3.96.84 \(\int \frac {e^{\frac {-5+5 x^2+e^x x^2+(-50 x-10 e^x x) \log (x)+(125 x+25 e^x x) \log ^2(x)}{5+e^x}} (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+(1000+400 e^x+40 e^{2 x}) \log (x)+(625+250 e^x+25 e^{2 x}) \log ^2(x))}{25+10 e^x+e^{2 x}} \, dx\)

Optimal. Leaf size=28 \[ e^{-\frac {5}{5+e^x}-x+x^2+x (-1+5 \log (x))^2} \]

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Rubi [F]  time = 12.23, 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 {\exp \left (\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}\right ) \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5 + 5*x^2 + E^x*x^2 + (-50*x - 10*E^x*x)*Log[x] + (125*x + 25*E^x*x)*Log[x]^2)/(5 + E^x))*(-250 + 50
*x + E^(2*x)*(-10 + 2*x) + E^x*(-95 + 20*x) + (1000 + 400*E^x + 40*E^(2*x))*Log[x] + (625 + 250*E^x + 25*E^(2*
x))*Log[x]^2))/(25 + 10*E^x + E^(2*x)),x]

[Out]

2*Defer[Int][E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)*x^(1 - 10*x), x] - 10*Defer[Int][E^(-5/(5 + E^x) + x^2 + 2
5*x*Log[x]^2)/x^(10*x), x] - 25*Defer[Int][E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)/((5 + E^x)^2*x^(10*x)), x] +
5*Defer[Int][E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)/((5 + E^x)*x^(10*x)), x] + 40*Defer[Int][(E^(-5/(5 + E^x)
+ x^2 + 25*x*Log[x]^2)*Log[x])/x^(10*x), x] + 25*Defer[Int][(E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)*Log[x]^2)
/x^(10*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \left (50 (-5+x)+2 e^{2 x} (-5+x)+5 e^x (-19+4 x)+40 \left (5+e^x\right )^2 \log (x)+25 \left (5+e^x\right )^2 \log ^2(x)\right )}{\left (5+e^x\right )^2} \, dx\\ &=\int \left (2 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x}-10 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}-\frac {25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2}+\frac {5 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x}+40 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x)+25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x)\right ) \, dx\\ &=2 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x} \, dx+5 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x} \, dx-10 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \, dx-25 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2} \, dx+25 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x) \, dx+40 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 28, normalized size = 1.00 \begin {gather*} e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5 + 5*x^2 + E^x*x^2 + (-50*x - 10*E^x*x)*Log[x] + (125*x + 25*E^x*x)*Log[x]^2)/(5 + E^x))*(-25
0 + 50*x + E^(2*x)*(-10 + 2*x) + E^x*(-95 + 20*x) + (1000 + 400*E^x + 40*E^(2*x))*Log[x] + (625 + 250*E^x + 25
*E^(2*x))*Log[x]^2))/(25 + 10*E^x + E^(2*x)),x]

[Out]

E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)/x^(10*x)

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fricas [A]  time = 0.49, size = 47, normalized size = 1.68 \begin {gather*} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x)^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x) - 5}{e^{x} + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1000)*log(x)+(2*x-10)*exp(x)^2+(20*x-
95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+
5))/(exp(x)^2+10*exp(x)+25),x, algorithm="fricas")

[Out]

e^((x^2*e^x + 25*(x*e^x + 5*x)*log(x)^2 + 5*x^2 - 10*(x*e^x + 5*x)*log(x) - 5)/(e^x + 5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (25 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \relax (x)^{2} + 2 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} + 5 \, {\left (4 \, x - 19\right )} e^{x} + 40 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \relax (x) + 50 \, x - 250\right )} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x)^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x) - 5}{e^{x} + 5}\right )}}{e^{\left (2 \, x\right )} + 10 \, e^{x} + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1000)*log(x)+(2*x-10)*exp(x)^2+(20*x-
95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+
5))/(exp(x)^2+10*exp(x)+25),x, algorithm="giac")

[Out]

integrate((25*(e^(2*x) + 10*e^x + 25)*log(x)^2 + 2*(x - 5)*e^(2*x) + 5*(4*x - 19)*e^x + 40*(e^(2*x) + 10*e^x +
25)*log(x) + 50*x - 250)*e^((x^2*e^x + 25*(x*e^x + 5*x)*log(x)^2 + 5*x^2 - 10*(x*e^x + 5*x)*log(x) - 5)/(e^x
+ 5))/(e^(2*x) + 10*e^x + 25), x)

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maple [A]  time = 0.05, size = 50, normalized size = 1.79

method	result	size
risch	\({\mathrm e}^{\frac {25 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+125 x \ln \relax (x )^{2}-10 x \,{\mathrm e}^{x} \ln \relax (x )+{\mathrm e}^{x} x^{2}-50 x \ln \relax (x )+5 x^{2}-5}{{\mathrm e}^{x}+5}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*exp(x)^2+250*exp(x)+625)*ln(x)^2+(40*exp(x)^2+400*exp(x)+1000)*ln(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(
x)+50*x-250)*exp(((25*exp(x)*x+125*x)*ln(x)^2+(-10*exp(x)*x-50*x)*ln(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5))/(exp(x
)^2+10*exp(x)+25),x,method=_RETURNVERBOSE)

[Out]

exp((25*x*exp(x)*ln(x)^2+125*x*ln(x)^2-10*x*exp(x)*ln(x)+exp(x)*x^2-50*x*ln(x)+5*x^2-5)/(exp(x)+5))

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maxima [B]  time = 0.62, size = 85, normalized size = 3.04 \begin {gather*} e^{\left (\frac {25 \, x e^{x} \log \relax (x)^{2}}{e^{x} + 5} + \frac {x^{2} e^{x}}{e^{x} + 5} - \frac {10 \, x e^{x} \log \relax (x)}{e^{x} + 5} + \frac {125 \, x \log \relax (x)^{2}}{e^{x} + 5} + \frac {5 \, x^{2}}{e^{x} + 5} - \frac {50 \, x \log \relax (x)}{e^{x} + 5} - \frac {5}{e^{x} + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1000)*log(x)+(2*x-10)*exp(x)^2+(20*x-
95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+
5))/(exp(x)^2+10*exp(x)+25),x, algorithm="maxima")

[Out]

e^(25*x*e^x*log(x)^2/(e^x + 5) + x^2*e^x/(e^x + 5) - 10*x*e^x*log(x)/(e^x + 5) + 125*x*log(x)^2/(e^x + 5) + 5*
x^2/(e^x + 5) - 50*x*log(x)/(e^x + 5) - 5/(e^x + 5))

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mupad [B]  time = 9.41, size = 72, normalized size = 2.57 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {125\,x\,{\ln \relax (x)}^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {5\,x^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2}{{\mathrm {e}}^x+5}}}{x^{10\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2*exp(x) - log(x)*(50*x + 10*x*exp(x)) + log(x)^2*(125*x + 25*x*exp(x)) + 5*x^2 - 5)/(exp(x) + 5))
*(50*x + exp(x)*(20*x - 95) + log(x)*(40*exp(2*x) + 400*exp(x) + 1000) + exp(2*x)*(2*x - 10) + log(x)^2*(25*ex
p(2*x) + 250*exp(x) + 625) - 250))/(exp(2*x) + 10*exp(x) + 25),x)

[Out]

(exp((x^2*exp(x))/(exp(x) + 5))*exp((125*x*log(x)^2)/(exp(x) + 5))*exp((5*x^2)/(exp(x) + 5))*exp(-5/(exp(x) +
5))*exp((25*x*exp(x)*log(x)^2)/(exp(x) + 5)))/x^(10*x)

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sympy [B]  time = 0.87, size = 49, normalized size = 1.75 \begin {gather*} e^{\frac {x^{2} e^{x} + 5 x^{2} + \left (- 10 x e^{x} - 50 x\right ) \log {\relax (x )} + \left (25 x e^{x} + 125 x\right ) \log {\relax (x )}^{2} - 5}{e^{x} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)**2+250*exp(x)+625)*ln(x)**2+(40*exp(x)**2+400*exp(x)+1000)*ln(x)+(2*x-10)*exp(x)**2+(20*
x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x)*ln(x)**2+(-10*exp(x)*x-50*x)*ln(x)+exp(x)*x**2+5*x**2-5)/(exp(
x)+5))/(exp(x)**2+10*exp(x)+25),x)

[Out]

exp((x**2*exp(x) + 5*x**2 + (-10*x*exp(x) - 50*x)*log(x) + (25*x*exp(x) + 125*x)*log(x)**2 - 5)/(exp(x) + 5))

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