3.37.91 \(\int \frac {e^{\frac {x}{\log (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x)}} ((25 x^2+10 x^3+x^4) \log ^2(x)+e^{\frac {7}{(5+x) \log (x)}} (35 x+7 x^2+7 x^2 \log (x))+\log ^2(-e^2+e^{\frac {7}{(5+x) \log (x)}}-x) (e^{\frac {7}{(5+x) \log (x)}} (-25-10 x-x^2) \log ^2(x)+(25 x+10 x^2+x^3+e^2 (25+10 x+x^2)) \log ^2(x))+\log (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x) (e^{\frac {7}{(5+x) \log (x)}} (25 x+10 x^2+x^3) \log ^2(x)+(-25 x^2-10 x^3-x^4+e^2 (-25 x-10 x^2-x^3)) \log ^2(x)))}{\log ^2(-e^2+e^{\frac {7}{(5+x) \log (x)}}-x) (e^{\frac {7}{(5+x) \log (x)}} (25 x^2+10 x^3+x^4) \log ^2(x)+(-25 x^3-10 x^4-x^5+e^2 (-25 x^2-10 x^3-x^4)) \log ^2(x))} \, dx\)
Optimal. Leaf size=33 \[ \frac {e^{\frac {x}{\log \left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right )}}}{x} \]
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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 {gather*} \frac {e^{\frac {x}{\log \left (-x+e^{\frac {7}{(x+5) \log (x)}}-e^2\right )}} \left (7 e^{\frac {7}{(x+5) \log (x)}} \left (x^2+x^2 \log (x)+5 x\right )+\left (x^4+10 x^3+25 x^2\right ) \log ^2(x)+\log \left (-x+e^{\frac {7}{(x+5) \log (x)}}-e^2\right ) \left (\left (x^3+10 x^2+25 x\right ) e^{\frac {7}{(x+5) \log (x)}} \log ^2(x)-\left (x^4+10 x^3+25 x^2+e^2 \left (x^3+10 x^2+25 x\right )\right ) \log ^2(x)\right )\right )}{\log ^2\left (-x+e^{\frac {7}{(x+5) \log (x)}}-e^2\right ) \left (\frac {1}{\log \left (-x+e^{\frac {7}{(x+5) \log (x)}}-e^2\right )}-\frac {x \left (7 e^{\frac {7}{(x+5) \log (x)}} \left (\frac {1}{x (x+5) \log ^2(x)}+\frac {1}{(x+5)^2 \log (x)}\right )+1\right )}{\left (x-e^{\frac {7}{(x+5) \log (x)}}+e^2\right ) \log ^2\left (-x+e^{\frac {7}{(x+5) \log (x)}}-e^2\right )}\right ) \left (\left (x^4+10 x^3+25 x^2\right ) e^{\frac {7}{(x+5) \log (x)}} \log ^2(x)-\left (x^5+10 x^4+25 x^3+e^2 \left (x^4+10 x^3+25 x^2\right )\right ) \log ^2(x)\right )} \end {gather*}
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 {gather*} \begin {aligned} \text {integral} &=\frac {e^{\frac {x}{\log \left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right )}} \left (\left (25 x^2+10 x^3+x^4\right ) \log ^2(x)+7 e^{\frac {7}{(5+x) \log (x)}} \left (5 x+x^2+x^2 \log (x)\right )+\log \left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right ) \left (e^{\frac {7}{(5+x) \log (x)}} \left (25 x+10 x^2+x^3\right ) \log ^2(x)-\left (25 x^2+10 x^3+x^4+e^2 \left (25 x+10 x^2+x^3\right )\right ) \log ^2(x)\right )\right )}{\log ^2\left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right ) \left (\frac {1}{\log \left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right )}-\frac {x \left (1+7 e^{\frac {7}{(5+x) \log (x)}} \left (\frac {1}{x (5+x) \log ^2(x)}+\frac {1}{(5+x)^2 \log (x)}\right )\right )}{\left (e^2-e^{\frac {7}{(5+x) \log (x)}}+x\right ) \log ^2\left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right )}\right ) \left (e^{\frac {7}{(5+x) \log (x)}} \left (25 x^2+10 x^3+x^4\right ) \log ^2(x)-\left (25 x^3+10 x^4+x^5+e^2 \left (25 x^2+10 x^3+x^4\right )\right ) \log ^2(x)\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.82, size = 33, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {x}{\log \left (-e^2+e^{\frac {7}{(5+x) \log (x)}}-x\right )}}}{x} \end {gather*}
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 {gather*} \frac {e^{\left (\frac {x}{\log \left (-x - e^{2} + e^{\left (\frac {7}{{\left (x + 5\right )} \log \relax (x)}\right )}\right )}\right )}}{x} \end {gather*}
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 {gather*} \text {Exception raised: TypeError} \end {gather*}
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
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\(\frac {{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{\frac {7}{\ln \relax (x ) \left (5+x \right )}}-{\mathrm e}^{2}-x \right )}}}{x}\) |
\(31\) |
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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 {gather*} \text {Exception raised: RuntimeError} \end {gather*}
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 {gather*} \frac {{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^{\frac {7}{5\,\ln \relax (x)+x\,\ln \relax (x)}}-{\mathrm {e}}^2-x\right )}}}{x} \end {gather*}
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 {gather*} \text {Timed out} \end {gather*}
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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