3.3.40 \(\int \frac {e^{5 e^{9-6 x+x^2+(6-2 x) \log ^2(-2+x)+\log ^4(-2+x)}+5 x^2} (2-x-20 x^2+10 x^3+e^{9-6 x+x^2+(6-2 x) \log ^2(-2+x)+\log ^4(-2+x)} (60 x-50 x^2+10 x^3+(60 x-20 x^2) \log (-2+x)+(20 x-10 x^2) \log ^2(-2+x)+20 x \log ^3(-2+x)))}{-2 x^2+x^3} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{5 \left (e^{\left (3-x+\log ^2(-2+x)\right )^2}+x^2\right )}}{x}+\log (4) \]

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Rubi [B]  time = 10.10, antiderivative size = 223, normalized size of antiderivative = 7.43, number of steps used = 2, number of rules used = 2, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {1593, 2288} \begin {gather*} \frac {\exp \left (5 \exp \left (x^2-6 x+\log ^4(x-2)+2 (3-x) \log ^2(x-2)+9\right )+5 x^2\right ) \left (-\left (x^3-5 x^2+\left (2 x-x^2\right ) \log ^2(x-2)+2 \left (3 x-x^2\right ) \log (x-2)+6 x+2 x \log ^3(x-2)\right ) \exp \left (x^2-6 x+\log ^4(x-2)+2 (3-x) \log ^2(x-2)+9\right )-x^3+2 x^2\right )}{(2-x) x^2 \left (x-\left (-x+\frac {2 \log ^3(x-2)}{2-x}+\log ^2(x-2)+\frac {2 (3-x) \log (x-2)}{2-x}+3\right ) \exp \left (x^2-6 x+\log ^4(x-2)+2 (3-x) \log ^2(x-2)+9\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5*E^(9 - 6*x + x^2 + (6 - 2*x)*Log[-2 + x]^2 + Log[-2 + x]^4) + 5*x^2)*(2 - x - 20*x^2 + 10*x^3 + E^(9
- 6*x + x^2 + (6 - 2*x)*Log[-2 + x]^2 + Log[-2 + x]^4)*(60*x - 50*x^2 + 10*x^3 + (60*x - 20*x^2)*Log[-2 + x]
+ (20*x - 10*x^2)*Log[-2 + x]^2 + 20*x*Log[-2 + x]^3)))/(-2*x^2 + x^3),x]

[Out]

(E^(5*E^(9 - 6*x + x^2 + 2*(3 - x)*Log[-2 + x]^2 + Log[-2 + x]^4) + 5*x^2)*(2*x^2 - x^3 - E^(9 - 6*x + x^2 + 2
*(3 - x)*Log[-2 + x]^2 + Log[-2 + x]^4)*(6*x - 5*x^2 + x^3 + 2*(3*x - x^2)*Log[-2 + x] + (2*x - x^2)*Log[-2 +
x]^2 + 2*x*Log[-2 + x]^3)))/((2 - x)*x^2*(x - E^(9 - 6*x + x^2 + 2*(3 - x)*Log[-2 + x]^2 + Log[-2 + x]^4)*(3 -
x + (2*(3 - x)*Log[-2 + x])/(2 - x) + Log[-2 + x]^2 + (2*Log[-2 + x]^3)/(2 - x))))

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 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 {\exp \left (5 \exp \left (9-6 x+x^2+(6-2 x) \log ^2(-2+x)+\log ^4(-2+x)\right )+5 x^2\right ) \left (2-x-20 x^2+10 x^3+\exp \left (9-6 x+x^2+(6-2 x) \log ^2(-2+x)+\log ^4(-2+x)\right ) \left (60 x-50 x^2+10 x^3+\left (60 x-20 x^2\right ) \log (-2+x)+\left (20 x-10 x^2\right ) \log ^2(-2+x)+20 x \log ^3(-2+x)\right )\right )}{(-2+x) x^2} \, dx\\ &=\frac {\exp \left (5 \exp \left (9-6 x+x^2+2 (3-x) \log ^2(-2+x)+\log ^4(-2+x)\right )+5 x^2\right ) \left (2 x^2-x^3-\exp \left (9-6 x+x^2+2 (3-x) \log ^2(-2+x)+\log ^4(-2+x)\right ) \left (6 x-5 x^2+x^3+2 \left (3 x-x^2\right ) \log (-2+x)+\left (2 x-x^2\right ) \log ^2(-2+x)+2 x \log ^3(-2+x)\right )\right )}{(2-x) x^2 \left (x-\exp \left (9-6 x+x^2+2 (3-x) \log ^2(-2+x)+\log ^4(-2+x)\right ) \left (3-x+\frac {2 (3-x) \log (-2+x)}{2-x}+\log ^2(-2+x)+\frac {2 \log ^3(-2+x)}{2-x}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 27, normalized size = 0.90 \begin {gather*} \frac {e^{5 \left (e^{\left (3-x+\log ^2(-2+x)\right )^2}+x^2\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5*E^(9 - 6*x + x^2 + (6 - 2*x)*Log[-2 + x]^2 + Log[-2 + x]^4) + 5*x^2)*(2 - x - 20*x^2 + 10*x^3
+ E^(9 - 6*x + x^2 + (6 - 2*x)*Log[-2 + x]^2 + Log[-2 + x]^4)*(60*x - 50*x^2 + 10*x^3 + (60*x - 20*x^2)*Log[-2
+ x] + (20*x - 10*x^2)*Log[-2 + x]^2 + 20*x*Log[-2 + x]^3)))/(-2*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 0.67, size = 39, normalized size = 1.30 \begin {gather*} \frac {e^{\left (5 \, x^{2} + 5 \, e^{\left (\log \left (x - 2\right )^{4} - 2 \, {\left (x - 3\right )} \log \left (x - 2\right )^{2} + x^{2} - 6 \, x + 9\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(x-2)^3+(-10*x^2+20*x)*log(x-2)^2+(-20*x^2+60*x)*log(x-2)+10*x^3-50*x^2+60*x)*exp(log(x-2)
^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+10*x^3-20*x^2-x+2)*exp(5*exp(log(x-2)^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+5*x^2)/
(x^3-2*x^2),x, algorithm="fricas")

[Out]

e^(5*x^2 + 5*e^(log(x - 2)^4 - 2*(x - 3)*log(x - 2)^2 + x^2 - 6*x + 9))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (10 \, x^{3} - 20 \, x^{2} + 10 \, {\left (2 \, x \log \left (x - 2\right )^{3} + x^{3} - {\left (x^{2} - 2 \, x\right )} \log \left (x - 2\right )^{2} - 5 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (x - 2\right ) + 6 \, x\right )} e^{\left (\log \left (x - 2\right )^{4} - 2 \, {\left (x - 3\right )} \log \left (x - 2\right )^{2} + x^{2} - 6 \, x + 9\right )} - x + 2\right )} e^{\left (5 \, x^{2} + 5 \, e^{\left (\log \left (x - 2\right )^{4} - 2 \, {\left (x - 3\right )} \log \left (x - 2\right )^{2} + x^{2} - 6 \, x + 9\right )}\right )}}{x^{3} - 2 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(x-2)^3+(-10*x^2+20*x)*log(x-2)^2+(-20*x^2+60*x)*log(x-2)+10*x^3-50*x^2+60*x)*exp(log(x-2)
^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+10*x^3-20*x^2-x+2)*exp(5*exp(log(x-2)^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+5*x^2)/
(x^3-2*x^2),x, algorithm="giac")

[Out]

integrate((10*x^3 - 20*x^2 + 10*(2*x*log(x - 2)^3 + x^3 - (x^2 - 2*x)*log(x - 2)^2 - 5*x^2 - 2*(x^2 - 3*x)*log
(x - 2) + 6*x)*e^(log(x - 2)^4 - 2*(x - 3)*log(x - 2)^2 + x^2 - 6*x + 9) - x + 2)*e^(5*x^2 + 5*e^(log(x - 2)^4
- 2*(x - 3)*log(x - 2)^2 + x^2 - 6*x + 9))/(x^3 - 2*x^2), x)

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maple [A]  time = 0.36, size = 28, normalized size = 0.93

method	result	size
risch	\(\frac {{\mathrm e}^{5 \,{\mathrm e}^{\left (3+\ln \left (x -2\right )^{2}-x \right )^{2}}+5 x^{2}}}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x*ln(x-2)^3+(-10*x^2+20*x)*ln(x-2)^2+(-20*x^2+60*x)*ln(x-2)+10*x^3-50*x^2+60*x)*exp(ln(x-2)^4+(6-2*x)
*ln(x-2)^2+x^2-6*x+9)+10*x^3-20*x^2-x+2)*exp(5*exp(ln(x-2)^4+(6-2*x)*ln(x-2)^2+x^2-6*x+9)+5*x^2)/(x^3-2*x^2),x
,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (10 \, x^{3} - 20 \, x^{2} + 10 \, {\left (2 \, x \log \left (x - 2\right )^{3} + x^{3} - {\left (x^{2} - 2 \, x\right )} \log \left (x - 2\right )^{2} - 5 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (x - 2\right ) + 6 \, x\right )} e^{\left (\log \left (x - 2\right )^{4} - 2 \, {\left (x - 3\right )} \log \left (x - 2\right )^{2} + x^{2} - 6 \, x + 9\right )} - x + 2\right )} e^{\left (5 \, x^{2} + 5 \, e^{\left (\log \left (x - 2\right )^{4} - 2 \, {\left (x - 3\right )} \log \left (x - 2\right )^{2} + x^{2} - 6 \, x + 9\right )}\right )}}{x^{3} - 2 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(x-2)^3+(-10*x^2+20*x)*log(x-2)^2+(-20*x^2+60*x)*log(x-2)+10*x^3-50*x^2+60*x)*exp(log(x-2)
^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+10*x^3-20*x^2-x+2)*exp(5*exp(log(x-2)^4+(6-2*x)*log(x-2)^2+x^2-6*x+9)+5*x^2)/
(x^3-2*x^2),x, algorithm="maxima")

[Out]

integrate((10*x^3 - 20*x^2 + 10*(2*x*log(x - 2)^3 + x^3 - (x^2 - 2*x)*log(x - 2)^2 - 5*x^2 - 2*(x^2 - 3*x)*log
(x - 2) + 6*x)*e^(log(x - 2)^4 - 2*(x - 3)*log(x - 2)^2 + x^2 - 6*x + 9) - x + 2)*e^(5*x^2 + 5*e^(log(x - 2)^4
- 2*(x - 3)*log(x - 2)^2 + x^2 - 6*x + 9))/(x^3 - 2*x^2), x)

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mupad [B]  time = 0.64, size = 49, normalized size = 1.63 \begin {gather*} \frac {{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{5\,{\mathrm {e}}^{{\ln \left (x-2\right )}^4}\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^9\,{\mathrm {e}}^{6\,{\ln \left (x-2\right )}^2}\,{\mathrm {e}}^{-2\,x\,{\ln \left (x-2\right )}^2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5*exp(log(x - 2)^4 - log(x - 2)^2*(2*x - 6) - 6*x + x^2 + 9) + 5*x^2)*(exp(log(x - 2)^4 - log(x - 2)
^2*(2*x - 6) - 6*x + x^2 + 9)*(60*x + log(x - 2)*(60*x - 20*x^2) + log(x - 2)^2*(20*x - 10*x^2) + 20*x*log(x -
2)^3 - 50*x^2 + 10*x^3) - x - 20*x^2 + 10*x^3 + 2))/(2*x^2 - x^3),x)

[Out]

(exp(5*x^2)*exp(5*exp(log(x - 2)^4)*exp(-6*x)*exp(x^2)*exp(9)*exp(6*log(x - 2)^2)*exp(-2*x*log(x - 2)^2)))/x

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sympy [A]  time = 2.96, size = 37, normalized size = 1.23 \begin {gather*} \frac {e^{5 x^{2} + 5 e^{x^{2} - 6 x + \left (6 - 2 x\right ) \log {\left (x - 2 \right )}^{2} + \log {\left (x - 2 \right )}^{4} + 9}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*ln(x-2)**3+(-10*x**2+20*x)*ln(x-2)**2+(-20*x**2+60*x)*ln(x-2)+10*x**3-50*x**2+60*x)*exp(ln(x-
2)**4+(6-2*x)*ln(x-2)**2+x**2-6*x+9)+10*x**3-20*x**2-x+2)*exp(5*exp(ln(x-2)**4+(6-2*x)*ln(x-2)**2+x**2-6*x+9)+
5*x**2)/(x**3-2*x**2),x)

[Out]

exp(5*x**2 + 5*exp(x**2 - 6*x + (6 - 2*x)*log(x - 2)**2 + log(x - 2)**4 + 9))/x

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