3.37.91 $\int \frac{e^{\frac{x}{\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)}}((25x^2+10x^3+x^4){\log }^2(x)+e^{\frac{7}{(5+x)\log (x)}}(35x+7x^2+7x^2\log (x))+{\log }^2(-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)(e^{\frac{7}{(5+x)\log (x)}}(-25-10x-x^2){\log }^2(x)+(25x+10x^2+x^3+e^2(25+10x+x^2)){\log }^2(x))+\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)(e^{\frac{7}{(5+x)\log (x)}}(25x+10x^2+x^3){\log }^2(x)+(-25x^2-10x^3-x^4+e^2(-25x-10x^2-x^3)){\log }^2(x)))}{{\log }^2(-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)(e^{\frac{7}{(5+x)\log (x)}}(25x^2+10x^3+x^4){\log }^2(x)+(-25x^3-10x^4-x^5+e^2(-25x^2-10x^3-x^4)){\log }^2(x))}dx$

Optimal. Leaf size=33 $$\frac{e^{\frac{x}{\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)}}}{x}$$

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Rubi [B]  time = 6.47, antiderivative size = 386, normalized size of antiderivative = 11.70, number of steps used = 1, number of rules used = 1, integrand size = 359, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.003, Rules used = {2288} $$\begin{array}{c}\frac{e^{\frac{x}{\log (-x+e^{\frac{7}{(x+5)\log (x)}}-e^2)}}(7e^{\frac{7}{(x+5)\log (x)}}(x^2+x^2\log (x)+5x)+(x^4+10x^3+25x^2){\log }^2(x)+\log (-x+e^{\frac{7}{(x+5)\log (x)}}-e^2)((x^3+10x^2+25x)e^{\frac{7}{(x+5)\log (x)}}{\log }^2(x)-(x^4+10x^3+25x^2+e^2(x^3+10x^2+25x)){\log }^2(x)))}{{\log }^2(-x+e^{\frac{7}{(x+5)\log (x)}}-e^2)(\frac{1}{\log (-x+e^{\frac{7}{(x+5)\log (x)}}-e^2)}-\frac{x(7e^{\frac{7}{(x+5)\log (x)}}(\frac{1}{x(x+5){\log }^2(x)}+\frac{1}{(x+5{)}^2\log (x)})+1)}{(x-e^{\frac{7}{(x+5)\log (x)}}+e^2){\log }^2(-x+e^{\frac{7}{(x+5)\log (x)}}-e^2)})((x^4+10x^3+25x^2)e^{\frac{7}{(x+5)\log (x)}}{\log }^2(x)-(x^5+10x^4+25x^3+e^2(x^4+10x^3+25x^2)){\log }^2(x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(x/Log[-E^2 + E^(7/((5 + x)*Log[x])) - x])*((25*x^2 + 10*x^3 + x^4)*Log[x]^2 + E^(7/((5 + x)*Log[x]))*(
35*x + 7*x^2 + 7*x^2*Log[x]) + Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^2*(E^(7/((5 + x)*Log[x]))*(-25 - 10*x -
x^2)*Log[x]^2 + (25*x + 10*x^2 + x^3 + E^2*(25 + 10*x + x^2))*Log[x]^2) + Log[-E^2 + E^(7/((5 + x)*Log[x])) -
x]*(E^(7/((5 + x)*Log[x]))*(25*x + 10*x^2 + x^3)*Log[x]^2 + (-25*x^2 - 10*x^3 - x^4 + E^2*(-25*x - 10*x^2 - x^
3))*Log[x]^2)))/(Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^2*(E^(7/((5 + x)*Log[x]))*(25*x^2 + 10*x^3 + x^4)*Log[
x]^2 + (-25*x^3 - 10*x^4 - x^5 + E^2*(-25*x^2 - 10*x^3 - x^4))*Log[x]^2)),x]

[Out]

(E^(x/Log[-E^2 + E^(7/((5 + x)*Log[x])) - x])*((25*x^2 + 10*x^3 + x^4)*Log[x]^2 + 7*E^(7/((5 + x)*Log[x]))*(5*
x + x^2 + x^2*Log[x]) + Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]*(E^(7/((5 + x)*Log[x]))*(25*x + 10*x^2 + x^3)*L
og[x]^2 - (25*x^2 + 10*x^3 + x^4 + E^2*(25*x + 10*x^2 + x^3))*Log[x]^2)))/(Log[-E^2 + E^(7/((5 + x)*Log[x])) -
x]^2*(Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^(-1) - (x*(1 + 7*E^(7/((5 + x)*Log[x]))*(1/(x*(5 + x)*Log[x]^2)
+ 1/((5 + x)^2*Log[x]))))/((E^2 - E^(7/((5 + x)*Log[x])) + x)*Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^2))*(E^(7
/((5 + x)*Log[x]))*(25*x^2 + 10*x^3 + x^4)*Log[x]^2 - (25*x^3 + 10*x^4 + x^5 + E^2*(25*x^2 + 10*x^3 + x^4))*Lo
g[x]^2))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{e^{\frac{x}{\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)}}((25x^2+10x^3+x^4){\log }^2(x)+7e^{\frac{7}{(5+x)\log (x)}}(5x+x^2+x^2\log (x))+\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)(e^{\frac{7}{(5+x)\log (x)}}(25x+10x^2+x^3){\log }^2(x)-(25x^2+10x^3+x^4+e^2(25x+10x^2+x^3)){\log }^2(x)))}{{\log }^2(-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)(\frac{1}{\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)}-\frac{x(1+7e^{\frac{7}{(5+x)\log (x)}}(\frac{1}{x(5+x){\log }^2(x)}+\frac{1}{(5+x{)}^2\log (x)}))}{(e^2-e^{\frac{7}{(5+x)\log (x)}}+x){\log }^2(-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)})(e^{\frac{7}{(5+x)\log (x)}}(25x^2+10x^3+x^4){\log }^2(x)-(25x^3+10x^4+x^5+e^2(25x^2+10x^3+x^4)){\log }^2(x))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.82, size = 33, normalized size = 1.00 $$\begin{array}{c}\frac{e^{\frac{x}{\log (-e^2+e^{\frac{7}{(5+x)\log (x)}}-x)}}}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/Log[-E^2 + E^(7/((5 + x)*Log[x])) - x])*((25*x^2 + 10*x^3 + x^4)*Log[x]^2 + E^(7/((5 + x)*Log[
x]))*(35*x + 7*x^2 + 7*x^2*Log[x]) + Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^2*(E^(7/((5 + x)*Log[x]))*(-25 - 1
0*x - x^2)*Log[x]^2 + (25*x + 10*x^2 + x^3 + E^2*(25 + 10*x + x^2))*Log[x]^2) + Log[-E^2 + E^(7/((5 + x)*Log[x
])) - x]*(E^(7/((5 + x)*Log[x]))*(25*x + 10*x^2 + x^3)*Log[x]^2 + (-25*x^2 - 10*x^3 - x^4 + E^2*(-25*x - 10*x^
2 - x^3))*Log[x]^2)))/(Log[-E^2 + E^(7/((5 + x)*Log[x])) - x]^2*(E^(7/((5 + x)*Log[x]))*(25*x^2 + 10*x^3 + x^4
)*Log[x]^2 + (-25*x^3 - 10*x^4 - x^5 + E^2*(-25*x^2 - 10*x^3 - x^4))*Log[x]^2)),x]

[Out]

E^(x/Log[-E^2 + E^(7/((5 + x)*Log[x])) - x])/x

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fricas [A]  time = 0.85, size = 30, normalized size = 0.91 $$\begin{array}{c}\frac{e^{\left(\frac{x}{\log (-x-e^2+e^{\left(\frac{7}{(x+5)\log (x)}\right)})}\right)}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-10*x-25)*log(x)^2*exp(7/log(x)/(5+x))+((x^2+10*x+25)*exp(2)+x^3+10*x^2+25*x)*log(x)^2)*log(e
xp(7/log(x)/(5+x))-exp(2)-x)^2+((x^3+10*x^2+25*x)*log(x)^2*exp(7/log(x)/(5+x))+((-x^3-10*x^2-25*x)*exp(2)-x^4-
10*x^3-25*x^2)*log(x)^2)*log(exp(7/log(x)/(5+x))-exp(2)-x)+(7*x^2*log(x)+7*x^2+35*x)*exp(7/log(x)/(5+x))+(x^4+
10*x^3+25*x^2)*log(x)^2)*exp(x/log(exp(7/log(x)/(5+x))-exp(2)-x))/((x^4+10*x^3+25*x^2)*log(x)^2*exp(7/log(x)/(
5+x))+((-x^4-10*x^3-25*x^2)*exp(2)-x^5-10*x^4-25*x^3)*log(x)^2)/log(exp(7/log(x)/(5+x))-exp(2)-x)^2,x, algorit
hm="fricas")

[Out]

e^(x/log(-x - e^2 + e^(7/((x + 5)*log(x)))))/x

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Exception raised: TypeError}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-10*x-25)*log(x)^2*exp(7/log(x)/(5+x))+((x^2+10*x+25)*exp(2)+x^3+10*x^2+25*x)*log(x)^2)*log(e
xp(7/log(x)/(5+x))-exp(2)-x)^2+((x^3+10*x^2+25*x)*log(x)^2*exp(7/log(x)/(5+x))+((-x^3-10*x^2-25*x)*exp(2)-x^4-
10*x^3-25*x^2)*log(x)^2)*log(exp(7/log(x)/(5+x))-exp(2)-x)+(7*x^2*log(x)+7*x^2+35*x)*exp(7/log(x)/(5+x))+(x^4+
10*x^3+25*x^2)*log(x)^2)*exp(x/log(exp(7/log(x)/(5+x))-exp(2)-x))/((x^4+10*x^3+25*x^2)*log(x)^2*exp(7/log(x)/(
5+x))+((-x^4-10*x^3-25*x^2)*exp(2)-x^5-10*x^4-25*x^3)*log(x)^2)/log(exp(7/log(x)/(5+x))-exp(2)-x)^2,x, algorit
hm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 5.48Unable to divide, perhaps due to rounding error%%%{117649,[0,20]%%%}+%%%{9411920,[0,19
]%%%}+%%%{3

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maple [A]  time = 0.08, size = 31, normalized size = 0.94

method	result	size
risch	$\frac{{\mathrm{e}}^{\frac{x}{\ln ({\mathrm{e}}^{\frac{7}{\ln (x)(5+x)}}-{\mathrm{e}}^2-x)}}}{x}$	$31$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2-10*x-25)*ln(x)^2*exp(7/ln(x)/(5+x))+((x^2+10*x+25)*exp(2)+x^3+10*x^2+25*x)*ln(x)^2)*ln(exp(7/ln(x)
/(5+x))-exp(2)-x)^2+((x^3+10*x^2+25*x)*ln(x)^2*exp(7/ln(x)/(5+x))+((-x^3-10*x^2-25*x)*exp(2)-x^4-10*x^3-25*x^2
)*ln(x)^2)*ln(exp(7/ln(x)/(5+x))-exp(2)-x)+(7*x^2*ln(x)+7*x^2+35*x)*exp(7/ln(x)/(5+x))+(x^4+10*x^3+25*x^2)*ln(
x)^2)*exp(x/ln(exp(7/ln(x)/(5+x))-exp(2)-x))/((x^4+10*x^3+25*x^2)*ln(x)^2*exp(7/ln(x)/(5+x))+((-x^4-10*x^3-25*
x^2)*exp(2)-x^5-10*x^4-25*x^3)*ln(x)^2)/ln(exp(7/ln(x)/(5+x))-exp(2)-x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(x/ln(exp(7/ln(x)/(5+x))-exp(2)-x))/x

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Exception raised: RuntimeError}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-10*x-25)*log(x)^2*exp(7/log(x)/(5+x))+((x^2+10*x+25)*exp(2)+x^3+10*x^2+25*x)*log(x)^2)*log(e
xp(7/log(x)/(5+x))-exp(2)-x)^2+((x^3+10*x^2+25*x)*log(x)^2*exp(7/log(x)/(5+x))+((-x^3-10*x^2-25*x)*exp(2)-x^4-
10*x^3-25*x^2)*log(x)^2)*log(exp(7/log(x)/(5+x))-exp(2)-x)+(7*x^2*log(x)+7*x^2+35*x)*exp(7/log(x)/(5+x))+(x^4+
10*x^3+25*x^2)*log(x)^2)*exp(x/log(exp(7/log(x)/(5+x))-exp(2)-x))/((x^4+10*x^3+25*x^2)*log(x)^2*exp(7/log(x)/(
5+x))+((-x^4-10*x^3-25*x^2)*exp(2)-x^5-10*x^4-25*x^3)*log(x)^2)/log(exp(7/log(x)/(5+x))-exp(2)-x)^2,x, algorit
hm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 3.62, size = 32, normalized size = 0.97 $$\begin{array}{c}\frac{{\mathrm{e}}^{\frac{x}{\ln ({\mathrm{e}}^{\frac{7}{5\ln (x)+x\ln (x)}}-{\mathrm{e}}^2-x)}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x/log(exp(7/(log(x)*(x + 5))) - x - exp(2)))*(exp(7/(log(x)*(x + 5)))*(35*x + 7*x^2*log(x) + 7*x^2)
+ log(x)^2*(25*x^2 + 10*x^3 + x^4) + log(exp(7/(log(x)*(x + 5))) - x - exp(2))^2*(log(x)^2*(25*x + exp(2)*(10*
x + x^2 + 25) + 10*x^2 + x^3) - exp(7/(log(x)*(x + 5)))*log(x)^2*(10*x + x^2 + 25)) - log(exp(7/(log(x)*(x + 5
))) - x - exp(2))*(log(x)^2*(exp(2)*(25*x + 10*x^2 + x^3) + 25*x^2 + 10*x^3 + x^4) - exp(7/(log(x)*(x + 5)))*l
og(x)^2*(25*x + 10*x^2 + x^3))))/(log(exp(7/(log(x)*(x + 5))) - x - exp(2))^2*(log(x)^2*(exp(2)*(25*x^2 + 10*x
^3 + x^4) + 25*x^3 + 10*x^4 + x^5) - exp(7/(log(x)*(x + 5)))*log(x)^2*(25*x^2 + 10*x^3 + x^4))),x)

[Out]

exp(x/log(exp(7/(5*log(x) + x*log(x))) - exp(2) - x))/x

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sympy [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**2-10*x-25)*ln(x)**2*exp(7/ln(x)/(5+x))+((x**2+10*x+25)*exp(2)+x**3+10*x**2+25*x)*ln(x)**2)*ln
(exp(7/ln(x)/(5+x))-exp(2)-x)**2+((x**3+10*x**2+25*x)*ln(x)**2*exp(7/ln(x)/(5+x))+((-x**3-10*x**2-25*x)*exp(2)
-x**4-10*x**3-25*x**2)*ln(x)**2)*ln(exp(7/ln(x)/(5+x))-exp(2)-x)+(7*x**2*ln(x)+7*x**2+35*x)*exp(7/ln(x)/(5+x))
+(x**4+10*x**3+25*x**2)*ln(x)**2)*exp(x/ln(exp(7/ln(x)/(5+x))-exp(2)-x))/((x**4+10*x**3+25*x**2)*ln(x)**2*exp(
7/ln(x)/(5+x))+((-x**4-10*x**3-25*x**2)*exp(2)-x**5-10*x**4-25*x**3)*ln(x)**2)/ln(exp(7/ln(x)/(5+x))-exp(2)-x)
**2,x)

[Out]

Timed out

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