3.72.42 \(\int \frac {e^{e^x-2 x+x^2} (e^{2 x} (4 x^2+x^3)+e^x (2 x-4 x^2+7 x^3+2 x^4)+(e^{2 x} (4 x+x^2)+e^x (-4-5 x+7 x^2+2 x^3)) \log (16+8 x+x^2))}{4 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {e^{e^x-x+x^2} \left (x+\log \left ((4+x)^2\right )\right )}{x} \]

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Rubi [F]  time = 10.71, 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 {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^x - 2*x + x^2)*(E^(2*x)*(4*x^2 + x^3) + E^x*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + (E^(2*x)*(4*x + x^2) + E
^x*(-4 - 5*x + 7*x^2 + 2*x^3))*Log[16 + 8*x + x^2]))/(4*x^2 + x^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \frac {e^{e^x+(-1+x) x} \left (x \left (2+4 \left (-1+e^x\right ) x+\left (7+e^x\right ) x^2+2 x^3\right )+(4+x) \left (-1+\left (-1+e^x\right ) x+2 x^2\right ) \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \left (\frac {e^{e^x+x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x}+\frac {e^{e^x+(-1+x) x} \left (2 x-4 x^2+7 x^3+2 x^4-4 \log \left ((4+x)^2\right )-5 x \log \left ((4+x)^2\right )+7 x^2 \log \left ((4+x)^2\right )+2 x^3 \log \left ((4+x)^2\right )\right )}{x^2 (4+x)}\right ) \, dx\\ &=\int \frac {e^{e^x+x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x} \, dx+\int \frac {e^{e^x+(-1+x) x} \left (2 x-4 x^2+7 x^3+2 x^4-4 \log \left ((4+x)^2\right )-5 x \log \left ((4+x)^2\right )+7 x^2 \log \left ((4+x)^2\right )+2 x^3 \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \left (e^{e^x+x+(-1+x) x}+\frac {e^{e^x+x+(-1+x) x} \log \left ((4+x)^2\right )}{x}\right ) \, dx+\int \frac {e^{e^x+(-1+x) x} \left (x \left (2-4 x+7 x^2+2 x^3\right )+\left (-4-5 x+7 x^2+2 x^3\right ) \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int e^{e^x+x+(-1+x) x} \, dx+\int \frac {e^{e^x+x+(-1+x) x} \log \left ((4+x)^2\right )}{x} \, dx+\int \left (\frac {e^{e^x+(-1+x) x} \left (2-4 x+7 x^2+2 x^3\right )}{x (4+x)}+\frac {e^{e^x+(-1+x) x} (-1+x) (1+2 x) \log \left ((4+x)^2\right )}{x^2}\right ) \, dx\\ &=\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx+\int \frac {e^{e^x+(-1+x) x} \left (2-4 x+7 x^2+2 x^3\right )}{x (4+x)} \, dx+\int \frac {e^{e^x+(-1+x) x} (-1+x) (1+2 x) \log \left ((4+x)^2\right )}{x^2} \, dx-\int \frac {2 \int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx\\ &=-\left (2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx\right )-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx+\int \left (-e^{e^x+(-1+x) x}+\frac {e^{e^x+(-1+x) x}}{2 x}+2 e^{e^x+(-1+x) x} x-\frac {e^{e^x+(-1+x) x}}{2 (4+x)}\right ) \, dx-\int \frac {4 \int e^{e^x+(-1+x) x} \, dx-2 \left (\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx+\int \frac {e^{e^x+(-1+x) x}}{x} \, dx\right )}{4+x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx-\int \left (\frac {2 \left (2 \int e^{e^x+(-1+x) x} \, dx-\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx\right )}{4+x}-\frac {2 \int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \frac {2 \int e^{e^x+(-1+x) x} \, dx-\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x} \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \left (\frac {2 \int e^{e^x+(-1+x) x} \, dx}{4+x}-\frac {\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x}\right ) \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x} \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-4 \int \frac {\int e^{e^x+(-1+x) x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.12, size = 23, normalized size = 0.96 \begin {gather*} \frac {e^{e^x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^x - 2*x + x^2)*(E^(2*x)*(4*x^2 + x^3) + E^x*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + (E^(2*x)*(4*x + x^
2) + E^x*(-4 - 5*x + 7*x^2 + 2*x^3))*Log[16 + 8*x + x^2]))/(4*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 1.05, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (x e^{x} + e^{x} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (x^{2} - 2 \, x + e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16)+(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4
*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp(x)-x^2+2*x),x, algorithm="fricas")

[Out]

(x*e^x + e^x*log(x^2 + 8*x + 16))*e^(x^2 - 2*x + e^x)/x

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giac [A]  time = 0.25, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (x e^{\left (x^{2} + e^{x}\right )} + e^{\left (x^{2} + e^{x}\right )} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16)+(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4
*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp(x)-x^2+2*x),x, algorithm="giac")

[Out]

(x*e^(x^2 + e^x) + e^(x^2 + e^x)*log(x^2 + 8*x + 16))*e^(-x)/x

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maple [C]  time = 0.17, size = 85, normalized size = 3.54

method	result	size
risch	\(\frac {\left (-i \pi \mathrm {csgn}\left (i \left (4+x \right )\right )^{2} \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{3}+2 x +4 \ln \left (4+x \right )\right ) {\mathrm e}^{x^{2}+{\mathrm e}^{x}-x}}{2 x}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*ln(x^2+8*x+16)+(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4*x^2+2*
x)*exp(x))/(x^3+4*x^2)/exp(-exp(x)-x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

1/2/x*(-I*Pi*csgn(I*(4+x))^2*csgn(I*(4+x)^2)+2*I*Pi*csgn(I*(4+x))*csgn(I*(4+x)^2)^2-I*Pi*csgn(I*(4+x)^2)^3+2*x
+4*ln(4+x))*exp(x^2+exp(x)-x)

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maxima [A]  time = 0.42, size = 22, normalized size = 0.92 \begin {gather*} \frac {{\left (x + 2 \, \log \left (x + 4\right )\right )} e^{\left (x^{2} - x + e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16)+(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4
*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp(x)-x^2+2*x),x, algorithm="maxima")

[Out]

(x + 2*log(x + 4))*e^(x^2 - x + e^x)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^x-2\,x+x^2}\,\left ({\mathrm {e}}^{2\,x}\,\left (x^3+4\,x^2\right )+{\mathrm {e}}^x\,\left (2\,x^4+7\,x^3-4\,x^2+2\,x\right )+\ln \left (x^2+8\,x+16\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (x^2+4\,x\right )-{\mathrm {e}}^x\,\left (-2\,x^3-7\,x^2+5\,x+4\right )\right )\right )}{x^3+4\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x) - 2*x + x^2)*(exp(2*x)*(4*x^2 + x^3) + exp(x)*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + log(8*x + x^2 +
16)*(exp(2*x)*(4*x + x^2) - exp(x)*(5*x - 7*x^2 - 2*x^3 + 4))))/(4*x^2 + x^3),x)

[Out]

int((exp(exp(x) - 2*x + x^2)*(exp(2*x)*(4*x^2 + x^3) + exp(x)*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + log(8*x + x^2 +
16)*(exp(2*x)*(4*x + x^2) - exp(x)*(5*x - 7*x^2 - 2*x^3 + 4))))/(4*x^2 + x^3), x)

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sympy [A]  time = 0.49, size = 31, normalized size = 1.29 \begin {gather*} \frac {\left (x e^{x} + e^{x} \log {\left (x^{2} + 8 x + 16 \right )}\right ) e^{x^{2} - 2 x + e^{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+4*x)*exp(x)**2+(2*x**3+7*x**2-5*x-4)*exp(x))*ln(x**2+8*x+16)+(x**3+4*x**2)*exp(x)**2+(2*x**4
+7*x**3-4*x**2+2*x)*exp(x))/(x**3+4*x**2)/exp(-exp(x)-x**2+2*x),x)

[Out]

(x*exp(x) + exp(x)*log(x**2 + 8*x + 16))*exp(x**2 - 2*x + exp(x))/x

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