3.79.27 \(\int \frac {e^2 (-e x-x^2)+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} (x+e^2 (e+x))}{e^2 (e x+x^2)} \, dx\)

Optimal. Leaf size=16 \[ -x+5 x (e+x)^{\frac {1}{e^2}} \log (4) \]

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Rubi [A]  time = 0.37, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 1593, 6688, 34} \begin {gather*} 5 x (x+e)^{\frac {1}{e^2}} \log (4)-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-(E*x) - x^2) + E^((Log[E + x] + E^2*Log[5*x*Log[4]])/E^2)*(x + E^2*(E + x)))/(E^2*(E*x + x^2)),x]

[Out]

-x + 5*x*(E + x)^E^(-2)*Log[4]

Rule 12

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

Rule 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=\frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e x+x^2} \, dx}{e^2}\\ &=\frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{x (e+x)} \, dx}{e^2}\\ &=\frac {\int \left (-e^2+5 (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \log (4)\right ) \, dx}{e^2}\\ &=-x+\frac {(5 \log (4)) \int (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \, dx}{e^2}\\ &=-x+5 x (e+x)^{\frac {1}{e^2}} \log (4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 15, normalized size = 0.94 \begin {gather*} x \left (-1+5 (e+x)^{\frac {1}{e^2}} \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-(E*x) - x^2) + E^((Log[E + x] + E^2*Log[5*x*Log[4]])/E^2)*(x + E^2*(E + x)))/(E^2*(E*x + x^2)
),x]

[Out]

x*(-1 + 5*(E + x)^E^(-2)*Log[4])

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fricas [A]  time = 0.89, size = 23, normalized size = 1.44 \begin {gather*} -x + e^{\left ({\left (e^{2} \log \left (10 \, x \log \relax (2)\right ) + \log \left (x + e\right )\right )} e^{\left (-2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+exp(1))*exp(2)+x)*exp((exp(2)*log(10*x*log(2))+log(x+exp(1)))/exp(2))+(-x*exp(1)-x^2)*exp(2))/(
x*exp(1)+x^2)/exp(2),x, algorithm="fricas")

[Out]

-x + e^((e^2*log(10*x*log(2)) + log(x + e))*e^(-2))

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giac [A]  time = 0.22, size = 23, normalized size = 1.44 \begin {gather*} {\left (10 \, {\left (x + e\right )}^{e^{\left (-2\right )}} x e^{2} \log \relax (2) - x e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+exp(1))*exp(2)+x)*exp((exp(2)*log(10*x*log(2))+log(x+exp(1)))/exp(2))+(-x*exp(1)-x^2)*exp(2))/(
x*exp(1)+x^2)/exp(2),x, algorithm="giac")

[Out]

(10*(x + e)^e^(-2)*x*e^2*log(2) - x*e^2)*e^(-2)

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maple [A]  time = 0.23, size = 17, normalized size = 1.06

method	result	size
risch	\(-x +10 x \ln \relax (2) \left (x +{\mathrm e}\right )^{{\mathrm e}^{-2}}\)	\(17\)
norman	\(-x +{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \relax (2)\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}\)	\(26\)
default	\({\mathrm e}^{-2} \left ({\mathrm e}^{2} {\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \relax (2)\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}-{\mathrm e}^{2} x \right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+exp(1))*exp(2)+x)*exp((exp(2)*ln(10*x*ln(2))+ln(x+exp(1)))/exp(2))+(-x*exp(1)-x^2)*exp(2))/(x*exp(1)+
x^2)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-x+10*x*ln(2)*(x+exp(1))^exp(-2)

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maxima [B]  time = 0.38, size = 173, normalized size = 10.81 \begin {gather*} {\left ({\left (e \log \left (x + e\right ) - x\right )} e^{2} + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 4\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 7\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1}\right )} \log \relax (2) + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 2\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 3\right )}}{e^{2} + 1}\right )} \log \relax (2) + 10 \, e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )} \log \relax (2) - e^{3} \log \left (x + e\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+exp(1))*exp(2)+x)*exp((exp(2)*log(10*x*log(2))+log(x+exp(1)))/exp(2))+(-x*exp(1)-x^2)*exp(2))/(
x*exp(1)+x^2)/exp(2),x, algorithm="maxima")

[Out]

((e*log(x + e) - x)*e^2 + 10*((x + e)*e^(e^(-2)*log(x + e) + 4)/(e^2 + 1) - e^(e^(-2)*log(x + e) + 7)/(e^2 + 1
) - e^(e^(-2)*log(x + e) + 5)/(e^2 + 1))*log(2) + 10*((x + e)*e^(e^(-2)*log(x + e) + 2)/(e^2 + 1) - e^(e^(-2)*
log(x + e) + 5)/(e^2 + 1) - e^(e^(-2)*log(x + e) + 3)/(e^2 + 1))*log(2) + 10*e^(e^(-2)*log(x + e) + 5)*log(2)
- e^3*log(x + e))*e^(-2)

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mupad [B]  time = 7.95, size = 15, normalized size = 0.94 \begin {gather*} x\,\left (10\,\ln \relax (2)\,{\left (x+\mathrm {e}\right )}^{{\mathrm {e}}^{-2}}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(10*log(2)*(x + exp(1))^exp(-2) - 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \left (- \frac {e^{3}}{x + e}\right )\, dx + \int \left (- \frac {x e^{2}}{x + e}\right )\, dx + \int \frac {10 x \left (x + e\right )^{e^{-2}} \log {\relax (2 )}}{x + e}\, dx + \int \frac {10 \left (x + e\right )^{e^{-2}} e^{3} \log {\relax (2 )}}{x + e}\, dx + \int \frac {10 x \left (x + e\right )^{e^{-2}} e^{2} \log {\relax (2 )}}{x + e}\, dx}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+exp(1))*exp(2)+x)*exp((exp(2)*ln(10*x*ln(2))+ln(x+exp(1)))/exp(2))+(-x*exp(1)-x**2)*exp(2))/(x*
exp(1)+x**2)/exp(2),x)

[Out]

(Integral(-exp(3)/(x + E), x) + Integral(-x*exp(2)/(x + E), x) + Integral(10*x*(x + E)**exp(-2)*log(2)/(x + E)
, x) + Integral(10*(x + E)**exp(-2)*exp(3)*log(2)/(x + E), x) + Integral(10*x*(x + E)**exp(-2)*exp(2)*log(2)/(
x + E), x))*exp(-2)

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