (4))-3 x \left (1+\frac {\log (16)}{3}\right )-2 x \log (2 x)-10 e^x x \log (2 x)+10 e^x x^2 \log (2 x)}{x \left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}\right ) \, dx\\ &=\int \frac {1-x+2 x^2}{x \left (2-x+x^2+\log (8 x)\right )} \, dx+\int \frac {-1-2 x^3+10 e^x x^4+20 e^x x^2 (1+\log (2))-10 e^x x (1+\log (4))-3 x \left (1+\frac {\log (16)}{3}\right )-2 x \log (2 x)-10 e^x x \log (2 x)+10 e^x x^2 \log (2 x)}{x \left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx\\ &=\log \left (2-x+x^2+\log (8 x)\right )+\int \left (\frac {1}{x \left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {2 x^2}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {2 \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x x^3}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {20 e^x x (1+\log (2))}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x (-1-\log (4))}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {-3-\log (16)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x x \log (2 x)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}\right ) \, dx\\ &=\log \left (2-x+x^2+\log (8 x)\right )+2 \int \frac {x^2}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+2 \int \frac {\log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x x^3}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x x \log (2 x)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+(20 (1+\log (2))) \int \frac {e^x x}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx-(10 (1+\log (4))) \int \frac {e^x}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+(-3-\log (16)) \int \frac {1}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+\int \frac {1}{x \left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.64, size = 58, normalized size = 2.00 \begin {gather*} -2 x+\log \left (1+50 e^{2 x}-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+25 e^{2 x} \log (4)+25 e^{2 x} \log (2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + E^x*(-10*x + 10*x^2) + E^(2*x)*(25 - 25*x + 50*x^2))/(x - 10*E^x*x^2 + E^(2*x)*(50*x - 25*x^
2 + 25*x^3 + 25*x*Log[4]) + 25*E^(2*x)*x*Log[2*x]),x]

[Out]

-2*x + Log[1 + 50*E^(2*x) - 10*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 + 25*E^(2*x)*Log[4] + 25*E^(2*x)*Log[2*x]
]

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fricas [A]  time = 0.74, size = 41, normalized size = 1.41 \begin {gather*} \log \left ({\left (25 \, {\left (x^{2} - x + 2 \, \log \relax (2) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 25 \, e^{\left (2 \, x\right )} \log \left (2 \, x\right ) + 1\right )} e^{\left (-2 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x)^2*log(2*x)+(50*x*log(2)+25*x^3-25*
x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm="fricas")

[Out]

log((25*(x^2 - x + 2*log(2) + 2)*e^(2*x) - 10*x*e^x + 25*e^(2*x)*log(2*x) + 1)*e^(-2*x))

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giac [A]  time = 0.33, size = 50, normalized size = 1.72 \begin {gather*} -2 \, x + \log \left (25 \, x^{2} e^{\left (2 \, x\right )} - 25 \, x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 75 \, e^{\left (2 \, x\right )} \log \relax (2) + 25 \, e^{\left (2 \, x\right )} \log \relax (x) + 50 \, e^{\left (2 \, x\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x)^2*log(2*x)+(50*x*log(2)+25*x^3-25*
x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm="giac")

[Out]

-2*x + log(25*x^2*e^(2*x) - 25*x*e^(2*x) - 10*x*e^x + 75*e^(2*x)*log(2) + 25*e^(2*x)*log(x) + 50*e^(2*x) + 1)

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maple [A]  time = 0.04, size = 50, normalized size = 1.72

method	result	size
risch	\(\ln \left (\ln \left (2 x \right )+\frac {\left (25 \,{\mathrm e}^{2 x} x^{2}+50 \ln \relax (2) {\mathrm e}^{2 x}-25 x \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x} x +50 \,{\mathrm e}^{2 x}+1\right ) {\mathrm e}^{-2 x}}{25}\right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x)^2*ln(2*x)+(50*x*ln(2)+25*x^3-25*x^2+50*x
)*exp(x)^2-10*exp(x)*x^2+x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(2*x)+1/25*(25*exp(2*x)*x^2+50*ln(2)*exp(2*x)-25*x*exp(2*x)-10*exp(x)*x+50*exp(2*x)+1)*exp(-2*x))

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maxima [B]  time = 0.75, size = 65, normalized size = 2.24 \begin {gather*} -2 \, x + \log \left (x^{2} - x + 3 \, \log \relax (2) + \log \relax (x) + 2\right ) + \log \left (\frac {25 \, {\left (x^{2} - x + 3 \, \log \relax (2) + \log \relax (x) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 1}{25 \, {\left (x^{2} - x + 3 \, \log \relax (2) + \log \relax (x) + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-25*x+25)*exp(x)^2+(10*x^2-10*x)*exp(x)-2*x)/(25*x*exp(x)^2*log(2*x)+(50*x*log(2)+25*x^3-25*
x^2+50*x)*exp(x)^2-10*exp(x)*x^2+x),x, algorithm="maxima")

[Out]

-2*x + log(x^2 - x + 3*log(2) + log(x) + 2) + log(1/25*(25*(x^2 - x + 3*log(2) + log(x) + 2)*e^(2*x) - 10*x*e^
x + 1)/(x^2 - x + 3*log(2) + log(x) + 2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {2\,x-{\mathrm {e}}^{2\,x}\,\left (50\,x^2-25\,x+25\right )+{\mathrm {e}}^x\,\left (10\,x-10\,x^2\right )}{x-10\,x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (50\,x+50\,x\,\ln \relax (2)-25\,x^2+25\,x^3\right )+25\,x\,\ln \left (2\,x\right )\,{\mathrm {e}}^{2\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - exp(2*x)*(50*x^2 - 25*x + 25) + exp(x)*(10*x - 10*x^2))/(x - 10*x^2*exp(x) + exp(2*x)*(50*x + 50*x
*log(2) - 25*x^2 + 25*x^3) + 25*x*log(2*x)*exp(2*x)),x)

[Out]

int(-(2*x - exp(2*x)*(50*x^2 - 25*x + 25) + exp(x)*(10*x - 10*x^2))/(x - 10*x^2*exp(x) + exp(2*x)*(50*x + 50*x
*log(2) - 25*x^2 + 25*x^3) + 25*x*log(2*x)*exp(2*x)), x)

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sympy [B]  time = 21.35, size = 80, normalized size = 2.76 \begin {gather*} - 2 x + \log {\left (- \frac {2 x e^{x}}{5 x^{2} - 5 x + 5 \log {\left (2 x \right )} + 10 \log {\relax (2 )} + 10} + e^{2 x} + \frac {1}{25 x^{2} - 25 x + 25 \log {\left (2 x \right )} + 50 \log {\relax (2 )} + 50} \right )} + \log {\left (x^{2} - x + \log {\left (2 x \right )} + 2 \log {\relax (2 )} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**2-25*x+25)*exp(x)**2+(10*x**2-10*x)*exp(x)-2*x)/(25*x*exp(x)**2*ln(2*x)+(50*x*ln(2)+25*x**3-
25*x**2+50*x)*exp(x)**2-10*exp(x)*x**2+x),x)

[Out]

-2*x + log(-2*x*exp(x)/(5*x**2 - 5*x + 5*log(2*x) + 10*log(2) + 10) + exp(2*x) + 1/(25*x**2 - 25*x + 25*log(2*
x) + 50*log(2) + 50)) + log(x**2 - x + log(2*x) + 2*log(2) + 2)

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