3.41.56 \(\int \frac {e^{\frac {1}{15} (12+5 e^{\frac {2-16 x+4 x^2+4 x \log (25+10 x+x^2)}{-4 x+x^2+x \log (25+10 x+x^2)}} x)+\frac {2-16 x+4 x^2+4 x \log (25+10 x+x^2)}{-4 x+x^2+x \log (25+10 x+x^2)}} (40+64 x-28 x^2-3 x^3+x^4+(-10-42 x+2 x^2+2 x^3) \log (25+10 x+x^2)+(5 x+x^2) \log ^2(25+10 x+x^2))}{240 x-72 x^2-9 x^3+3 x^4+(-120 x+6 x^2+6 x^3) \log (25+10 x+x^2)+(15 x+3 x^2) \log ^2(25+10 x+x^2)} \, dx\)

Optimal. Leaf size=38 \[ e^{\frac {4}{5}+\frac {1}{3} e^{4-\frac {2}{-x+x \left (5-x-\log \left ((5+x)^2\right )\right )}} x} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((12 + 5*E^((2 - 16*x + 4*x^2 + 4*x*Log[25 + 10*x + x^2])/(-4*x + x^2 + x*Log[25 + 10*x + x^2]))*x)/15
+ (2 - 16*x + 4*x^2 + 4*x*Log[25 + 10*x + x^2])/(-4*x + x^2 + x*Log[25 + 10*x + x^2]))*(40 + 64*x - 28*x^2 - 3
*x^3 + x^4 + (-10 - 42*x + 2*x^2 + 2*x^3)*Log[25 + 10*x + x^2] + (5*x + x^2)*Log[25 + 10*x + x^2]^2))/(240*x -
72*x^2 - 9*x^3 + 3*x^4 + (-120*x + 6*x^2 + 6*x^3)*Log[25 + 10*x + x^2] + (15*x + 3*x^2)*Log[25 + 10*x + x^2]^
2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.92, size = 58, normalized size = 1.53 \begin {gather*} e^{\frac {4}{5}+\frac {1}{3} e^{\frac {2 \left (1-8 x+2 x^2\right )}{x \left (-4+x+\log \left ((5+x)^2\right )\right )}} x \left ((5+x)^2\right )^{\frac {4}{-4+x+\log \left ((5+x)^2\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((12 + 5*E^((2 - 16*x + 4*x^2 + 4*x*Log[25 + 10*x + x^2])/(-4*x + x^2 + x*Log[25 + 10*x + x^2]))*
x)/15 + (2 - 16*x + 4*x^2 + 4*x*Log[25 + 10*x + x^2])/(-4*x + x^2 + x*Log[25 + 10*x + x^2]))*(40 + 64*x - 28*x
^2 - 3*x^3 + x^4 + (-10 - 42*x + 2*x^2 + 2*x^3)*Log[25 + 10*x + x^2] + (5*x + x^2)*Log[25 + 10*x + x^2]^2))/(2
40*x - 72*x^2 - 9*x^3 + 3*x^4 + (-120*x + 6*x^2 + 6*x^3)*Log[25 + 10*x + x^2] + (15*x + 3*x^2)*Log[25 + 10*x +
x^2]^2),x]

[Out]

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

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fricas [B]  time = 0.57, size = 159, normalized size = 4.18 \begin {gather*} e^{\left (\frac {72 \, x^{2} + 5 \, {\left (x^{3} + x^{2} \log \left (x^{2} + 10 \, x + 25\right ) - 4 \, x^{2}\right )} e^{\left (\frac {2 \, {\left (2 \, x^{2} + 2 \, x \log \left (x^{2} + 10 \, x + 25\right ) - 8 \, x + 1\right )}}{x^{2} + x \log \left (x^{2} + 10 \, x + 25\right ) - 4 \, x}\right )} + 72 \, x \log \left (x^{2} + 10 \, x + 25\right ) - 288 \, x + 30}{15 \, {\left (x^{2} + x \log \left (x^{2} + 10 \, x + 25\right ) - 4 \, x\right )}} - \frac {2 \, {\left (2 \, x^{2} + 2 \, x \log \left (x^{2} + 10 \, x + 25\right ) - 8 \, x + 1\right )}}{x^{2} + x \log \left (x^{2} + 10 \, x + 25\right ) - 4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x)*log(x^2+10*x+25)^2+(2*x^3+2*x^2-42*x-10)*log(x^2+10*x+25)+x^4-3*x^3-28*x^2+64*x+40)*exp((
4*x*log(x^2+10*x+25)+4*x^2-16*x+2)/(x*log(x^2+10*x+25)+x^2-4*x))*exp(1/3*x*exp((4*x*log(x^2+10*x+25)+4*x^2-16*
x+2)/(x*log(x^2+10*x+25)+x^2-4*x))+4/5)/((3*x^2+15*x)*log(x^2+10*x+25)^2+(6*x^3+6*x^2-120*x)*log(x^2+10*x+25)+
3*x^4-9*x^3-72*x^2+240*x),x, algorithm="fricas")

[Out]

e^(1/15*(72*x^2 + 5*(x^3 + x^2*log(x^2 + 10*x + 25) - 4*x^2)*e^(2*(2*x^2 + 2*x*log(x^2 + 10*x + 25) - 8*x + 1)
/(x^2 + x*log(x^2 + 10*x + 25) - 4*x)) + 72*x*log(x^2 + 10*x + 25) - 288*x + 30)/(x^2 + x*log(x^2 + 10*x + 25)
- 4*x) - 2*(2*x^2 + 2*x*log(x^2 + 10*x + 25) - 8*x + 1)/(x^2 + x*log(x^2 + 10*x + 25) - 4*x))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x)*log(x^2+10*x+25)^2+(2*x^3+2*x^2-42*x-10)*log(x^2+10*x+25)+x^4-3*x^3-28*x^2+64*x+40)*exp((
4*x*log(x^2+10*x+25)+4*x^2-16*x+2)/(x*log(x^2+10*x+25)+x^2-4*x))*exp(1/3*x*exp((4*x*log(x^2+10*x+25)+4*x^2-16*
x+2)/(x*log(x^2+10*x+25)+x^2-4*x))+4/5)/((3*x^2+15*x)*log(x^2+10*x+25)^2+(6*x^3+6*x^2-120*x)*log(x^2+10*x+25)+
3*x^4-9*x^3-72*x^2+240*x),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.12, size = 49, normalized size = 1.29

method	result	size
risch	\({\mathrm e}^{\frac {x \,{\mathrm e}^{\frac {4 x \ln \left (x^{2}+10 x +25\right )+4 x^{2}-16 x +2}{x \left (\ln \left (x^{2}+10 x +25\right )+x -4\right )}}}{3}+\frac {4}{5}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+5*x)*ln(x^2+10*x+25)^2+(2*x^3+2*x^2-42*x-10)*ln(x^2+10*x+25)+x^4-3*x^3-28*x^2+64*x+40)*exp((4*x*ln(x
^2+10*x+25)+4*x^2-16*x+2)/(x*ln(x^2+10*x+25)+x^2-4*x))*exp(1/3*x*exp((4*x*ln(x^2+10*x+25)+4*x^2-16*x+2)/(x*ln(
x^2+10*x+25)+x^2-4*x))+4/5)/((3*x^2+15*x)*ln(x^2+10*x+25)^2+(6*x^3+6*x^2-120*x)*ln(x^2+10*x+25)+3*x^4-9*x^3-72
*x^2+240*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*x*exp(2*(2*x*ln(x^2+10*x+25)+2*x^2-8*x+1)/x/(ln(x^2+10*x+25)+x-4))+4/5)

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maxima [A]  time = 1.06, size = 49, normalized size = 1.29 \begin {gather*} e^{\left (\frac {1}{3} \, x e^{\left (-\frac {1}{x {\left (\log \left (x + 5\right ) - 2\right )} + 2 \, \log \left (x + 5\right )^{2} - 8 \, \log \left (x + 5\right ) + 8} + \frac {1}{x {\left (\log \left (x + 5\right ) - 2\right )}} + 4\right )} + \frac {4}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x)*log(x^2+10*x+25)^2+(2*x^3+2*x^2-42*x-10)*log(x^2+10*x+25)+x^4-3*x^3-28*x^2+64*x+40)*exp((
4*x*log(x^2+10*x+25)+4*x^2-16*x+2)/(x*log(x^2+10*x+25)+x^2-4*x))*exp(1/3*x*exp((4*x*log(x^2+10*x+25)+4*x^2-16*
x+2)/(x*log(x^2+10*x+25)+x^2-4*x))+4/5)/((3*x^2+15*x)*log(x^2+10*x+25)^2+(6*x^3+6*x^2-120*x)*log(x^2+10*x+25)+
3*x^4-9*x^3-72*x^2+240*x),x, algorithm="maxima")

[Out]

e^(1/3*x*e^(-1/(x*(log(x + 5) - 2) + 2*log(x + 5)^2 - 8*log(x + 5) + 8) + 1/(x*(log(x + 5) - 2)) + 4) + 4/5)

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mupad [B]  time = 3.72, size = 90, normalized size = 2.37 \begin {gather*} {\mathrm {e}}^{4/5}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-\frac {16}{x+\ln \left (x^2+10\,x+25\right )-4}}\,{\mathrm {e}}^{\frac {2}{x\,\ln \left (x^2+10\,x+25\right )-4\,x+x^2}}\,{\mathrm {e}}^{\frac {4\,x}{x+\ln \left (x^2+10\,x+25\right )-4}}\,{\left (x^2+10\,x+25\right )}^{\frac {4}{x+\ln \left (x^2+10\,x+25\right )-4}}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*x*log(10*x + x^2 + 25) - 16*x + 4*x^2 + 2)/(x*log(10*x + x^2 + 25) - 4*x + x^2))*exp((x*exp((4*x*l
og(10*x + x^2 + 25) - 16*x + 4*x^2 + 2)/(x*log(10*x + x^2 + 25) - 4*x + x^2)))/3 + 4/5)*(64*x + log(10*x + x^2
+ 25)^2*(5*x + x^2) - log(10*x + x^2 + 25)*(42*x - 2*x^2 - 2*x^3 + 10) - 28*x^2 - 3*x^3 + x^4 + 40))/(240*x +
log(10*x + x^2 + 25)*(6*x^2 - 120*x + 6*x^3) + log(10*x + x^2 + 25)^2*(15*x + 3*x^2) - 72*x^2 - 9*x^3 + 3*x^4
),x)

[Out]

exp(4/5)*exp((x*exp(-16/(x + log(10*x + x^2 + 25) - 4))*exp(2/(x*log(10*x + x^2 + 25) - 4*x + x^2))*exp((4*x)/
(x + log(10*x + x^2 + 25) - 4))*(10*x + x^2 + 25)^(4/(x + log(10*x + x^2 + 25) - 4)))/3)

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sympy [A]  time = 11.73, size = 51, normalized size = 1.34 \begin {gather*} e^{\frac {x e^{\frac {4 x^{2} + 4 x \log {\left (x^{2} + 10 x + 25 \right )} - 16 x + 2}{x^{2} + x \log {\left (x^{2} + 10 x + 25 \right )} - 4 x}}}{3} + \frac {4}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+5*x)*ln(x**2+10*x+25)**2+(2*x**3+2*x**2-42*x-10)*ln(x**2+10*x+25)+x**4-3*x**3-28*x**2+64*x+40
)*exp((4*x*ln(x**2+10*x+25)+4*x**2-16*x+2)/(x*ln(x**2+10*x+25)+x**2-4*x))*exp(1/3*x*exp((4*x*ln(x**2+10*x+25)+
4*x**2-16*x+2)/(x*ln(x**2+10*x+25)+x**2-4*x))+4/5)/((3*x**2+15*x)*ln(x**2+10*x+25)**2+(6*x**3+6*x**2-120*x)*ln
(x**2+10*x+25)+3*x**4-9*x**3-72*x**2+240*x),x)

[Out]

exp(x*exp((4*x**2 + 4*x*log(x**2 + 10*x + 25) - 16*x + 2)/(x**2 + x*log(x**2 + 10*x + 25) - 4*x))/3 + 4/5)

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