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

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

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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 = {} $$\begin{array}{c}\text{\$Aborted}\end{array}$$

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 $$\begin{array}{c}{e}^{\frac{4}{5}+\frac{1}{3}{e}^{\frac{2(1-8x+2x^2)}{x(-4+x+\log ((5+x{)}^2))}}x{((5+x{)}^2)}^{\frac{4}{-4+x+\log ((5+x{)}^2)}}}\end{array}$$

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 $$\begin{array}{c}{e}^{(\frac{72x^2+5(x^3+x^2\log (x^2+10x+25)-4x^2)e^{\left(\frac{2(2x^2+2x\log (x^2+10x+25)-8x+1)}{x^2+x\log (x^2+10x+25)-4x}\right)}+72x\log (x^2+10x+25)-288x+30}{15(x^2+x\log (x^2+10x+25)-4x)}-\frac{2(2x^2+2x\log (x^2+10x+25)-8x+1)}{x^2+x\log (x^2+10x+25)-4x})}\end{array}$$

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 $$\begin{array}{c}\text{Timed out}\end{array}$$

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	${\mathrm{e}}^{\frac{x{\mathrm{e}}^{\frac{4x\ln (x^2+10x+25)+4x^2-16x+2}{x(\ln (x^2+10x+25)+x-4)}}}{3}+\frac{4}{5}}$	$49$

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 $$\begin{array}{c}{e}^{(\frac{1}{3}x{e}^{(-\frac{1}{x(\log (x+5)-2)+2\log {(x+5)}^2-8\log (x+5)+8}+\frac{1}{x(\log (x+5)-2)}+4)}+\frac{4}{5})}\end{array}$$

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 $$\begin{array}{c}{\mathrm{e}}^{4/5}{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-\frac{16}{x+\ln (x^2+10x+25)-4}}{\mathrm{e}}^{\frac{2}{x\ln (x^2+10x+25)-4x+x^2}}{\mathrm{e}}^{\frac{4x}{x+\ln (x^2+10x+25)-4}}{(x^2+10x+25)}^{\frac{4}{x+\ln (x^2+10x+25)-4}}}{3}}\end{array}$$

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 $$\begin{array}{c}{e}^{\frac{x{e}^{\frac{4x^2+4x\log (x^2+10x+25)-16x+2}{x^2+x\log (x^2+10x+25)-4x}}}{3}+\frac{4}{5}}\end{array}$$

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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