3.55.77 \(\int \frac {-10+e^{e^5+3 x} (e^4 (-2-6 x+3 x^2)+e^2 (-4-12 x+6 x^2) \log (5)+(-2-6 x+3 x^2) \log ^2(5))}{4-4 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {x \left (5+e^{e^5+3 x} \left (e^2+\log (5)\right )^2\right )}{-2+x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.54, antiderivative size = 54, normalized size of antiderivative = 2.00, number of steps used = 10, number of rules used = 7, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.101, Rules used = {27, 6688, 6742, 2199, 2194, 2177, 2178} \begin {gather*} -\frac {10}{2-x}+e^{3 x+e^5} \left (e^2+\log (5)\right )^2-\frac {2 e^{3 x+e^5} \left (e^2+\log (5)\right )^2}{2-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + E^(E^5 + 3*x)*(E^4*(-2 - 6*x + 3*x^2) + E^2*(-4 - 12*x + 6*x^2)*Log[5] + (-2 - 6*x + 3*x^2)*Log[5]^
2))/(4 - 4*x + x^2),x]

[Out]

-10/(2 - x) + E^(E^5 + 3*x)*(E^2 + Log[5])^2 - (2*E^(E^5 + 3*x)*(E^2 + Log[5])^2)/(2 - x)

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{(-2+x)^2} \, dx\\ &=\int \frac {-10+e^{e^5+3 x} \left (-2-6 x+3 x^2\right ) \left (e^2+\log (5)\right )^2}{(2-x)^2} \, dx\\ &=\int \left (-\frac {10}{(-2+x)^2}+\frac {e^{e^5+3 x} \left (-2-6 x+3 x^2\right ) \left (e^2+\log (5)\right )^2}{(-2+x)^2}\right ) \, dx\\ &=-\frac {10}{2-x}+\left (e^2+\log (5)\right )^2 \int \frac {e^{e^5+3 x} \left (-2-6 x+3 x^2\right )}{(-2+x)^2} \, dx\\ &=-\frac {10}{2-x}+\left (e^2+\log (5)\right )^2 \int \left (3 e^{e^5+3 x}-\frac {2 e^{e^5+3 x}}{(-2+x)^2}+\frac {6 e^{e^5+3 x}}{-2+x}\right ) \, dx\\ &=-\frac {10}{2-x}-\left (2 \left (e^2+\log (5)\right )^2\right ) \int \frac {e^{e^5+3 x}}{(-2+x)^2} \, dx+\left (3 \left (e^2+\log (5)\right )^2\right ) \int e^{e^5+3 x} \, dx+\left (6 \left (e^2+\log (5)\right )^2\right ) \int \frac {e^{e^5+3 x}}{-2+x} \, dx\\ &=-\frac {10}{2-x}+e^{e^5+3 x} \left (e^2+\log (5)\right )^2-\frac {2 e^{e^5+3 x} \left (e^2+\log (5)\right )^2}{2-x}+6 e^{6+e^5} \text {Ei}(-3 (2-x)) \left (e^2+\log (5)\right )^2-\left (6 \left (e^2+\log (5)\right )^2\right ) \int \frac {e^{e^5+3 x}}{-2+x} \, dx\\ &=-\frac {10}{2-x}+e^{e^5+3 x} \left (e^2+\log (5)\right )^2-\frac {2 e^{e^5+3 x} \left (e^2+\log (5)\right )^2}{2-x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 27, normalized size = 1.00 \begin {gather*} \frac {10+e^{e^5+3 x} x \left (e^2+\log (5)\right )^2}{-2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^(E^5 + 3*x)*(E^4*(-2 - 6*x + 3*x^2) + E^2*(-4 - 12*x + 6*x^2)*Log[5] + (-2 - 6*x + 3*x^2)*L
og[5]^2))/(4 - 4*x + x^2),x]

[Out]

(10 + E^(E^5 + 3*x)*x*(E^2 + Log[5])^2)/(-2 + x)

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 34, normalized size = 1.26 \begin {gather*} \frac {{\left (2 \, x e^{2} \log \relax (5) + x \log \relax (5)^{2} + x e^{4}\right )} e^{\left (3 \, x + e^{5}\right )} + 10}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x-2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x
^2-4*x+4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.19, size = 44, normalized size = 1.63 \begin {gather*} \frac {x e^{\left (3 \, x + e^{5}\right )} \log \relax (5)^{2} + 2 \, x e^{\left (3 \, x + e^{5} + 2\right )} \log \relax (5) + x e^{\left (3 \, x + e^{5} + 4\right )} + 10}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x-2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x
^2-4*x+4),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 7.27, size = 33, normalized size = 1.22

method	result	size
norman	\(\frac {\left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2} \ln \relax (5)+\ln \relax (5)^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{5}+3 x}+10}{x -2}\)	\(33\)
risch	\(\frac {10}{x -2}+\frac {\left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2} \ln \relax (5)+\ln \relax (5)^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{5}+3 x}}{x -2}\)	\(36\)
derivativedivides	\(6 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{4} {\mathrm e}^{10} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{{\mathrm e}^{5}+3 x} \ln \relax (5)^{2}-\frac {30}{-3 x +6}-6 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left ({\mathrm e}^{5}+6\right )}{-3 x +6}-\left ({\mathrm e}^{5}+7\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{4} \left ({\mathrm e}^{{\mathrm e}^{5}+3 x}+\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left (12 \,{\mathrm e}^{5}+{\mathrm e}^{10}+36\right )}{-3 x +6}-\left ({\mathrm e}^{10}+14 \,{\mathrm e}^{5}+48\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+2 \,{\mathrm e}^{{\mathrm e}^{5}+3 x} {\mathrm e}^{2} \ln \relax (5)-6 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )-\frac {12 \,{\mathrm e}^{2} \ln \relax (5) {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-2 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left ({\mathrm e}^{5}+6\right )}{-3 x +6}-\left ({\mathrm e}^{5}+7\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )-\frac {6 \ln \relax (5)^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}\)	\(422\)
default	\(6 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{4} {\mathrm e}^{10} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{{\mathrm e}^{5}+3 x} \ln \relax (5)^{2}-\frac {30}{-3 x +6}-6 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left ({\mathrm e}^{5}+6\right )}{-3 x +6}-\left ({\mathrm e}^{5}+7\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+{\mathrm e}^{4} \left ({\mathrm e}^{{\mathrm e}^{5}+3 x}+\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left (12 \,{\mathrm e}^{5}+{\mathrm e}^{10}+36\right )}{-3 x +6}-\left ({\mathrm e}^{10}+14 \,{\mathrm e}^{5}+48\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )+2 \,{\mathrm e}^{{\mathrm e}^{5}+3 x} {\mathrm e}^{2} \ln \relax (5)-6 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-{\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )-\frac {12 \,{\mathrm e}^{2} \ln \relax (5) {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-2 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}+3 x} \left ({\mathrm e}^{5}+6\right )}{-3 x +6}-\left ({\mathrm e}^{5}+7\right ) {\mathrm e}^{{\mathrm e}^{5}+6} \expIntegralEi \left (1, -3 x +6\right )\right )-\frac {6 \ln \relax (5)^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}\)	\(422\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2-6*x-2)*ln(5)^2+(6*x^2-12*x-4)*exp(2)*ln(5)+(3*x^2-6*x-2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4
),x,method=_RETURNVERBOSE)

[Out]

((ln(5)^2+2*exp(2)*ln(5)+exp(2)^2)*x*exp(exp(5)+3*x)+10)/(x-2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, e^{\left (e^{5} + 6\right )} E_{2}\left (-3 \, x + 6\right ) \log \relax (5)^{2}}{x - 2} + 2 \, \int \frac {e^{\left (3 \, x + e^{5}\right )}}{x^{2} - 4 \, x + 4}\,{d x} \log \relax (5)^{2} + \frac {{\left (e^{\left (e^{5}\right )} \log \relax (5)^{2} + 2 \, e^{\left (e^{5} + 2\right )} \log \relax (5) + e^{\left (e^{5} + 4\right )}\right )} x e^{\left (3 \, x\right )}}{x - 2} + \frac {10}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x-2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x
^2-4*x+4),x, algorithm="maxima")

[Out]

2*e^(e^5 + 6)*exp_integral_e(2, -3*x + 6)*log(5)^2/(x - 2) + 2*integrate(e^(3*x + e^5)/(x^2 - 4*x + 4), x)*log
(5)^2 + (e^(e^5)*log(5)^2 + 2*e^(e^5 + 2)*log(5) + e^(e^5 + 4))*x*e^(3*x)/(x - 2) + 10/(x - 2)

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 26, normalized size = 0.96 \begin {gather*} \frac {5\,x+x\,{\mathrm {e}}^{3\,x+{\mathrm {e}}^5}\,{\left ({\mathrm {e}}^2+\ln \relax (5)\right )}^2}{x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3*x + exp(5))*(exp(4)*(6*x - 3*x^2 + 2) + log(5)^2*(6*x - 3*x^2 + 2) + exp(2)*log(5)*(12*x - 6*x^2 +
4)) + 10)/(x^2 - 4*x + 4),x)

[Out]

(5*x + x*exp(3*x + exp(5))*(exp(2) + log(5))^2)/(x - 2)

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 37, normalized size = 1.37 \begin {gather*} \frac {\left (x \log {\relax (5 )}^{2} + 2 x e^{2} \log {\relax (5 )} + x e^{4}\right ) e^{3 x + e^{5}}}{x - 2} + \frac {10}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2-6*x-2)*ln(5)**2+(6*x**2-12*x-4)*exp(2)*ln(5)+(3*x**2-6*x-2)*exp(2)**2)*exp(exp(5)+3*x)-10)
/(x**2-4*x+4),x)

[Out]

(x*log(5)**2 + 2*x*exp(2)*log(5) + x*exp(4))*exp(3*x + exp(5))/(x - 2) + 10/(x - 2)

________________________________________________________________________________________