3.29.86 \(\int \frac {e^{7-3 x^2} (18 x-18 e^2 x)}{-1+e^3 (11-12 e^2)+e^{7-3 x^2} (-3+3 e^2)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (-4+e^{4-3 x^2}-\frac {5+\frac {5}{e^3}}{15 \left (-1+e^2\right )}\right ) \]

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Rubi [B]  time = 0.13, antiderivative size = 69, normalized size of antiderivative = 2.30, number of steps used = 3, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 12, 6684} \begin {gather*} \log \left (e^{-3 x^2} \left (e^{3 x^2}-e^{3 x^2+1}+e^{3 x^2+2}-12 e^{3 x^2+3}+12 e^{3 x^2+4}+3 (1-e) e^7\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(7 - 3*x^2)*(18*x - 18*E^2*x))/(-1 + E^3*(11 - 12*E^2) + E^(7 - 3*x^2)*(-3 + 3*E^2)),x]

[Out]

Log[(3*(1 - E)*E^7 + E^(3*x^2) - E^(1 + 3*x^2) + E^(2 + 3*x^2) - 12*E^(3 + 3*x^2) + 12*E^(4 + 3*x^2))/E^(3*x^2
)]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{7-3 x^2} \left (18-18 e^2\right ) x}{-1+e^3 \left (11-12 e^2\right )+e^{7-3 x^2} \left (-3+3 e^2\right )} \, dx\\ &=\left (18 \left (1-e^2\right )\right ) \int \frac {e^{7-3 x^2} x}{-1+e^3 \left (11-12 e^2\right )+e^{7-3 x^2} \left (-3+3 e^2\right )} \, dx\\ &=\log \left (e^{-3 x^2} \left (3 (1-e) e^7+e^{3 x^2}-e^{1+3 x^2}+e^{2+3 x^2}-12 e^{3+3 x^2}+12 e^{4+3 x^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 41, normalized size = 1.37 \begin {gather*} \log \left (1-e+e^2-12 e^3+12 e^4+3 e^{7-3 x^2}-3 e^{8-3 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(7 - 3*x^2)*(18*x - 18*E^2*x))/(-1 + E^3*(11 - 12*E^2) + E^(7 - 3*x^2)*(-3 + 3*E^2)),x]

[Out]

Log[1 - E + E^2 - 12*E^3 + 12*E^4 + 3*E^(7 - 3*x^2) - 3*E^(8 - 3*x^2)]

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fricas [A]  time = 0.97, size = 31, normalized size = 1.03 \begin {gather*} \log \left (3 \, {\left (e - 1\right )} e^{\left (-3 \, x^{2} + 7\right )} - 12 \, e^{4} + 12 \, e^{3} - e^{2} + e - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*exp(2)*x+18*x)*exp(3)*exp(-3*x^2+4)/((3*exp(2)-3)*exp(3)*exp(-3*x^2+4)+(-12*exp(2)+11)*exp(3)-1
),x, algorithm="fricas")

[Out]

log(3*(e - 1)*e^(-3*x^2 + 7) - 12*e^4 + 12*e^3 - e^2 + e - 1)

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giac [A]  time = 0.19, size = 32, normalized size = 1.07 \begin {gather*} \log \left ({\left | -12 \, e^{5} + 11 \, e^{3} + 3 \, e^{\left (-3 \, x^{2} + 9\right )} - 3 \, e^{\left (-3 \, x^{2} + 7\right )} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*exp(2)*x+18*x)*exp(3)*exp(-3*x^2+4)/((3*exp(2)-3)*exp(3)*exp(-3*x^2+4)+(-12*exp(2)+11)*exp(3)-1
),x, algorithm="giac")

[Out]

log(abs(-12*e^5 + 11*e^3 + 3*e^(-3*x^2 + 9) - 3*e^(-3*x^2 + 7) - 1))

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maple [A]  time = 0.10, size = 39, normalized size = 1.30

method	result	size
risch	\(-4+\ln \left ({\mathrm e}^{-3 x^{2}+4}-\frac {\left (12 \,{\mathrm e}^{4}-12 \,{\mathrm e}^{3}+{\mathrm e}^{2}-{\mathrm e}+1\right ) {\mathrm e}^{-3}}{3 \left ({\mathrm e}-1\right )}\right )\)	\(39\)
norman	\(\ln \left (3 \,{\mathrm e}^{3} {\mathrm e}^{-3 x^{2}+4} {\mathrm e}^{2}-12 \,{\mathrm e}^{2} {\mathrm e}^{3}-3 \,{\mathrm e}^{3} {\mathrm e}^{-3 x^{2}+4}+11 \,{\mathrm e}^{3}-1\right )\)	\(40\)
default	\(18 \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{4} \ln \left ({\mathrm e}^{x^{2}}\right )}{6 \,{\mathrm e}^{9}-6 \,{\mathrm e}^{7}}-\frac {{\mathrm e}^{4} \ln \left (12 \,{\mathrm e}^{3 x^{2}} {\mathrm e}^{5}-11 \,{\mathrm e}^{3} {\mathrm e}^{3 x^{2}}+{\mathrm e}^{3 x^{2}}-3 \,{\mathrm e}^{9}+3 \,{\mathrm e}^{7}\right )}{6 \left (3 \,{\mathrm e}^{9}-3 \,{\mathrm e}^{7}\right )}-\frac {{\mathrm e}^{2} {\mathrm e}^{4} \ln \left ({\mathrm e}^{x^{2}}\right )}{2 \left (3 \,{\mathrm e}^{9}-3 \,{\mathrm e}^{7}\right )}+\frac {{\mathrm e}^{2} {\mathrm e}^{4} \ln \left (12 \,{\mathrm e}^{3 x^{2}} {\mathrm e}^{5}-11 \,{\mathrm e}^{3} {\mathrm e}^{3 x^{2}}+{\mathrm e}^{3 x^{2}}-3 \,{\mathrm e}^{9}+3 \,{\mathrm e}^{7}\right )}{18 \,{\mathrm e}^{9}-18 \,{\mathrm e}^{7}}\right )\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-18*exp(2)*x+18*x)*exp(3)*exp(-3*x^2+4)/((3*exp(2)-3)*exp(3)*exp(-3*x^2+4)+(-12*exp(2)+11)*exp(3)-1),x,me
thod=_RETURNVERBOSE)

[Out]

-4+ln(exp(-3*x^2+4)-1/3*(12*exp(4)-12*exp(3)+exp(2)-exp(1)+1)*exp(-3)/(exp(1)-1))

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maxima [B]  time = 0.37, size = 80, normalized size = 2.67 \begin {gather*} \frac {e^{2} \log \left (12 \, e^{5} - 11 \, e^{3} - 3 \, e^{\left (-3 \, x^{2} + 9\right )} + 3 \, e^{\left (-3 \, x^{2} + 7\right )} + 1\right )}{e^{2} - 1} - \frac {\log \left (12 \, e^{5} - 11 \, e^{3} - 3 \, e^{\left (-3 \, x^{2} + 9\right )} + 3 \, e^{\left (-3 \, x^{2} + 7\right )} + 1\right )}{e^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*exp(2)*x+18*x)*exp(3)*exp(-3*x^2+4)/((3*exp(2)-3)*exp(3)*exp(-3*x^2+4)+(-12*exp(2)+11)*exp(3)-1
),x, algorithm="maxima")

[Out]

e^2*log(12*e^5 - 11*e^3 - 3*e^(-3*x^2 + 9) + 3*e^(-3*x^2 + 7) + 1)/(e^2 - 1) - log(12*e^5 - 11*e^3 - 3*e^(-3*x
^2 + 9) + 3*e^(-3*x^2 + 7) + 1)/(e^2 - 1)

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mupad [B]  time = 0.46, size = 37, normalized size = 1.23 \begin {gather*} \ln \left ({\mathrm {e}}^2-\mathrm {e}-12\,{\mathrm {e}}^3+12\,{\mathrm {e}}^4+3\,{\mathrm {e}}^{7-3\,x^2}-3\,{\mathrm {e}}^{8-3\,x^2}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*exp(4 - 3*x^2)*(18*x - 18*x*exp(2)))/(exp(3)*(12*exp(2) - 11) - exp(3)*exp(4 - 3*x^2)*(3*exp(2) -
3) + 1),x)

[Out]

log(exp(2) - exp(1) - 12*exp(3) + 12*exp(4) + 3*exp(7 - 3*x^2) - 3*exp(8 - 3*x^2) + 1)

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sympy [A]  time = 0.24, size = 37, normalized size = 1.23 \begin {gather*} \log {\left (e^{4 - 3 x^{2}} + \frac {- 12 e^{4} - e^{2} - 1 + e + 12 e^{3}}{- 3 e^{3} + 3 e^{4}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*exp(2)*x+18*x)*exp(3)*exp(-3*x**2+4)/((3*exp(2)-3)*exp(3)*exp(-3*x**2+4)+(-12*exp(2)+11)*exp(3)
-1),x)

[Out]

log(exp(4 - 3*x**2) + (-12*exp(4) - exp(2) - 1 + E + 12*exp(3))/(-3*exp(3) + 3*exp(4)))

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