3.94.88 \(\int \frac {e^{\frac {e}{\log (\frac {3+e^{8+x}}{e^8})}} (e^{8+x} (-24 e-e^3 x)+(3 e^2+e^{10+x}) \log ^2(\frac {3+e^{8+x}}{e^8}))}{(3+e^{8+x}) \log ^2(\frac {3+e^{8+x}}{e^8})} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {e}{\log \left (\frac {3}{e^8}+e^x\right )}} \left (24+e^2 x\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 1, number of rules used = 1, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} \left (e^2 x+24\right ) e^{\frac {e}{\log \left (\frac {e^{x+8}+3}{e^8}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E/Log[(3 + E^(8 + x))/E^8])*(E^(8 + x)*(-24*E - E^3*x) + (3*E^2 + E^(10 + x))*Log[(3 + E^(8 + x))/E^8]
^2))/((3 + E^(8 + x))*Log[(3 + E^(8 + x))/E^8]^2),x]

[Out]

E^(E/Log[(3 + E^(8 + x))/E^8])*(24 + E^2*x)

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} &=e^{\frac {e}{\log \left (\frac {3+e^{8+x}}{e^8}\right )}} \left (24+e^2 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 24, normalized size = 1.00 \begin {gather*} e^{\frac {e}{-8+\log \left (3+e^{8+x}\right )}} \left (24+e^2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E/Log[(3 + E^(8 + x))/E^8])*(E^(8 + x)*(-24*E - E^3*x) + (3*E^2 + E^(10 + x))*Log[(3 + E^(8 + x)
)/E^8]^2))/((3 + E^(8 + x))*Log[(3 + E^(8 + x))/E^8]^2),x]

[Out]

E^(E/(-8 + Log[3 + E^(8 + x)]))*(24 + E^2*x)

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fricas [A]  time = 0.66, size = 26, normalized size = 1.08 \begin {gather*} {\left (x e^{2} + 24\right )} e^{\left (\frac {e}{\log \left ({\left (3 \, e^{2} + e^{\left (x + 10\right )}\right )} e^{\left (-10\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*exp(4)^2*exp(x)+3*exp(2))*log((exp(4)^2*exp(x)+3)/exp(4)^2)^2+(-x*exp(1)*exp(2)-24*exp(1))*
exp(4)^2*exp(x))*exp(exp(1)/log((exp(4)^2*exp(x)+3)/exp(4)^2))/(exp(4)^2*exp(x)+3)/log((exp(4)^2*exp(x)+3)/exp
(4)^2)^2,x, algorithm="fricas")

[Out]

(x*e^2 + 24)*e^(e/log((3*e^2 + e^(x + 10))*e^(-10)))

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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(((exp(2)*exp(4)^2*exp(x)+3*exp(2))*log((exp(4)^2*exp(x)+3)/exp(4)^2)^2+(-x*exp(1)*exp(2)-24*exp(1))*
exp(4)^2*exp(x))*exp(exp(1)/log((exp(4)^2*exp(x)+3)/exp(4)^2))/(exp(4)^2*exp(x)+3)/log((exp(4)^2*exp(x)+3)/exp
(4)^2)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 18.59Unable to divide, perhaps due to rounding error%%%{1,[1,17,9,10,0,80,1]%%%}+%%%{21,[1
,17,9,9,0,7

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maple [A]  time = 0.07, size = 24, normalized size = 1.00

method	result	size
risch	\(\left (24+{\mathrm e}^{2} x \right ) {\mathrm e}^{\frac {{\mathrm e}}{\ln \left (\left ({\mathrm e}^{x +8}+3\right ) {\mathrm e}^{-8}\right )}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)*exp(4)^2*exp(x)+3*exp(2))*ln((exp(4)^2*exp(x)+3)/exp(4)^2)^2+(-x*exp(1)*exp(2)-24*exp(1))*exp(4)^
2*exp(x))*exp(exp(1)/ln((exp(4)^2*exp(x)+3)/exp(4)^2))/(exp(4)^2*exp(x)+3)/ln((exp(4)^2*exp(x)+3)/exp(4)^2)^2,
x,method=_RETURNVERBOSE)

[Out]

(24+exp(2)*x)*exp(exp(1)/ln((exp(x+8)+3)*exp(-8)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x e^{\left (\frac {e}{\log \left (e^{\left (x + 8\right )} + 3\right ) - 8} + 2\right )} + 24 \, e^{\left (\frac {e}{\log \left (e^{\left (x + 8\right )} + 3\right ) - 8}\right )} + \int e^{\left (\frac {e}{\log \left (e^{\left (x + 8\right )} + 3\right ) - 8} + 2\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)*exp(4)^2*exp(x)+3*exp(2))*log((exp(4)^2*exp(x)+3)/exp(4)^2)^2+(-x*exp(1)*exp(2)-24*exp(1))*
exp(4)^2*exp(x))*exp(exp(1)/log((exp(4)^2*exp(x)+3)/exp(4)^2))/(exp(4)^2*exp(x)+3)/log((exp(4)^2*exp(x)+3)/exp
(4)^2)^2,x, algorithm="maxima")

[Out]

x*e^(e/(log(e^(x + 8) + 3) - 8) + 2) + 24*e^(e/(log(e^(x + 8) + 3) - 8)) + integrate(e^(e/(log(e^(x + 8) + 3)
- 8) + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {\mathrm {e}}{\ln \left ({\mathrm {e}}^{-8}\,\left ({\mathrm {e}}^8\,{\mathrm {e}}^x+3\right )\right )}}\,\left ({\ln \left ({\mathrm {e}}^{-8}\,\left ({\mathrm {e}}^8\,{\mathrm {e}}^x+3\right )\right )}^2\,\left ({\mathrm {e}}^{x+10}+3\,{\mathrm {e}}^2\right )-{\mathrm {e}}^{x+8}\,\left (24\,\mathrm {e}+x\,{\mathrm {e}}^3\right )\right )}{{\ln \left ({\mathrm {e}}^{-8}\,\left ({\mathrm {e}}^8\,{\mathrm {e}}^x+3\right )\right )}^2\,\left ({\mathrm {e}}^{x+8}+3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(1)/log(exp(-8)*(exp(8)*exp(x) + 3)))*(log(exp(-8)*(exp(8)*exp(x) + 3))^2*(3*exp(2) + exp(10)*exp(
x)) - exp(8)*exp(x)*(24*exp(1) + x*exp(3))))/(log(exp(-8)*(exp(8)*exp(x) + 3))^2*(exp(8)*exp(x) + 3)),x)

[Out]

int((exp(exp(1)/log(exp(-8)*(exp(8)*exp(x) + 3)))*(log(exp(-8)*(exp(8)*exp(x) + 3))^2*(exp(x + 10) + 3*exp(2))
- exp(x + 8)*(24*exp(1) + x*exp(3))))/(log(exp(-8)*(exp(8)*exp(x) + 3))^2*(exp(x + 8) + 3)), 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(((exp(2)*exp(4)**2*exp(x)+3*exp(2))*ln((exp(4)**2*exp(x)+3)/exp(4)**2)**2+(-x*exp(1)*exp(2)-24*exp(1
))*exp(4)**2*exp(x))*exp(exp(1)/ln((exp(4)**2*exp(x)+3)/exp(4)**2))/(exp(4)**2*exp(x)+3)/ln((exp(4)**2*exp(x)+
3)/exp(4)**2)**2,x)

[Out]

Timed out

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