3.28.67 \(\int \frac {(\frac {-4+4 x}{x})^{-5 x} (-1-4 x+e^{x^2} (\frac {-4+4 x}{x})^{5 x} (-4 x+4 x^2)+(5 x-5 x^2) \log (\frac {-4+4 x}{x}))}{-2+2 x} \, dx\)

Optimal. Leaf size=25 \[ e^{x^2}+\frac {1}{2} x \left (5-\frac {4+x}{x}\right )^{-5 x} \]

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Rubi [F]  time = 1.50, 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 {\left (\frac {-4+4 x}{x}\right )^{-5 x} \left (-1-4 x+e^{x^2} \left (\frac {-4+4 x}{x}\right )^{5 x} \left (-4 x+4 x^2\right )+\left (5 x-5 x^2\right ) \log \left (\frac {-4+4 x}{x}\right )\right )}{-2+2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 4*x + E^x^2*((-4 + 4*x)/x)^(5*x)*(-4*x + 4*x^2) + (5*x - 5*x^2)*Log[(-4 + 4*x)/x])/((-2 + 2*x)*((-4
+ 4*x)/x)^(5*x)),x]

[Out]

E^x^2 - 2*Defer[Int][(4 - 4/x)^(-5*x), x] - (5*Defer[Int][1/((4 - 4/x)^(5*x)*(-1 + x)), x])/2 - (5*Log[4 - 4/x
]*Defer[Int][x/(4 - 4/x)^(5*x), x])/2 + (5*Defer[Int][Defer[Int][x/(4 - 4/x)^(5*x), x]/(-1 + x), x])/2 - (5*De
fer[Int][Defer[Int][x/(4 - 4/x)^(5*x), x]/x, x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4-\frac {4}{x}\right )^{-5 x} \left (1+4 x-e^{x^2} \left (\frac {-4+4 x}{x}\right )^{5 x} \left (-4 x+4 x^2\right )-\left (5 x-5 x^2\right ) \log \left (\frac {-4+4 x}{x}\right )\right )}{2-2 x} \, dx\\ &=\int \frac {\left (4-\frac {4}{x}\right )^{-5 x} \left (1+4 x-4 e^{x^2} \left (4-\frac {4}{x}\right )^{5 x} (-1+x) x+5 (-1+x) x \log \left (4-\frac {4}{x}\right )\right )}{2-2 x} \, dx\\ &=\int \left (2 e^{x^2} x+\frac {\left (4-\frac {4}{x}\right )^{-5 x} \left (-1-4 x+5 x \log \left (4-\frac {4}{x}\right )-5 x^2 \log \left (4-\frac {4}{x}\right )\right )}{2 (-1+x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\left (4-\frac {4}{x}\right )^{-5 x} \left (-1-4 x+5 x \log \left (4-\frac {4}{x}\right )-5 x^2 \log \left (4-\frac {4}{x}\right )\right )}{-1+x} \, dx+2 \int e^{x^2} x \, dx\\ &=e^{x^2}+\frac {1}{2} \int \left (\frac {\left (4-\frac {4}{x}\right )^{-5 x} (-1-4 x)}{-1+x}-5 \left (4-\frac {4}{x}\right )^{-5 x} x \log \left (4-\frac {4}{x}\right )\right ) \, dx\\ &=e^{x^2}+\frac {1}{2} \int \frac {\left (4-\frac {4}{x}\right )^{-5 x} (-1-4 x)}{-1+x} \, dx-\frac {5}{2} \int \left (4-\frac {4}{x}\right )^{-5 x} x \log \left (4-\frac {4}{x}\right ) \, dx\\ &=e^{x^2}+\frac {1}{2} \int \left (-4 \left (4-\frac {4}{x}\right )^{-5 x}-\frac {5 \left (4-\frac {4}{x}\right )^{-5 x}}{-1+x}\right ) \, dx+\frac {5}{2} \int \frac {\int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx}{(-1+x) x} \, dx-\frac {1}{2} \left (5 \log \left (4-\frac {4}{x}\right )\right ) \int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx\\ &=e^{x^2}-2 \int \left (4-\frac {4}{x}\right )^{-5 x} \, dx-\frac {5}{2} \int \frac {\left (4-\frac {4}{x}\right )^{-5 x}}{-1+x} \, dx+\frac {5}{2} \int \left (\frac {\int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx}{-1+x}-\frac {\int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx}{x}\right ) \, dx-\frac {1}{2} \left (5 \log \left (4-\frac {4}{x}\right )\right ) \int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx\\ &=e^{x^2}-2 \int \left (4-\frac {4}{x}\right )^{-5 x} \, dx-\frac {5}{2} \int \frac {\left (4-\frac {4}{x}\right )^{-5 x}}{-1+x} \, dx+\frac {5}{2} \int \frac {\int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx}{-1+x} \, dx-\frac {5}{2} \int \frac {\int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx}{x} \, dx-\frac {1}{2} \left (5 \log \left (4-\frac {4}{x}\right )\right ) \int \left (4-\frac {4}{x}\right )^{-5 x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.08, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (2 e^{x^2}+\left (4-\frac {4}{x}\right )^{-5 x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 4*x + E^x^2*((-4 + 4*x)/x)^(5*x)*(-4*x + 4*x^2) + (5*x - 5*x^2)*Log[(-4 + 4*x)/x])/((-2 + 2*x)
*((-4 + 4*x)/x)^(5*x)),x]

[Out]

(2*E^x^2 + x/(4 - 4/x)^(5*x))/2

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fricas [A]  time = 0.54, size = 36, normalized size = 1.44 \begin {gather*} \frac {2 \, \left (\frac {4 \, {\left (x - 1\right )}}{x}\right )^{5 \, x} e^{\left (x^{2}\right )} + x}{2 \, \left (\frac {4 \, {\left (x - 1\right )}}{x}\right )^{5 \, x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-4*x)*exp(x^2)*exp(5*x*log((4*x-4)/x))+(-5*x^2+5*x)*log((4*x-4)/x)-4*x-1)/(2*x-2)/exp(5*x*log
((4*x-4)/x)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (x^{2} - x\right )} \left (\frac {4 \, {\left (x - 1\right )}}{x}\right )^{5 \, x} e^{\left (x^{2}\right )} - 5 \, {\left (x^{2} - x\right )} \log \left (\frac {4 \, {\left (x - 1\right )}}{x}\right ) - 4 \, x - 1}{2 \, {\left (x - 1\right )} \left (\frac {4 \, {\left (x - 1\right )}}{x}\right )^{5 \, x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-4*x)*exp(x^2)*exp(5*x*log((4*x-4)/x))+(-5*x^2+5*x)*log((4*x-4)/x)-4*x-1)/(2*x-2)/exp(5*x*log
((4*x-4)/x)),x, algorithm="giac")

[Out]

integrate(1/2*(4*(x^2 - x)*(4*(x - 1)/x)^(5*x)*e^(x^2) - 5*(x^2 - x)*log(4*(x - 1)/x) - 4*x - 1)/((x - 1)*(4*(
x - 1)/x)^(5*x)), x)

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maple [A]  time = 0.50, size = 25, normalized size = 1.00

method	result	size
default	\(\frac {x \,{\mathrm e}^{-5 x \ln \left (\frac {4 x -4}{x}\right )}}{2}+{\mathrm e}^{x^{2}}\)	\(25\)
risch	\({\mathrm e}^{x^{2}}+\frac {x \,{\mathrm e}^{-\frac {5 x \left (-i \pi \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right )-2 \ln \relax (x )+2 \ln \left (x -1\right )+4 \ln \relax (2)\right )}{2}}}{2}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2-4*x)*exp(x^2)*exp(5*x*ln((4*x-4)/x))+(-5*x^2+5*x)*ln((4*x-4)/x)-4*x-1)/(2*x-2)/exp(5*x*ln((4*x-4)/
x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.78, size = 38, normalized size = 1.52 \begin {gather*} \frac {x e^{\left (-5 \, x \log \left (x - 1\right ) + 5 \, x \log \relax (x)\right )} + 2 \, e^{\left (x^{2} + 10 \, x \log \relax (2)\right )}}{2 \cdot 2^{10 \, x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-4*x)*exp(x^2)*exp(5*x*log((4*x-4)/x))+(-5*x^2+5*x)*log((4*x-4)/x)-4*x-1)/(2*x-2)/exp(5*x*log
((4*x-4)/x)),x, algorithm="maxima")

[Out]

1/2*(x*e^(-5*x*log(x - 1) + 5*x*log(x)) + 2*e^(x^2 + 10*x*log(2)))/2^(10*x)

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mupad [B]  time = 2.01, size = 21, normalized size = 0.84 \begin {gather*} {\mathrm {e}}^{x^2}+\frac {x}{2\,{\left (4-\frac {4}{x}\right )}^{5\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5*x*log((4*x - 4)/x))*(4*x - log((4*x - 4)/x)*(5*x - 5*x^2) + exp(5*x*log((4*x - 4)/x))*exp(x^2)*(4
*x - 4*x^2) + 1))/(2*x - 2),x)

[Out]

exp(x^2) + x/(2*(4 - 4/x)^(5*x))

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sympy [A]  time = 0.65, size = 20, normalized size = 0.80 \begin {gather*} \frac {x e^{- 5 x \log {\left (\frac {4 x - 4}{x} \right )}}}{2} + e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2-4*x)*exp(x**2)*exp(5*x*ln((4*x-4)/x))+(-5*x**2+5*x)*ln((4*x-4)/x)-4*x-1)/(2*x-2)/exp(5*x*ln
((4*x-4)/x)),x)

[Out]

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

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