3.44.15 \(\int \frac {8 e^{3+e^{e^3}}-27 x+(e^{3+e^{e^3}}-3 x) \log (\frac {-e^{3+e^{e^3}}+3 x}{2 x})}{e^{3+e^{e^3}}-3 x} \, dx\)

Optimal. Leaf size=28 \[ x \left (9+\frac {1}{x}+\log \left (\frac {1}{2} \left (3-\frac {e^{3+e^{e^3}}}{x}\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6742, 43, 2448, 263, 31} \begin {gather*} 9 x+x \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^(3 + E^E^3) - 27*x + (E^(3 + E^E^3) - 3*x)*Log[(-E^(3 + E^E^3) + 3*x)/(2*x)])/(E^(3 + E^E^3) - 3*x),x
]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{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} &=\int \left (\frac {8 e^{3+e^{e^3}}-27 x}{e^{3+e^{e^3}}-3 x}+\log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right )\right ) \, dx\\ &=\int \frac {8 e^{3+e^{e^3}}-27 x}{e^{3+e^{e^3}}-3 x} \, dx+\int \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right ) \, dx\\ &=x \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right )-\frac {1}{2} e^{3+e^{e^3}} \int \frac {1}{\left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right ) x} \, dx+\int \left (9-\frac {e^{3+e^{e^3}}}{e^{3+e^{e^3}}-3 x}\right ) \, dx\\ &=9 x+x \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right )+\frac {1}{3} e^{3+e^{e^3}} \log \left (e^{3+e^{e^3}}-3 x\right )-\frac {1}{2} e^{3+e^{e^3}} \int \frac {1}{-\frac {1}{2} e^{3+e^{e^3}}+\frac {3 x}{2}} \, dx\\ &=9 x+x \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.14, size = 81, normalized size = 2.89 \begin {gather*} \frac {1}{3} \left (27 x-\left (e^{3+e^{e^3}}-3 x\right ) \log \left (\frac {3}{2}-\frac {e^{3+e^{e^3}}}{2 x}\right )+e^{3+e^{e^3}} \log \left (e^{3+e^{e^3}}-3 x\right )-e^{3+e^{e^3}} \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^(3 + E^E^3) - 27*x + (E^(3 + E^E^3) - 3*x)*Log[(-E^(3 + E^E^3) + 3*x)/(2*x)])/(E^(3 + E^E^3) -
3*x),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 24, normalized size = 0.86 \begin {gather*} x \log \left (\frac {3 \, x - e^{\left (e^{\left (e^{3}\right )} + 3\right )}}{2 \, x}\right ) + 9 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(exp(3))+3)-3*x)*log(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+8*exp(exp(exp(3))+3)-27*x)/(exp(exp(e
xp(3))+3)-3*x),x, algorithm="fricas")

[Out]

x*log(1/2*(3*x - e^(e^(e^3) + 3))/x) + 9*x

________________________________________________________________________________________

giac [B]  time = 0.31, size = 68, normalized size = 2.43 \begin {gather*} -\frac {{\left (e^{\left (2 \, e^{\left (e^{3}\right )} + 6\right )} \log \left (\frac {3 \, x - e^{\left (e^{\left (e^{3}\right )} + 3\right )}}{2 \, x}\right ) + 9 \, e^{\left (2 \, e^{\left (e^{3}\right )} + 6\right )}\right )} e^{\left (-e^{\left (e^{3}\right )} - 3\right )}}{\frac {3 \, x - e^{\left (e^{\left (e^{3}\right )} + 3\right )}}{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(exp(3))+3)-3*x)*log(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+8*exp(exp(exp(3))+3)-27*x)/(exp(exp(e
xp(3))+3)-3*x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.13, size = 25, normalized size = 0.89

method	result	size
norman	\(x \ln \left (\frac {-{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}+3 x}{2 x}\right )+9 x\)	\(25\)
risch	\(x \ln \left (\frac {-{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}+3 x}{2 x}\right )+9 x\)	\(25\)
derivativedivides	\(2 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3} \left (\frac {\ln \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right ) \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{3}}-3} x}{3}+\frac {9 \,{\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{3}}-3} x}{2}+\frac {\ln \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right )}{6}\right )\)	\(75\)
default	\(2 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3} \left (\frac {\ln \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right ) \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{3}}-3} x}{3}+\frac {9 \,{\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{3}}-3} x}{2}+\frac {\ln \left (\frac {3}{2}-\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}+3}}{2 x}\right )}{6}\right )\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(exp(exp(3))+3)-3*x)*ln(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+8*exp(exp(exp(3))+3)-27*x)/(exp(exp(exp(3))+
3)-3*x),x,method=_RETURNVERBOSE)

[Out]

x*ln(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+9*x

________________________________________________________________________________________

maxima [B]  time = 0.49, size = 61, normalized size = 2.18 \begin {gather*} -x {\left (\log \relax (2) - 9\right )} + \frac {1}{3} \, {\left (3 \, x + 8 \, e^{\left (e^{\left (e^{3}\right )} + 3\right )}\right )} \log \left (3 \, x - e^{\left (e^{\left (e^{3}\right )} + 3\right )}\right ) - \frac {8}{3} \, e^{\left (e^{\left (e^{3}\right )} + 3\right )} \log \left (3 \, x - e^{\left (e^{\left (e^{3}\right )} + 3\right )}\right ) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(exp(3))+3)-3*x)*log(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+8*exp(exp(exp(3))+3)-27*x)/(exp(exp(e
xp(3))+3)-3*x),x, algorithm="maxima")

[Out]

-x*(log(2) - 9) + 1/3*(3*x + 8*e^(e^(e^3) + 3))*log(3*x - e^(e^(e^3) + 3)) - 8/3*e^(e^(e^3) + 3)*log(3*x - e^(
e^(e^3) + 3)) - x*log(x)

________________________________________________________________________________________

mupad [B]  time = 3.84, size = 22, normalized size = 0.79 \begin {gather*} x\,\left (\ln \left (\frac {3\,x-{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}}\,{\mathrm {e}}^3}{2\,x}\right )+9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((27*x - 8*exp(exp(exp(3)) + 3) + log(((3*x)/2 - exp(exp(exp(3)) + 3)/2)/x)*(3*x - exp(exp(exp(3)) + 3)))/(
3*x - exp(exp(exp(3)) + 3)),x)

[Out]

x*(log((3*x - exp(exp(exp(3)))*exp(3))/(2*x)) + 9)

________________________________________________________________________________________

sympy [A]  time = 0.18, size = 22, normalized size = 0.79 \begin {gather*} x \log {\left (\frac {\frac {3 x}{2} - \frac {e^{3 + e^{e^{3}}}}{2}}{x} \right )} + 9 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(exp(3))+3)-3*x)*ln(1/2*(-exp(exp(exp(3))+3)+3*x)/x)+8*exp(exp(exp(3))+3)-27*x)/(exp(exp(ex
p(3))+3)-3*x),x)

[Out]

x*log((3*x/2 - exp(3 + exp(exp(3)))/2)/x) + 9*x

________________________________________________________________________________________