3.68.66 \(\int \frac {-8 e^{e^6}-32 x^2+(64 e x-64 x^2) \log (-e+x)}{e-x} \, dx\)

Optimal. Leaf size=19 \[ 8 \left (e^{e^6}+4 x^2\right ) \log (-e+x) \]

________________________________________________________________________________________

Rubi [B]  time = 0.15, antiderivative size = 42, normalized size of antiderivative = 2.21, number of steps used = 7, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6742, 697, 2395, 43} \begin {gather*} 32 x^2 \log (x-e)+8 \left (4 e^2+e^{e^6}\right ) \log (e-x)-32 e^2 \log (e-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*E^E^6 - 32*x^2 + (64*E*x - 64*x^2)*Log[-E + x])/(E - x),x]

[Out]

-32*E^2*Log[E - x] + 8*(4*E^2 + E^E^6)*Log[E - x] + 32*x^2*Log[-E + x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 \left (-\frac {8 \left (e^{e^6}+4 x^2\right )}{e-x}+64 x \log (-e+x)\right ) \, dx\\ &=-\left (8 \int \frac {e^{e^6}+4 x^2}{e-x} \, dx\right )+64 \int x \log (-e+x) \, dx\\ &=32 x^2 \log (-e+x)-8 \int \left (-4 e+\frac {4 e^2+e^{e^6}}{e-x}-4 x\right ) \, dx-32 \int \frac {x^2}{-e+x} \, dx\\ &=32 e x+16 x^2+8 \left (4 e^2+e^{e^6}\right ) \log (e-x)+32 x^2 \log (-e+x)-32 \int \left (e-\frac {e^2}{e-x}+x\right ) \, dx\\ &=-32 e^2 \log (e-x)+8 \left (4 e^2+e^{e^6}\right ) \log (e-x)+32 x^2 \log (-e+x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 27, normalized size = 1.42 \begin {gather*} -8 \left (-e^{e^6} \log (e-x)-4 x^2 \log (-e+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^E^6 - 32*x^2 + (64*E*x - 64*x^2)*Log[-E + x])/(E - x),x]

[Out]

-8*(-(E^E^6*Log[E - x]) - 4*x^2*Log[-E + x])

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 18, normalized size = 0.95 \begin {gather*} 8 \, {\left (4 \, x^{2} + e^{\left (e^{6}\right )}\right )} \log \left (x - e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(1)-64*x^2)*log(x-exp(1))-8*exp(exp(3)^2)-32*x^2)/(exp(1)-x),x, algorithm="fricas")

[Out]

8*(4*x^2 + e^(e^6))*log(x - e)

________________________________________________________________________________________

giac [A]  time = 0.24, size = 25, normalized size = 1.32 \begin {gather*} 32 \, x^{2} \log \left (x - e\right ) + 8 \, e^{\left (e^{6}\right )} \log \left (x - e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(1)-64*x^2)*log(x-exp(1))-8*exp(exp(3)^2)-32*x^2)/(exp(1)-x),x, algorithm="giac")

[Out]

32*x^2*log(x - e) + 8*e^(e^6)*log(x - e)

________________________________________________________________________________________

maple [A]  time = 0.11, size = 26, normalized size = 1.37

method	result	size
risch	\(32 \ln \left (x -{\mathrm e}\right ) x^{2}+8 \,{\mathrm e}^{{\mathrm e}^{6}} \ln \left (x -{\mathrm e}\right )\)	\(26\)
norman	\(32 \ln \left (x -{\mathrm e}\right ) x^{2}+8 \,{\mathrm e}^{{\mathrm e}^{6}} \ln \left (x -{\mathrm e}\right )\)	\(28\)
derivativedivides	\(64 \,{\mathrm e} \left (\left (x -{\mathrm e}\right ) \ln \left (x -{\mathrm e}\right )-x +{\mathrm e}\right )+32 \ln \left (x -{\mathrm e}\right ) \left (x -{\mathrm e}\right )^{2}+32 \ln \left (x -{\mathrm e}\right ) {\mathrm e}^{2}+64 \,{\mathrm e} \left (x -{\mathrm e}\right )+8 \,{\mathrm e}^{{\mathrm e}^{6}} \ln \left (x -{\mathrm e}\right )\)	\(80\)
default	\(64 \,{\mathrm e} \left (\left (x -{\mathrm e}\right ) \ln \left (x -{\mathrm e}\right )-x +{\mathrm e}\right )+32 \ln \left (x -{\mathrm e}\right ) \left (x -{\mathrm e}\right )^{2}+32 \ln \left (x -{\mathrm e}\right ) {\mathrm e}^{2}+64 \,{\mathrm e} \left (x -{\mathrm e}\right )+8 \,{\mathrm e}^{{\mathrm e}^{6}} \ln \left (x -{\mathrm e}\right )\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x*exp(1)-64*x^2)*ln(x-exp(1))-8*exp(exp(3)^2)-32*x^2)/(exp(1)-x),x,method=_RETURNVERBOSE)

[Out]

32*ln(x-exp(1))*x^2+8*exp(exp(6))*ln(x-exp(1))

________________________________________________________________________________________

maxima [B]  time = 0.38, size = 125, normalized size = 6.58 \begin {gather*} -64 \, {\left (e \log \left (x - e\right ) + x\right )} e \log \left (x - e\right ) - 32 \, e^{2} \log \left (x - e\right )^{2} + 32 \, {\left (e \log \left (x - e\right )^{2} + 2 \, e \log \left (x - e\right ) + 2 \, x\right )} e - 64 \, x e + 32 \, {\left (x^{2} + 2 \, x e + 2 \, e^{2} \log \left (x - e\right )\right )} \log \left (x - e\right ) - 64 \, e^{2} \log \left (x - e\right ) + 8 \, e^{\left (e^{6}\right )} \log \left (x - e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(1)-64*x^2)*log(x-exp(1))-8*exp(exp(3)^2)-32*x^2)/(exp(1)-x),x, algorithm="maxima")

[Out]

-64*(e*log(x - e) + x)*e*log(x - e) - 32*e^2*log(x - e)^2 + 32*(e*log(x - e)^2 + 2*e*log(x - e) + 2*x)*e - 64*
x*e + 32*(x^2 + 2*x*e + 2*e^2*log(x - e))*log(x - e) - 64*e^2*log(x - e) + 8*e^(e^6)*log(x - e)

________________________________________________________________________________________

mupad [B]  time = 0.40, size = 18, normalized size = 0.95 \begin {gather*} 8\,\ln \left (x-\mathrm {e}\right )\,\left (4\,x^2+{\mathrm {e}}^{{\mathrm {e}}^6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(exp(6)) - log(x - exp(1))*(64*x*exp(1) - 64*x^2) + 32*x^2)/(x - exp(1)),x)

[Out]

8*log(x - exp(1))*(exp(exp(6)) + 4*x^2)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 24, normalized size = 1.26 \begin {gather*} 32 x^{2} \log {\left (x - e \right )} + 8 e^{e^{6}} \log {\left (x - e \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(1)-64*x**2)*ln(x-exp(1))-8*exp(exp(3)**2)-32*x**2)/(exp(1)-x),x)

[Out]

32*x**2*log(x - E) + 8*exp(exp(6))*log(x - E)

________________________________________________________________________________________