3.59.68 \(\int \frac {e^{\frac {e^{4 e^x}}{-5+x}} ((25-10 x+x^2) \log (3)+e^x (25-10 x+x^2) \log (3)+e^{4 e^x} ((e^5-x) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x (-1+e^5 (20-4 x)-20 x+4 x^2) \log (3)))}{25-10 x+x^2} \, dx\)

Optimal. Leaf size=31 \[ \left (-2+e^{\frac {e^{4 e^x}}{-5+x}} \left (-e^5+e^x+x\right )\right ) \log (3) \]

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Rubi [B]  time = 0.50, antiderivative size = 120, normalized size of antiderivative = 3.87, number of steps used = 2, number of rules used = 2, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {27, 2288} \begin {gather*} \frac {e^{4 e^x-\frac {e^{4 e^x}}{5-x}} \left (e^x \left (-4 x^2+20 x-4 e^5 (5-x)+1\right ) \log (3)+4 e^{2 x} (5-x) \log (3)-\left (e^5-x\right ) \log (3)\right )}{\left (\frac {4 e^{x+4 e^x}}{5-x}+\frac {e^{4 e^x}}{(5-x)^2}\right ) (5-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^(4*E^x)/(-5 + x))*((25 - 10*x + x^2)*Log[3] + E^x*(25 - 10*x + x^2)*Log[3] + E^(4*E^x)*((E^5 - x)*Lo
g[3] + E^(2*x)*(-20 + 4*x)*Log[3] + E^x*(-1 + E^5*(20 - 4*x) - 20*x + 4*x^2)*Log[3])))/(25 - 10*x + x^2),x]

[Out]

(E^(4*E^x - E^(4*E^x)/(5 - x))*(4*E^(2*x)*(5 - x)*Log[3] - (E^5 - x)*Log[3] + E^x*(1 - 4*E^5*(5 - x) + 20*x -
4*x^2)*Log[3]))/((E^(4*E^x)/(5 - x)^2 + (4*E^(4*E^x + x))/(5 - x))*(5 - x)^2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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} &=\int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{(-5+x)^2} \, dx\\ &=\frac {e^{4 e^x-\frac {e^{4 e^x}}{5-x}} \left (4 e^{2 x} (5-x) \log (3)-\left (e^5-x\right ) \log (3)+e^x \left (1-4 e^5 (5-x)+20 x-4 x^2\right ) \log (3)\right )}{\left (\frac {e^{4 e^x}}{(5-x)^2}+\frac {4 e^{4 e^x+x}}{5-x}\right ) (5-x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 28, normalized size = 0.90 \begin {gather*} e^{\frac {e^{4 e^x}}{-5+x}} \left (-e^5+e^x+x\right ) \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(4*E^x)/(-5 + x))*((25 - 10*x + x^2)*Log[3] + E^x*(25 - 10*x + x^2)*Log[3] + E^(4*E^x)*((E^5 -
x)*Log[3] + E^(2*x)*(-20 + 4*x)*Log[3] + E^x*(-1 + E^5*(20 - 4*x) - 20*x + 4*x^2)*Log[3])))/(25 - 10*x + x^2)
,x]

[Out]

E^(E^(4*E^x)/(-5 + x))*(-E^5 + E^x + x)*Log[3]

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fricas [A]  time = 0.60, size = 28, normalized size = 0.90 \begin {gather*} {\left ({\left (x - e^{5}\right )} \log \relax (3) + e^{x} \log \relax (3)\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3)*exp(x)+(exp(5)-x)*log(3))*exp(exp(
x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2-10*x+25)*log(3))*exp(exp(exp(x))^4/(x-5))/(x^2-10*x+25),x, algorithm="f
ricas")

[Out]

((x - e^5)*log(3) + e^x*log(3))*e^(e^(4*e^x)/(x - 5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x^{2} - 10 \, x + 25\right )} e^{x} \log \relax (3) + {\left (4 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} \log \relax (3) + {\left (4 \, x^{2} - 4 \, {\left (x - 5\right )} e^{5} - 20 \, x - 1\right )} e^{x} \log \relax (3) - {\left (x - e^{5}\right )} \log \relax (3)\right )} e^{\left (4 \, e^{x}\right )} + {\left (x^{2} - 10 \, x + 25\right )} \log \relax (3)\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )}}{x^{2} - 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3)*exp(x)+(exp(5)-x)*log(3))*exp(exp(
x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2-10*x+25)*log(3))*exp(exp(exp(x))^4/(x-5))/(x^2-10*x+25),x, algorithm="g
iac")

[Out]

integrate(((x^2 - 10*x + 25)*e^x*log(3) + (4*(x - 5)*e^(2*x)*log(3) + (4*x^2 - 4*(x - 5)*e^5 - 20*x - 1)*e^x*l
og(3) - (x - e^5)*log(3))*e^(4*e^x) + (x^2 - 10*x + 25)*log(3))*e^(e^(4*e^x)/(x - 5))/(x^2 - 10*x + 25), x)

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maple [A]  time = 0.38, size = 27, normalized size = 0.87

method	result	size
risch	\(-\left ({\mathrm e}^{5}-x -{\mathrm e}^{x}\right ) \ln \relax (3) {\mathrm e}^{\frac {{\mathrm e}^{4 \,{\mathrm e}^{x}}}{x -5}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x-20)*ln(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*ln(3)*exp(x)+(exp(5)-x)*ln(3))*exp(exp(x))^4+(x^
2-10*x+25)*ln(3)*exp(x)+(x^2-10*x+25)*ln(3))*exp(exp(exp(x))^4/(x-5))/(x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

-(exp(5)-x-exp(x))*ln(3)*exp(exp(4*exp(x))/(x-5))

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maxima [A]  time = 0.64, size = 29, normalized size = 0.94 \begin {gather*} {\left (x \log \relax (3) - e^{5} \log \relax (3) + e^{x} \log \relax (3)\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3)*exp(x)+(exp(5)-x)*log(3))*exp(exp(
x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2-10*x+25)*log(3))*exp(exp(exp(x))^4/(x-5))/(x^2-10*x+25),x, algorithm="m
axima")

[Out]

(x*log(3) - e^5*log(3) + e^x*log(3))*e^(e^(4*e^x)/(x - 5))

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mupad [B]  time = 4.89, size = 29, normalized size = 0.94 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}}{x-5}}\,\left (x\,\ln \relax (3)-{\mathrm {e}}^5\,\ln \relax (3)+{\mathrm {e}}^x\,\ln \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(4*exp(x))/(x - 5))*(log(3)*(x^2 - 10*x + 25) - exp(4*exp(x))*(log(3)*(x - exp(5)) + exp(x)*log(3)
*(20*x - 4*x^2 + exp(5)*(4*x - 20) + 1) - exp(2*x)*log(3)*(4*x - 20)) + exp(x)*log(3)*(x^2 - 10*x + 25)))/(x^2
- 10*x + 25),x)

[Out]

exp(exp(4*exp(x))/(x - 5))*(x*log(3) - exp(5)*log(3) + exp(x)*log(3))

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sympy [A]  time = 23.42, size = 29, normalized size = 0.94 \begin {gather*} \left (x \log {\relax (3 )} + e^{x} \log {\relax (3 )} - e^{5} \log {\relax (3 )}\right ) e^{\frac {e^{4 e^{x}}}{x - 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-20)*ln(3)*exp(x)**2+((-4*x+20)*exp(5)+4*x**2-20*x-1)*ln(3)*exp(x)+(exp(5)-x)*ln(3))*exp(exp(x
))**4+(x**2-10*x+25)*ln(3)*exp(x)+(x**2-10*x+25)*ln(3))*exp(exp(exp(x))**4/(x-5))/(x**2-10*x+25),x)

[Out]

(x*log(3) + exp(x)*log(3) - exp(5)*log(3))*exp(exp(4*exp(x))/(x - 5))

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