3.95.36 \(\int \frac {e^{-5+8 x-3 x^2-e^{10} (2 x-x^2)} (-3+24 x-18 x^2+e^{10} (-6 x+6 x^2)+(24-18 x+e^{10} (-6+6 x)) \log (3))}{x^2+2 x \log (3)+\log ^2(3)} \, dx\)

Optimal. Leaf size=25 \[ \frac {3 e^{-5+\left (-4+e^{10}\right ) (-2+x) x+x^2}}{x+\log (3)} \]

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Rubi [B]  time = 0.38, antiderivative size = 90, normalized size of antiderivative = 3.60, number of steps used = 2, number of rules used = 2, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {27, 2288} \begin {gather*} \frac {3 e^{-3 x^2-e^{10} \left (2 x-x^2\right )+8 x-5} \left (-3 x^2-e^{10} \left (x-x^2\right )+4 x+\left (-e^{10} (1-x)-3 x+4\right ) \log (3)\right )}{\left (-e^{10} (1-x)-3 x+4\right ) (x+\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-5 + 8*x - 3*x^2 - E^10*(2*x - x^2))*(-3 + 24*x - 18*x^2 + E^10*(-6*x + 6*x^2) + (24 - 18*x + E^10*(-6
+ 6*x))*Log[3]))/(x^2 + 2*x*Log[3] + Log[3]^2),x]

[Out]

(3*E^(-5 + 8*x - 3*x^2 - E^10*(2*x - x^2))*(4*x - 3*x^2 - E^10*(x - x^2) + (4 - E^10*(1 - x) - 3*x)*Log[3]))/(
(4 - E^10*(1 - x) - 3*x)*(x + Log[3])^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^{-5+8 x-3 x^2-e^{10} \left (2 x-x^2\right )} \left (-3+24 x-18 x^2+e^{10} \left (-6 x+6 x^2\right )+\left (24-18 x+e^{10} (-6+6 x)\right ) \log (3)\right )}{(x+\log (3))^2} \, dx\\ &=\frac {3 e^{-5+8 x-3 x^2-e^{10} \left (2 x-x^2\right )} \left (4 x-3 x^2-e^{10} \left (x-x^2\right )+\left (4-e^{10} (1-x)-3 x\right ) \log (3)\right )}{\left (4-e^{10} (1-x)-3 x\right ) (x+\log (3))^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 85, normalized size = 3.40 \begin {gather*} \frac {3 e^{-5+8 x+e^{10} (-2+x) x-3 x^2} \left (2 \left (-3+e^{10}\right ) x^2+8 \log (3)+x \left (8-6 \log (3)+e^{10} (-2+\log (9))\right )-e^{10} \log (9)\right )}{\left (8+e^{10} (-2+x)-6 x+e^{10} x\right ) (x+\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-5 + 8*x - 3*x^2 - E^10*(2*x - x^2))*(-3 + 24*x - 18*x^2 + E^10*(-6*x + 6*x^2) + (24 - 18*x + E^
10*(-6 + 6*x))*Log[3]))/(x^2 + 2*x*Log[3] + Log[3]^2),x]

[Out]

(3*E^(-5 + 8*x + E^10*(-2 + x)*x - 3*x^2)*(2*(-3 + E^10)*x^2 + 8*Log[3] + x*(8 - 6*Log[3] + E^10*(-2 + Log[9])
) - E^10*Log[9]))/((8 + E^10*(-2 + x) - 6*x + E^10*x)*(x + Log[3])^2)

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fricas [A]  time = 0.93, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, e^{\left (-3 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{10} + 8 \, x - 5\right )}}{x + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x-6)*exp(5)^2-18*x+24)*log(3)+(6*x^2-6*x)*exp(5)^2-18*x^2+24*x-3)/(log(3)^2+2*x*log(3)+x^2)/exp
((-x^2+2*x)*exp(5)^2+3*x^2-8*x+5),x, algorithm="fricas")

[Out]

3*e^(-3*x^2 + (x^2 - 2*x)*e^10 + 8*x - 5)/(x + log(3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left (6 \, x^{2} - 2 \, {\left (x^{2} - x\right )} e^{10} - 2 \, {\left ({\left (x - 1\right )} e^{10} - 3 \, x + 4\right )} \log \relax (3) - 8 \, x + 1\right )} e^{\left (-3 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{10} + 8 \, x - 5\right )}}{x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x-6)*exp(5)^2-18*x+24)*log(3)+(6*x^2-6*x)*exp(5)^2-18*x^2+24*x-3)/(log(3)^2+2*x*log(3)+x^2)/exp
((-x^2+2*x)*exp(5)^2+3*x^2-8*x+5),x, algorithm="giac")

[Out]

integrate(-3*(6*x^2 - 2*(x^2 - x)*e^10 - 2*((x - 1)*e^10 - 3*x + 4)*log(3) - 8*x + 1)*e^(-3*x^2 + (x^2 - 2*x)*
e^10 + 8*x - 5)/(x^2 + 2*x*log(3) + log(3)^2), x)

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maple [A]  time = 0.63, size = 31, normalized size = 1.24

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{x^{2} {\mathrm e}^{10}-2 x \,{\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\)	\(31\)
norman	\(\frac {3 \,{\mathrm e}^{-\left (-x^{2}+2 x \right ) {\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\)	\(36\)
gosper	\(\frac {3 \,{\mathrm e}^{x^{2} {\mathrm e}^{10}-2 x \,{\mathrm e}^{10}-3 x^{2}+8 x -5}}{\ln \relax (3)+x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x-6)*exp(5)^2-18*x+24)*ln(3)+(6*x^2-6*x)*exp(5)^2-18*x^2+24*x-3)/(ln(3)^2+2*x*ln(3)+x^2)/exp((-x^2+2*
x)*exp(5)^2+3*x^2-8*x+5),x,method=_RETURNVERBOSE)

[Out]

3/(ln(3)+x)*exp(x^2*exp(10)-2*x*exp(10)-3*x^2+8*x-5)

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maxima [A]  time = 0.54, size = 35, normalized size = 1.40 \begin {gather*} \frac {3 \, e^{\left (x^{2} e^{10} - 3 \, x^{2} - 2 \, x e^{10} + 8 \, x\right )}}{x e^{5} + e^{5} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x-6)*exp(5)^2-18*x+24)*log(3)+(6*x^2-6*x)*exp(5)^2-18*x^2+24*x-3)/(log(3)^2+2*x*log(3)+x^2)/exp
((-x^2+2*x)*exp(5)^2+3*x^2-8*x+5),x, algorithm="maxima")

[Out]

3*e^(x^2*e^10 - 3*x^2 - 2*x*e^10 + 8*x)/(x*e^5 + e^5*log(3))

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mupad [B]  time = 8.81, size = 33, normalized size = 1.32 \begin {gather*} \frac {3\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{10}}}{x+\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8*x - exp(10)*(2*x - x^2) - 3*x^2 - 5)*(exp(10)*(6*x - 6*x^2) - 24*x + 18*x^2 - log(3)*(exp(10)*(6*x
- 6) - 18*x + 24) + 3))/(2*x*log(3) + log(3)^2 + x^2),x)

[Out]

(3*exp(x^2*exp(10))*exp(8*x)*exp(-5)*exp(-3*x^2)*exp(-2*x*exp(10)))/(x + log(3))

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sympy [A]  time = 0.24, size = 27, normalized size = 1.08 \begin {gather*} \frac {3 e^{- 3 x^{2} + 8 x - \left (- x^{2} + 2 x\right ) e^{10} - 5}}{x + \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x-6)*exp(5)**2-18*x+24)*ln(3)+(6*x**2-6*x)*exp(5)**2-18*x**2+24*x-3)/(ln(3)**2+2*x*ln(3)+x**2)/
exp((-x**2+2*x)*exp(5)**2+3*x**2-8*x+5),x)

[Out]

3*exp(-3*x**2 + 8*x - (-x**2 + 2*x)*exp(10) - 5)/(x + log(3))

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