3.7.6 \(\int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x (6 x+5 x^2+x^3+(3 x^2+x^3) \log (2 x))}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (-e^{-2+2^x x^{2+x}}+x+\log (3+x)\right ) \]

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Rubi [F]  time = 6.79, 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 {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )\right )}{(3+x) \left (e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)\right )} \, dx\\ &=e^2 \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{(3+x) \left (e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)\right )} \, dx\\ &=e^2 \int \left (\frac {2^x e^{-2+2^x x^{2+x}} x^{1+x} (2+x+x \log (2 x))}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}+\frac {4+x}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )}\right ) \, dx\\ &=e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{1+x} (2+x+x \log (2 x))}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx+e^2 \int \frac {4+x}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )} \, dx\\ &=e^2 \int \left (-\frac {1}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}+\frac {1}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )}\right ) \, dx+e^2 \int \left (\frac {2^{1+x} e^{-2+2^x x^{2+x}} x^{1+x}}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}-\frac {2^x e^{-2+2^x x^{2+x}} x^{2+x}}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)}-\frac {2^x e^{-2+2^x x^{2+x}} x^{2+x} \log (2 x)}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)}\right ) \, dx\\ &=-\left (e^2 \int \frac {1}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx\right )+e^2 \int \frac {2^{1+x} e^{-2+2^x x^{2+x}} x^{1+x}}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx-e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{2+x}}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)} \, dx+e^2 \int \frac {1}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )} \, dx-e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{2+x} \log (2 x)}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 23, normalized size = 1.05 \begin {gather*} \log \left (-x + e^{\left (\left (2 \, x\right )^{x} x^{2} - 2\right )} - \log \left (x + 3\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 3.07, size = 73, normalized size = 3.32 \begin {gather*} -\left (2 \, x\right )^{x} x^{3} \log \left (2 \, x\right ) - 2 \, \left (2 \, x\right )^{x} x^{2} \log \left (2 \, x\right ) + \left (2 \, x\right )^{x} x^{2} + \log \left (-2 \, x e^{2} + 2 \, e^{2} \log \relax (2) - 2 \, e^{2} \log \left (2 \, x + 6\right ) + 2 \, e^{\left (\left (2 \, x\right )^{x} x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 26, normalized size = 1.18

method	result	size
risch	\(2+\ln \left (-x +{\mathrm e}^{x^{2} \left (2 x \right )^{x}-2}-\ln \left (3+x \right )\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 30, normalized size = 1.36 \begin {gather*} \log \left (-x e^{2} - e^{2} \log \left (x + 3\right ) + e^{\left (x^{2} e^{\left (x \log \relax (2) + x \log \relax (x)\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.23, size = 21, normalized size = 0.95 \begin {gather*} \ln \left (x+\ln \left (x+3\right )-{\mathrm {e}}^{2^x\,x^{x+2}-2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.56, size = 22, normalized size = 1.00 \begin {gather*} \log {\left (- x + e^{x^{2} e^{x \log {\left (2 x \right )}} - 2} - \log {\left (x + 3 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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