3.69.48 \(\int \frac {3-x-2 e^{3/2} x^2+e^{\frac {3}{2}+x} (-3 x+x^2)+e^{3/2} (6 x-2 x^2) \log (-3+x)}{e^{3/2} (-3 x+x^2)} \, dx\)

Optimal. Leaf size=20 \[ e^x-2 x \log (-3+x)-\frac {\log (x)}{e^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.93, antiderivative size = 32, normalized size of antiderivative = 1.60, number of steps used = 11, number of rules used = 7, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {12, 1593, 6742, 2194, 893, 2389, 2295} \begin {gather*} e^x-6 \log (3-x)+2 (3-x) \log (x-3)-\frac {\log (x)}{e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, 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} &=\frac {\int \frac {3-x-2 e^{3/2} x^2+e^{\frac {3}{2}+x} \left (-3 x+x^2\right )+e^{3/2} \left (6 x-2 x^2\right ) \log (-3+x)}{-3 x+x^2} \, dx}{e^{3/2}}\\ &=\frac {\int \frac {3-x-2 e^{3/2} x^2+e^{\frac {3}{2}+x} \left (-3 x+x^2\right )+e^{3/2} \left (6 x-2 x^2\right ) \log (-3+x)}{(-3+x) x} \, dx}{e^{3/2}}\\ &=\frac {\int \left (e^{\frac {3}{2}+x}+\frac {3-x-2 e^{3/2} x^2+6 e^{3/2} x \log (-3+x)-2 e^{3/2} x^2 \log (-3+x)}{(-3+x) x}\right ) \, dx}{e^{3/2}}\\ &=\frac {\int e^{\frac {3}{2}+x} \, dx}{e^{3/2}}+\frac {\int \frac {3-x-2 e^{3/2} x^2+6 e^{3/2} x \log (-3+x)-2 e^{3/2} x^2 \log (-3+x)}{(-3+x) x} \, dx}{e^{3/2}}\\ &=e^x+\frac {\int \left (\frac {3-x-2 e^{3/2} x^2}{(-3+x) x}-2 e^{3/2} \log (-3+x)\right ) \, dx}{e^{3/2}}\\ &=e^x-2 \int \log (-3+x) \, dx+\frac {\int \frac {3-x-2 e^{3/2} x^2}{(-3+x) x} \, dx}{e^{3/2}}\\ &=e^x-2 \operatorname {Subst}(\int \log (x) \, dx,x,-3+x)+\frac {\int \left (-2 e^{3/2}-\frac {6 e^{3/2}}{-3+x}-\frac {1}{x}\right ) \, dx}{e^{3/2}}\\ &=e^x-6 \log (3-x)+2 (3-x) \log (-3+x)-\frac {\log (x)}{e^{3/2}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.22, size = 47, normalized size = 2.35 \begin {gather*} \frac {e^{\frac {3}{2}+x}-6 e^{3/2} \log (3-x)+2 e^{3/2} (3-x) \log (-3+x)-\log (x)}{e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - x - 2*E^(3/2)*x^2 + E^(3/2 + x)*(-3*x + x^2) + E^(3/2)*(6*x - 2*x^2)*Log[-3 + x])/(E^(3/2)*(-3*
x + x^2)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 22, normalized size = 1.10 \begin {gather*} -{\left (2 \, x e^{\frac {3}{2}} \log \left (x - 3\right ) - e^{\left (x + \frac {3}{2}\right )} + \log \relax (x)\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.24, size = 22, normalized size = 1.10 \begin {gather*} -{\left (2 \, x e^{\frac {3}{2}} \log \left (x - 3\right ) - e^{\left (x + \frac {3}{2}\right )} + \log \relax (x)\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 17, normalized size = 0.85

method	result	size
risch	\({\mathrm e}^{x}-\ln \relax (x ) {\mathrm e}^{-\frac {3}{2}}-2 \ln \left (x -3\right ) x\)	\(17\)
norman	\({\mathrm e}^{x}-\ln \relax (x ) {\mathrm e}^{-\frac {3}{2}}-2 \ln \left (x -3\right ) x\)	\(19\)
default	\({\mathrm e}^{-\frac {3}{2}} \left ({\mathrm e}^{\frac {3}{2}} {\mathrm e}^{x}-2 \,{\mathrm e}^{\frac {3}{2}} \ln \left (x -3\right ) x -6 \,{\mathrm e}^{\frac {3}{2}}-\ln \relax (x )\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+6*x)*exp(3/2)*ln(x-3)+(x^2-3*x)*exp(3/2)*exp(x)-2*x^2*exp(3/2)+3-x)/(x^2-3*x)/exp(3/2),x,method=_
RETURNVERBOSE)

[Out]

exp(x)-ln(x)*exp(-3/2)-2*ln(x-3)*x

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (2 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} e^{\frac {3}{2}} \log \left (x - 3\right ) - 3 \, e^{\frac {3}{2}} \log \left (x - 3\right )^{2} - {\left (3 \, \log \left (x - 3\right )^{2} + 2 \, x + 6 \, \log \left (x - 3\right )\right )} e^{\frac {3}{2}} + 2 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} e^{\frac {3}{2}} - 3 \, e^{\frac {9}{2}} E_{1}\left (-x + 3\right ) - \frac {x e^{\left (x + \frac {3}{2}\right )}}{x - 3} - 3 \, \int \frac {e^{\left (x + \frac {3}{2}\right )}}{x^{2} - 6 \, x + 9}\,{d x} + \log \relax (x)\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(x + 3*log(x - 3))*e^(3/2)*log(x - 3) - 3*e^(3/2)*log(x - 3)^2 - (3*log(x - 3)^2 + 2*x + 6*log(x - 3))*e^(
3/2) + 2*(x + 3*log(x - 3))*e^(3/2) - 3*e^(9/2)*exp_integral_e(1, -x + 3) - x*e^(x + 3/2)/(x - 3) - 3*integrat
e(e^(x + 3/2)/(x^2 - 6*x + 9), x) + log(x))*e^(-3/2)

________________________________________________________________________________________

mupad [B]  time = 0.26, size = 16, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^x-2\,x\,\ln \left (x-3\right )-{\mathrm {e}}^{-\frac {3}{2}}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3/2)*(x + 2*x^2*exp(3/2) + exp(3/2)*exp(x)*(3*x - x^2) - log(x - 3)*exp(3/2)*(6*x - 2*x^2) - 3))/(3*
x - x^2),x)

[Out]

exp(x) - 2*x*log(x - 3) - exp(-3/2)*log(x)

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 19, normalized size = 0.95 \begin {gather*} - 2 x \log {\left (x - 3 \right )} + e^{x} - \frac {\log {\relax (x )}}{e^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+6*x)*exp(3/2)*ln(x-3)+(x**2-3*x)*exp(3/2)*exp(x)-2*x**2*exp(3/2)+3-x)/(x**2-3*x)/exp(3/2),
x)

[Out]

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

________________________________________________________________________________________