3.22.68 \(\int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx\)

Optimal. Leaf size=29 \[ \frac {7}{10 \left (e^{3-e^2-e^{x+\log ^2(3)} x}+\log (3)\right )} \]

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Rubi [A]  time = 1.18, antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {7 e^3}{10 \log (3) \left (\log (3) e^{x e^{x+\log ^2(3)}+e^2}+e^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 - E^2 + x - E^(x + Log[3]^2)*x + Log[3]^2)*(7 + 7*x))/(10*E^(6 - 2*E^2 - 2*E^(x + Log[3]^2)*x) + 20*
E^(3 - E^2 - E^(x + Log[3]^2)*x)*Log[3] + 10*Log[3]^2),x]

[Out]

(-7*E^3)/(10*Log[3]*(E^3 + E^(E^2 + E^(x + Log[3]^2)*x)*Log[3]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {7 \exp \left (x+e^{x+\log ^2(3)} x+3 \left (1+\frac {1}{3} \left (e^2+\log ^2(3)\right )\right )\right ) (1+x)}{10 \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )^2} \, dx\\ &=\frac {7}{10} \int \frac {\exp \left (x+e^{x+\log ^2(3)} x+3 \left (1+\frac {1}{3} \left (e^2+\log ^2(3)\right )\right )\right ) (1+x)}{\left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )^2} \, dx\\ &=-\frac {7 e^3}{10 \log (3) \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 36, normalized size = 1.24 \begin {gather*} -\frac {7 e^3}{10 \log (3) \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 - E^2 + x - E^(x + Log[3]^2)*x + Log[3]^2)*(7 + 7*x))/(10*E^(6 - 2*E^2 - 2*E^(x + Log[3]^2)*x)
+ 20*E^(3 - E^2 - E^(x + Log[3]^2)*x)*Log[3] + 10*Log[3]^2),x]

[Out]

(-7*E^3)/(10*Log[3]*(E^3 + E^(E^2 + E^(x + Log[3]^2)*x)*Log[3]))

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fricas [A]  time = 1.05, size = 44, normalized size = 1.52 \begin {gather*} \frac {7 \, e^{\left (\log \relax (3)^{2} + x\right )}}{10 \, {\left (e^{\left (\log \relax (3)^{2} + x\right )} \log \relax (3) + e^{\left (-x e^{\left (\log \relax (3)^{2} + x\right )} + \log \relax (3)^{2} + x - e^{2} + 3\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp(-x*exp(log(3)^2+x)-exp(2)+3)^2+20*l
og(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3)+10*log(3)^2),x, algorithm="fricas")

[Out]

7/10*e^(log(3)^2 + x)/(e^(log(3)^2 + x)*log(3) + e^(-x*e^(log(3)^2 + x) + log(3)^2 + x - e^2 + 3))

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giac [A]  time = 0.24, size = 39, normalized size = 1.34 \begin {gather*} -\frac {7 \, e^{\left (-e^{2} + 6\right )}}{10 \, {\left (e^{\left (x e^{\left (\log \relax (3)^{2} + x\right )} + 3\right )} \log \relax (3)^{2} + e^{\left (-e^{2} + 6\right )} \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp(-x*exp(log(3)^2+x)-exp(2)+3)^2+20*l
og(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3)+10*log(3)^2),x, algorithm="giac")

[Out]

-7/10*e^(-e^2 + 6)/(e^(x*e^(log(3)^2 + x) + 3)*log(3)^2 + e^(-e^2 + 6)*log(3))

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maple [A]  time = 0.24, size = 25, normalized size = 0.86

method	result	size
norman	\(\frac {7}{10 \left (\ln \relax (3)+{\mathrm e}^{-x \,{\mathrm e}^{\ln \relax (3)^{2}+x}-{\mathrm e}^{2}+3}\right )}\)	\(25\)
risch	\(\frac {7}{10 \left (\ln \relax (3)+{\mathrm e}^{-x \,{\mathrm e}^{\ln \relax (3)^{2}+x}-{\mathrm e}^{2}+3}\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x+7)*exp(ln(3)^2+x)*exp(-x*exp(ln(3)^2+x)-exp(2)+3)/(10*exp(-x*exp(ln(3)^2+x)-exp(2)+3)^2+20*ln(3)*exp(
-x*exp(ln(3)^2+x)-exp(2)+3)+10*ln(3)^2),x,method=_RETURNVERBOSE)

[Out]

7/10/(ln(3)+exp(-x*exp(ln(3)^2+x)-exp(2)+3))

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maxima [A]  time = 0.76, size = 30, normalized size = 1.03 \begin {gather*} -\frac {7 \, e^{3}}{10 \, {\left (e^{\left (x e^{\left (\log \relax (3)^{2} + x\right )} + e^{2}\right )} \log \relax (3)^{2} + e^{3} \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp(-x*exp(log(3)^2+x)-exp(2)+3)^2+20*l
og(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3)+10*log(3)^2),x, algorithm="maxima")

[Out]

-7/10*e^3/(e^(x*e^(log(3)^2 + x) + e^2)*log(3)^2 + e^3*log(3))

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mupad [B]  time = 0.17, size = 29, normalized size = 1.00 \begin {gather*} \frac {7}{10\,\left (\ln \relax (3)+{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{{\ln \relax (3)}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(3)^2)*exp(3 - x*exp(x + log(3)^2) - exp(2))*(7*x + 7))/(10*exp(6 - 2*x*exp(x + log(3)^2) - 2*
exp(2)) + 20*exp(3 - x*exp(x + log(3)^2) - exp(2))*log(3) + 10*log(3)^2),x)

[Out]

7/(10*(log(3) + exp(-exp(2))*exp(-x*exp(log(3)^2)*exp(x))*exp(3)))

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sympy [A]  time = 0.18, size = 24, normalized size = 0.83 \begin {gather*} \frac {7}{10 e^{- x e^{x + \log {\relax (3 )}^{2}} - e^{2} + 3} + 10 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((7*x+7)*exp(ln(3)**2+x)*exp(-x*exp(ln(3)**2+x)-exp(2)+3)/(10*exp(-x*exp(ln(3)**2+x)-exp(2)+3)**2+20*
ln(3)*exp(-x*exp(ln(3)**2+x)-exp(2)+3)+10*ln(3)**2),x)

[Out]

7/(10*exp(-x*exp(x + log(3)**2) - exp(2) + 3) + 10*log(3))

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