3.103.83 \(\int \frac {5 x^2+e^{2 e^x+4 x} (-1+4 x+2 e^x x)+8 \log ^7(x)-\log ^8(x)+e^{e^x+2 x} (8 \log ^3(x)+(-2+4 x+2 e^x x) \log ^4(x))}{x^2} \, dx\)

Optimal. Leaf size=27 \[ x+4 (4+x)+\frac {\left (e^{e^x+2 x}+\log ^4(x)\right )^2}{x} \]

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Rubi [F]  time = 1.65, 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 {5 x^2+e^{2 e^x+4 x} \left (-1+4 x+2 e^x x\right )+8 \log ^7(x)-\log ^8(x)+e^{e^x+2 x} \left (8 \log ^3(x)+\left (-2+4 x+2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x^2 + E^(2*E^x + 4*x)*(-1 + 4*x + 2*E^x*x) + 8*Log[x]^7 - Log[x]^8 + E^(E^x + 2*x)*(8*Log[x]^3 + (-2 +
4*x + 2*E^x*x)*Log[x]^4))/x^2,x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^{2 e^x+5 x}}{x}+\frac {e^{2 \left (e^x+2 x\right )} (-1+4 x)}{x^2}+\frac {2 e^{e^x+3 x} \log ^4(x)}{x}+\frac {2 e^{e^x+2 x} \log ^3(x) (4-\log (x)+2 x \log (x))}{x^2}+\frac {5 x^2+8 \log ^7(x)-\log ^8(x)}{x^2}\right ) \, dx\\ &=2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \frac {e^{e^x+2 x} \log ^3(x) (4-\log (x)+2 x \log (x))}{x^2} \, dx+\int \frac {e^{2 \left (e^x+2 x\right )} (-1+4 x)}{x^2} \, dx+\int \frac {5 x^2+8 \log ^7(x)-\log ^8(x)}{x^2} \, dx\\ &=2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \left (\frac {4 e^{e^x+2 x} \log ^3(x)}{x^2}+\frac {e^{e^x+2 x} (-1+2 x) \log ^4(x)}{x^2}\right ) \, dx+\int \left (-\frac {e^{2 \left (e^x+2 x\right )}}{x^2}+\frac {4 e^{2 \left (e^x+2 x\right )}}{x}\right ) \, dx+\int \left (5+\frac {8 \log ^7(x)}{x^2}-\frac {\log ^8(x)}{x^2}\right ) \, dx\\ &=5 x+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \frac {e^{e^x+2 x} (-1+2 x) \log ^4(x)}{x^2} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx+8 \int \frac {\log ^7(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx-\int \frac {\log ^8(x)}{x^2} \, dx\\ &=5 x-\frac {8 \log ^7(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+2 \int \left (-\frac {e^{e^x+2 x} \log ^4(x)}{x^2}+\frac {2 e^{e^x+2 x} \log ^4(x)}{x}\right ) \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-8 \int \frac {\log ^7(x)}{x^2} \, dx+56 \int \frac {\log ^6(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {56 \log ^6(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-56 \int \frac {\log ^6(x)}{x^2} \, dx+336 \int \frac {\log ^5(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {336 \log ^5(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-336 \int \frac {\log ^5(x)}{x^2} \, dx+1680 \int \frac {\log ^4(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {1680 \log ^4(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-1680 \int \frac {\log ^4(x)}{x^2} \, dx+6720 \int \frac {\log ^3(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {6720 \log ^3(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-6720 \int \frac {\log ^3(x)}{x^2} \, dx+20160 \int \frac {\log ^2(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x-\frac {20160 \log ^2(x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-20160 \int \frac {\log ^2(x)}{x^2} \, dx+40320 \int \frac {\log (x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=-\frac {40320}{x}+5 x-\frac {40320 \log (x)}{x}+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-40320 \int \frac {\log (x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ &=5 x+\frac {\log ^8(x)}{x}+2 \int \frac {e^{2 e^x+5 x}}{x} \, dx-2 \int \frac {e^{e^x+2 x} \log ^4(x)}{x^2} \, dx+2 \int \frac {e^{e^x+3 x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{2 \left (e^x+2 x\right )}}{x} \, dx+4 \int \frac {e^{e^x+2 x} \log ^4(x)}{x} \, dx+8 \int \frac {e^{e^x+2 x} \log ^3(x)}{x^2} \, dx-\int \frac {e^{2 \left (e^x+2 x\right )}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 40, normalized size = 1.48 \begin {gather*} \frac {e^{2 e^x+4 x}+5 x^2+2 e^{e^x+2 x} \log ^4(x)+\log ^8(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + E^(2*E^x + 4*x)*(-1 + 4*x + 2*E^x*x) + 8*Log[x]^7 - Log[x]^8 + E^(E^x + 2*x)*(8*Log[x]^3 +
(-2 + 4*x + 2*E^x*x)*Log[x]^4))/x^2,x]

[Out]

(E^(2*E^x + 4*x) + 5*x^2 + 2*E^(E^x + 2*x)*Log[x]^4 + Log[x]^8)/x

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fricas [A]  time = 0.97, size = 36, normalized size = 1.33 \begin {gather*} \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + 5 \, x^{2} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+4*x-1)*exp(exp(x)+2*x)^2+((2*exp(x)*x+4*x-2)*log(x)^4+8*log(x)^3)*exp(exp(x)+2*x)-log(x
)^8+8*log(x)^7+5*x^2)/x^2,x, algorithm="fricas")

[Out]

(log(x)^8 + 2*e^(2*x + e^x)*log(x)^4 + 5*x^2 + e^(4*x + 2*e^x))/x

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giac [A]  time = 0.23, size = 36, normalized size = 1.33 \begin {gather*} \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + 5 \, x^{2} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+4*x-1)*exp(exp(x)+2*x)^2+((2*exp(x)*x+4*x-2)*log(x)^4+8*log(x)^3)*exp(exp(x)+2*x)-log(x
)^8+8*log(x)^7+5*x^2)/x^2,x, algorithm="giac")

[Out]

(log(x)^8 + 2*e^(2*x + e^x)*log(x)^4 + 5*x^2 + e^(4*x + 2*e^x))/x

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maple [A]  time = 0.07, size = 42, normalized size = 1.56

method	result	size
risch	\(\frac {\ln \relax (x )^{8}}{x}+5 x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}+4 x}}{x}+\frac {2 \ln \relax (x )^{4} {\mathrm e}^{{\mathrm e}^{x}+2 x}}{x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x+4*x-1)*exp(exp(x)+2*x)^2+((2*exp(x)*x+4*x-2)*ln(x)^4+8*ln(x)^3)*exp(exp(x)+2*x)-ln(x)^8+8*ln(
x)^7+5*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/x*ln(x)^8+5*x+exp(2*exp(x)+4*x)/x+2/x*ln(x)^4*exp(exp(x)+2*x)

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maxima [A]  time = 0.43, size = 35, normalized size = 1.30 \begin {gather*} 5 \, x + \frac {\log \relax (x)^{8} + 2 \, e^{\left (2 \, x + e^{x}\right )} \log \relax (x)^{4} + e^{\left (4 \, x + 2 \, e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+4*x-1)*exp(exp(x)+2*x)^2+((2*exp(x)*x+4*x-2)*log(x)^4+8*log(x)^3)*exp(exp(x)+2*x)-log(x
)^8+8*log(x)^7+5*x^2)/x^2,x, algorithm="maxima")

[Out]

5*x + (log(x)^8 + 2*e^(2*x + e^x)*log(x)^4 + e^(4*x + 2*e^x))/x

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mupad [B]  time = 6.78, size = 41, normalized size = 1.52 \begin {gather*} 5\,x+\frac {{\mathrm {e}}^{4\,x+2\,{\mathrm {e}}^x}}{x}+\frac {{\ln \relax (x)}^8}{x}+\frac {2\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^x}\,{\ln \relax (x)}^4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x + 2*exp(x))*(4*x + 2*x*exp(x) - 1) + 8*log(x)^7 - log(x)^8 + exp(2*x + exp(x))*(8*log(x)^3 + log(
x)^4*(4*x + 2*x*exp(x) - 2)) + 5*x^2)/x^2,x)

[Out]

5*x + exp(4*x + 2*exp(x))/x + log(x)^8/x + (2*exp(2*x + exp(x))*log(x)^4)/x

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sympy [B]  time = 0.44, size = 41, normalized size = 1.52 \begin {gather*} 5 x + \frac {\log {\relax (x )}^{8}}{x} + \frac {2 x e^{2 x + e^{x}} \log {\relax (x )}^{4} + x e^{4 x + 2 e^{x}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+4*x-1)*exp(exp(x)+2*x)**2+((2*exp(x)*x+4*x-2)*ln(x)**4+8*ln(x)**3)*exp(exp(x)+2*x)-ln(x
)**8+8*ln(x)**7+5*x**2)/x**2,x)

[Out]

5*x + log(x)**8/x + (2*x*exp(2*x + exp(x))*log(x)**4 + x*exp(4*x + 2*exp(x)))/x**2

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