3.93.62 \(\int \frac {240+e^{2-x} (60+60 x)+e^x (240-30 x+e^{2-x} (60+60 x))+(15+15 e^x) \log (1+2 e^x+e^{2 x})}{x^2+e^x x^2} \, dx\)

Optimal. Leaf size=25 \[ -\frac {15 \left (4 \left (4+e^{2-x}\right )+\log \left (\left (1+e^x\right )^2\right )\right )}{x} \]

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Rubi [F]  time = 0.42, 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 {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(240 + E^(2 - x)*(60 + 60*x) + E^x*(240 - 30*x + E^(2 - x)*(60 + 60*x)) + (15 + 15*E^x)*Log[1 + 2*E^x + E^
(2*x)])/(x^2 + E^x*x^2),x]

[Out]

-240/x - (60*E^(2 - x))/x - (15*Log[(1 + E^x)^2])/x - 30*Log[x] + 30*Defer[Int][1/((1 + E^x)*x), x] + 30*Defer
[Int][E^x/(x + E^x*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {30}{\left (1+e^x\right ) x}+\frac {60 e^{2-x} (1+x)}{x^2}-\frac {15 \left (-16+2 x-\log \left (\left (1+e^x\right )^2\right )\right )}{x^2}\right ) \, dx\\ &=-\left (15 \int \frac {-16+2 x-\log \left (\left (1+e^x\right )^2\right )}{x^2} \, dx\right )+30 \int \frac {1}{\left (1+e^x\right ) x} \, dx+60 \int \frac {e^{2-x} (1+x)}{x^2} \, dx\\ &=-\frac {60 e^{2-x}}{x}-15 \int \left (\frac {2 (-8+x)}{x^2}-\frac {\log \left (\left (1+e^x\right )^2\right )}{x^2}\right ) \, dx+30 \int \frac {1}{\left (1+e^x\right ) x} \, dx\\ &=-\frac {60 e^{2-x}}{x}+15 \int \frac {\log \left (\left (1+e^x\right )^2\right )}{x^2} \, dx-30 \int \frac {-8+x}{x^2} \, dx+30 \int \frac {1}{\left (1+e^x\right ) x} \, dx\\ &=-\frac {60 e^{2-x}}{x}-\frac {15 \log \left (\left (1+e^x\right )^2\right )}{x}+15 \int \frac {2 e^x}{x+e^x x} \, dx-30 \int \left (-\frac {8}{x^2}+\frac {1}{x}\right ) \, dx+30 \int \frac {1}{\left (1+e^x\right ) x} \, dx\\ &=-\frac {240}{x}-\frac {60 e^{2-x}}{x}-\frac {15 \log \left (\left (1+e^x\right )^2\right )}{x}-30 \log (x)+30 \int \frac {1}{\left (1+e^x\right ) x} \, dx+30 \int \frac {e^x}{x+e^x x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.09, size = 30, normalized size = 1.20 \begin {gather*} \frac {15 \left (2 \left (-8-2 e^{2-x}+x\right )-\log \left (\left (1+e^x\right )^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240 + E^(2 - x)*(60 + 60*x) + E^x*(240 - 30*x + E^(2 - x)*(60 + 60*x)) + (15 + 15*E^x)*Log[1 + 2*E^
x + E^(2*x)])/(x^2 + E^x*x^2),x]

[Out]

(15*(2*(-8 - 2*E^(2 - x) + x) - Log[(1 + E^x)^2]))/x

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fricas [A]  time = 0.83, size = 32, normalized size = 1.28 \begin {gather*} -\frac {15 \, {\left (e^{x} \log \left (e^{\left (2 \, x\right )} + 2 \, e^{x} + 1\right ) + 4 \, e^{2} + 16 \, e^{x}\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)*exp(x)+(60*x+60)*exp(2-x)+240
)/(exp(x)*x^2+x^2),x, algorithm="fricas")

[Out]

-15*(e^x*log(e^(2*x) + 2*e^x + 1) + 4*e^2 + 16*e^x)*e^(-x)/x

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giac [A]  time = 0.19, size = 26, normalized size = 1.04 \begin {gather*} -\frac {30 \, {\left (e^{x} \log \left (e^{x} + 1\right ) + 2 \, e^{2} + 8 \, e^{x}\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)*exp(x)+(60*x+60)*exp(2-x)+240
)/(exp(x)*x^2+x^2),x, algorithm="giac")

[Out]

-30*(e^x*log(e^x + 1) + 2*e^2 + 8*e^x)*e^(-x)/x

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maple [A]  time = 0.24, size = 33, normalized size = 1.32

method	result	size
norman	\(\frac {\left (-15 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}+1\right )-60 \,{\mathrm e}^{2}-240 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\)	\(33\)
default	\(\frac {\left (\left (-15 \ln \left (\left ({\mathrm e}^{x}+1\right )^{2}\right )+30 \ln \left ({\mathrm e}^{x}+1\right )-240\right ) {\mathrm e}^{x}-30 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+1\right )-60 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-x}}{x}\)	\(44\)
risch	\(-\frac {30 \ln \left ({\mathrm e}^{x}+1\right )}{x}-\frac {15 \left (-i {\mathrm e}^{x} \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+1\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right )+2 i {\mathrm e}^{x} \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+1\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right )^{2}-i {\mathrm e}^{x} \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right )^{3}+8 \,{\mathrm e}^{2}+32 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{2 x}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*exp(x)+15)*ln(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)*exp(x)+(60*x+60)*exp(2-x)+240)/(exp(
x)*x^2+x^2),x,method=_RETURNVERBOSE)

[Out]

(-15*exp(x)*ln(exp(x)^2+2*exp(x)+1)-60*exp(2)-240*exp(x))/exp(x)/x

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maxima [A]  time = 0.39, size = 20, normalized size = 0.80 \begin {gather*} -\frac {30 \, {\left (2 \, e^{\left (-x + 2\right )} + \log \left (e^{x} + 1\right ) + 8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)*exp(x)+(60*x+60)*exp(2-x)+240
)/(exp(x)*x^2+x^2),x, algorithm="maxima")

[Out]

-30*(2*e^(-x + 2) + log(e^x + 1) + 8)/x

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mupad [B]  time = 7.29, size = 102, normalized size = 4.08 \begin {gather*} -\frac {240\,{\mathrm {e}}^{3\,x}+60\,{\mathrm {e}}^2+{\mathrm {e}}^x\,\left (120\,{\mathrm {e}}^2+240\right )+15\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (60\,{\mathrm {e}}^2+480\right )+30\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^{2\,x}+15\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^{3\,x}}{2\,x\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^{3\,x}+x\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2 - x)*(60*x + 60) + log(exp(2*x) + 2*exp(x) + 1)*(15*exp(x) + 15) + exp(x)*(exp(2 - x)*(60*x + 60) -
30*x + 240) + 240)/(x^2*exp(x) + x^2),x)

[Out]

-(240*exp(3*x) + 60*exp(2) + exp(x)*(120*exp(2) + 240) + 15*log(exp(2*x) + 2*exp(x) + 1)*exp(x) + exp(2*x)*(60
*exp(2) + 480) + 30*log(exp(2*x) + 2*exp(x) + 1)*exp(2*x) + 15*log(exp(2*x) + 2*exp(x) + 1)*exp(3*x))/(2*x*exp
(2*x) + x*exp(3*x) + x*exp(x))

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sympy [A]  time = 0.25, size = 31, normalized size = 1.24 \begin {gather*} - \frac {15 \log {\left (e^{2 x} + 2 e^{x} + 1 \right )}}{x} - \frac {240}{x} - \frac {60 e^{2} e^{- x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(x)+15)*ln(exp(x)**2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)*exp(x)+(60*x+60)*exp(2-x)+240
)/(exp(x)*x**2+x**2),x)

[Out]

-15*log(exp(2*x) + 2*exp(x) + 1)/x - 240/x - 60*exp(2)*exp(-x)/x

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