3.82.79 \(\int \frac {e^{\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}} (1-3 x-x^2+8 x^3-4 x^4+e^x (-3 x-x^2+2 x^3))}{1-3 x+4 x^3} \, dx\)

Optimal. Leaf size=23 \[ e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \]

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Rubi [F]  time = 22.47, 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 {\exp \left (\frac {1+e^x-x-2 x^2+(-1+2 x) \log (1+x)}{-1+2 x}\right ) \left (1-3 x-x^2+8 x^3-4 x^4+e^x \left (-3 x-x^2+2 x^3\right )\right )}{1-3 x+4 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((1 + E^x - x - 2*x^2 + (-1 + 2*x)*Log[1 + x])/(-1 + 2*x))*(1 - 3*x - x^2 + 8*x^3 - 4*x^4 + E^x*(-3*x -
x^2 + 2*x^3)))/(1 - 3*x + 4*x^3),x]

[Out]

Defer[Int][E^((1 + E^x - 2*x)/(-1 + 2*x)), x]/2 + Defer[Int][E^((1 + E^x - x - 2*x^2)/(-1 + 2*x)), x] + Defer[
Int][E^((1 + E^x - x - 2*x^2)/(-1 + 2*x))/(1 - 2*x)^2, x] + Defer[Int][E^((1 + E^x - 2*x)/(-1 + 2*x))*x, x]/2
+ Defer[Int][E^((1 + E^x - x - 2*x^2)/(-1 + 2*x))*x, x] - Defer[Int][E^((1 + E^x - x - 2*x^2)/(-1 + 2*x))*x^2,
x] - (3*Defer[Int][E^((1 + E^x - 2*x)/(-1 + 2*x))/(-1 + 2*x)^2, x])/2 - Defer[Int][E^((1 + E^x - x - 2*x^2)/(
-1 + 2*x))/(-1 + 2*x)^2, x] - Defer[Int][E^((1 + E^x - 2*x)/(-1 + 2*x))/(-1 + 2*x), x]/2 - Defer[Int][E^((1 +
E^x - 2*x - Log[1 + x] + 2*x*Log[1 + x])/(-1 + 2*x))/(-1 + 2*x), x]/2

Rubi steps

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

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Mathematica [A]  time = 1.38, size = 23, normalized size = 1.00 \begin {gather*} e^{-1-x+\frac {e^x}{-1+2 x}} x (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 + E^x - x - 2*x^2 + (-1 + 2*x)*Log[1 + x])/(-1 + 2*x))*(1 - 3*x - x^2 + 8*x^3 - 4*x^4 + E^x*(
-3*x - x^2 + 2*x^3)))/(1 - 3*x + 4*x^3),x]

[Out]

E^(-1 - x + E^x/(-1 + 2*x))*x*(1 + x)

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fricas [A]  time = 1.11, size = 35, normalized size = 1.52 \begin {gather*} x e^{\left (-\frac {2 \, x^{2} - {\left (2 \, x - 1\right )} \log \left (x + 1\right ) + x - e^{x} - 1}{2 \, x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((2*x-1)*log(x+1)+exp(x)-2*x^2-x+1)/(2*x-1))/(4*x
^3-3*x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.53, size = 36, normalized size = 1.57 \begin {gather*} {\left (x^{2} e^{\left (\frac {e^{x}}{2 \, x - 1}\right )} + x e^{\left (\frac {e^{x}}{2 \, x - 1}\right )}\right )} e^{\left (-x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((2*x-1)*log(x+1)+exp(x)-2*x^2-x+1)/(2*x-1))/(4*x
^3-3*x+1),x, algorithm="giac")

[Out]

(x^2*e^(e^x/(2*x - 1)) + x*e^(e^x/(2*x - 1)))*e^(-x - 1)

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maple [A]  time = 0.16, size = 37, normalized size = 1.61

method	result	size
risch	\(x \,{\mathrm e}^{\frac {2 \ln \left (x +1\right ) x -2 x^{2}+{\mathrm e}^{x}-\ln \left (x +1\right )-x +1}{2 x -1}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((2*x-1)*ln(x+1)+exp(x)-2*x^2-x+1)/(2*x-1))/(4*x^3-3*x+
1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (4 \, x^{4} - 8 \, x^{3} + x^{2} - {\left (2 \, x^{3} - x^{2} - 3 \, x\right )} e^{x} + 3 \, x - 1\right )} e^{\left (-\frac {2 \, x^{2} - {\left (2 \, x - 1\right )} \log \left (x + 1\right ) + x - e^{x} - 1}{2 \, x - 1}\right )}}{4 \, x^{3} - 3 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-x^2-3*x)*exp(x)-4*x^4+8*x^3-x^2-3*x+1)*exp(((2*x-1)*log(x+1)+exp(x)-2*x^2-x+1)/(2*x-1))/(4*x
^3-3*x+1),x, algorithm="maxima")

[Out]

-integrate((4*x^4 - 8*x^3 + x^2 - (2*x^3 - x^2 - 3*x)*e^x + 3*x - 1)*e^(-(2*x^2 - (2*x - 1)*log(x + 1) + x - e
^x - 1)/(2*x - 1))/(4*x^3 - 3*x + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^x-x-2\,x^2+\ln \left (x+1\right )\,\left (2\,x-1\right )+1}{2\,x-1}}\,\left (3\,x+{\mathrm {e}}^x\,\left (-2\,x^3+x^2+3\,x\right )+x^2-8\,x^3+4\,x^4-1\right )}{4\,x^3-3\,x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(x) - x - 2*x^2 + log(x + 1)*(2*x - 1) + 1)/(2*x - 1))*(3*x + exp(x)*(3*x + x^2 - 2*x^3) + x^2 -
8*x^3 + 4*x^4 - 1))/(4*x^3 - 3*x + 1),x)

[Out]

int(-(exp((exp(x) - x - 2*x^2 + log(x + 1)*(2*x - 1) + 1)/(2*x - 1))*(3*x + exp(x)*(3*x + x^2 - 2*x^3) + x^2 -
8*x^3 + 4*x^4 - 1))/(4*x^3 - 3*x + 1), x)

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sympy [A]  time = 14.31, size = 29, normalized size = 1.26 \begin {gather*} x e^{\frac {- 2 x^{2} - x + \left (2 x - 1\right ) \log {\left (x + 1 \right )} + e^{x} + 1}{2 x - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-x**2-3*x)*exp(x)-4*x**4+8*x**3-x**2-3*x+1)*exp(((2*x-1)*ln(x+1)+exp(x)-2*x**2-x+1)/(2*x-1))
/(4*x**3-3*x+1),x)

[Out]

x*exp((-2*x**2 - x + (2*x - 1)*log(x + 1) + exp(x) + 1)/(2*x - 1))

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