3.30.35 \(\int \frac {e^x (-7-e^{2-2 x}+6 e^{1-x}+\log (4))+e^x (-7-e^{2-2 x}+6 e^{1-x}+\log (4)) \log (x)+e^x (-7 x+e^{2-2 x} x+x \log (4)) \log (x) \log (x \log (x))}{x \log (x)} \, dx\)

Optimal. Leaf size=26 \[ e^x \left (2-\left (-3+e^{1-x}\right )^2+\log (4)\right ) \log (x \log (x)) \]

________________________________________________________________________________________

Rubi [A]  time = 1.02, antiderivative size = 42, normalized size of antiderivative = 1.62, number of steps used = 7, number of rules used = 4, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6742, 2365, 43, 2288} \begin {gather*} 6 e \log (x)+6 e \log (\log (x))-e^{2-x} \log (x \log (x))-e^x (7-\log (4)) \log (x \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-7 - E^(2 - 2*x) + 6*E^(1 - x) + Log[4]) + E^x*(-7 - E^(2 - 2*x) + 6*E^(1 - x) + Log[4])*Log[x] + E^
x*(-7*x + E^(2 - 2*x)*x + x*Log[4])*Log[x]*Log[x*Log[x]])/(x*Log[x]),x]

[Out]

6*E*Log[x] + 6*E*Log[Log[x]] - E^(2 - x)*Log[x*Log[x]] - E^x*(7 - Log[4])*Log[x*Log[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 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]

Rule 2365

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(c_.)*(x_)^(n_.)]*(e_.))^(q_.))/(x_), x_Symbol]
:> Dist[1/n, Subst[Int[(a + b*x)^p*(d + e*x)^q, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, e, n, p, q}, 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 {6 e (1+\log (x))}{x \log (x)}+\frac {e^{2-x} (-1-\log (x)+x \log (x) \log (x \log (x)))}{x \log (x)}+\frac {e^x (-7+\log (4)) (1+\log (x)+x \log (x) \log (x \log (x)))}{x \log (x)}\right ) \, dx\\ &=(6 e) \int \frac {1+\log (x)}{x \log (x)} \, dx+(-7+\log (4)) \int \frac {e^x (1+\log (x)+x \log (x) \log (x \log (x)))}{x \log (x)} \, dx+\int \frac {e^{2-x} (-1-\log (x)+x \log (x) \log (x \log (x)))}{x \log (x)} \, dx\\ &=-e^{2-x} \log (x \log (x))-e^x (7-\log (4)) \log (x \log (x))+(6 e) \operatorname {Subst}\left (\int \frac {1+x}{x} \, dx,x,\log (x)\right )\\ &=-e^{2-x} \log (x \log (x))-e^x (7-\log (4)) \log (x \log (x))+(6 e) \operatorname {Subst}\left (\int \left (1+\frac {1}{x}\right ) \, dx,x,\log (x)\right )\\ &=6 e \log (x)+6 e \log (\log (x))-e^{2-x} \log (x \log (x))-e^x (7-\log (4)) \log (x \log (x))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 35, normalized size = 1.35 \begin {gather*} 6 e \log (x)+6 e \log (\log (x))-e^x \left (7+e^{2-2 x}-\log (4)\right ) \log (x \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-7 - E^(2 - 2*x) + 6*E^(1 - x) + Log[4]) + E^x*(-7 - E^(2 - 2*x) + 6*E^(1 - x) + Log[4])*Log[x
] + E^x*(-7*x + E^(2 - 2*x)*x + x*Log[4])*Log[x]*Log[x*Log[x]])/(x*Log[x]),x]

[Out]

6*E*Log[x] + 6*E*Log[Log[x]] - E^x*(7 + E^(2 - 2*x) - Log[4])*Log[x*Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 32, normalized size = 1.23 \begin {gather*} {\left ({\left (2 \, \log \relax (2) - 7\right )} e^{\left (2 \, x\right )} - e^{2} + 6 \, e^{\left (x + 1\right )}\right )} e^{\left (-x\right )} \log \left (x \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(-x+1)^2+2*x*log(2)-7*x)*exp(x)*log(x)*log(x*log(x))+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*ex
p(x)*log(x)+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*exp(x))/x/log(x),x, algorithm="fricas")

[Out]

((2*log(2) - 7)*e^(2*x) - e^2 + 6*e^(x + 1))*e^(-x)*log(x*log(x))

________________________________________________________________________________________

giac [B]  time = 0.23, size = 65, normalized size = 2.50 \begin {gather*} 2 \, e^{x} \log \relax (2) \log \relax (x) + 2 \, e^{x} \log \relax (2) \log \left (\log \relax (x)\right ) + 6 \, e \log \relax (x) - 7 \, e^{x} \log \relax (x) - e^{\left (-x + 2\right )} \log \relax (x) + 6 \, e \log \left (\log \relax (x)\right ) - 7 \, e^{x} \log \left (\log \relax (x)\right ) - e^{\left (-x + 2\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(-x+1)^2+2*x*log(2)-7*x)*exp(x)*log(x)*log(x*log(x))+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*ex
p(x)*log(x)+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*exp(x))/x/log(x),x, algorithm="giac")

[Out]

2*e^x*log(2)*log(x) + 2*e^x*log(2)*log(log(x)) + 6*e*log(x) - 7*e^x*log(x) - e^(-x + 2)*log(x) + 6*e*log(log(x
)) - 7*e^x*log(log(x)) - e^(-x + 2)*log(log(x))

________________________________________________________________________________________

maple [C]  time = 0.16, size = 340, normalized size = 13.08

method	result	size
risch	\(-\left (-2 \ln \relax (2) {\mathrm e}^{2 x}+{\mathrm e}^{2}+7 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x} \ln \left (\ln \relax (x )\right )+\frac {\left (-7 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} {\mathrm e}^{2 x}-i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}+2 i \pi \ln \relax (2) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} {\mathrm e}^{2 x}+i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )-i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}+7 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right ) {\mathrm e}^{2 x}-2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} {\mathrm e}^{2 x}+2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} {\mathrm e}^{2 x}+7 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} {\mathrm e}^{2 x}+i \pi \,{\mathrm e}^{2} \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3}-2 i \pi \ln \relax (2) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right ) {\mathrm e}^{2 x}-7 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} {\mathrm e}^{2 x}+4 \ln \relax (2) {\mathrm e}^{2 x} \ln \relax (x )+12 \ln \relax (x ) {\mathrm e}^{x +1}-2 \,{\mathrm e}^{2} \ln \relax (x )+12 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{x +1}-14 \,{\mathrm e}^{2 x} \ln \relax (x )\right ) {\mathrm e}^{-x}}{2}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(1-x)^2+2*x*ln(2)-7*x)*exp(x)*ln(x)*ln(x*ln(x))+(-exp(1-x)^2+6*exp(1-x)+2*ln(2)-7)*exp(x)*ln(x)+(-e
xp(1-x)^2+6*exp(1-x)+2*ln(2)-7)*exp(x))/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

-(-2*ln(2)*exp(2*x)+exp(2)+7*exp(2*x))*exp(-x)*ln(ln(x))+1/2*(-7*I*Pi*csgn(I*x)*csgn(I*x*ln(x))^2*exp(2*x)-I*P
i*exp(2)*csgn(I*ln(x))*csgn(I*x*ln(x))^2+2*I*Pi*ln(2)*csgn(I*ln(x))*csgn(I*x*ln(x))^2*exp(2*x)+I*Pi*exp(2)*csg
n(I*x)*csgn(I*ln(x))*csgn(I*x*ln(x))-I*Pi*exp(2)*csgn(I*x)*csgn(I*x*ln(x))^2+7*I*Pi*csgn(I*x)*csgn(I*ln(x))*cs
gn(I*x*ln(x))*exp(2*x)-2*I*Pi*ln(2)*csgn(I*x*ln(x))^3*exp(2*x)+2*I*Pi*ln(2)*csgn(I*x)*csgn(I*x*ln(x))^2*exp(2*
x)+7*I*Pi*csgn(I*x*ln(x))^3*exp(2*x)+I*Pi*exp(2)*csgn(I*x*ln(x))^3-2*I*Pi*ln(2)*csgn(I*x)*csgn(I*ln(x))*csgn(I
*x*ln(x))*exp(2*x)-7*I*Pi*csgn(I*ln(x))*csgn(I*x*ln(x))^2*exp(2*x)+4*ln(2)*exp(2*x)*ln(x)+12*ln(x)*exp(x+1)-2*
exp(2)*ln(x)+12*ln(ln(x))*exp(x+1)-14*exp(2*x)*ln(x))*exp(-x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\rm Ei}\left (-x\right ) e^{2} + {\left ({\left (2 \, \log \relax (2) - 7\right )} e^{\left (2 \, x\right )} \log \relax (x) - e^{2} \log \relax (x) + {\left ({\left (2 \, \log \relax (2) - 7\right )} e^{\left (2 \, x\right )} - e^{2}\right )} \log \left (\log \relax (x)\right )\right )} e^{\left (-x\right )} - {\left (2 \, \log \relax (2) - 7\right )} \int \frac {e^{x}}{x}\,{d x} + 2 \, {\rm Ei}\relax (x) \log \relax (2) + 6 \, e \log \relax (x) + 6 \, e \log \left (\log \relax (x)\right ) - 7 \, {\rm Ei}\relax (x) + \int \frac {e^{\left (-x + 2\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(-x+1)^2+2*x*log(2)-7*x)*exp(x)*log(x)*log(x*log(x))+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*ex
p(x)*log(x)+(-exp(-x+1)^2+6*exp(-x+1)+2*log(2)-7)*exp(x))/x/log(x),x, algorithm="maxima")

[Out]

-Ei(-x)*e^2 + ((2*log(2) - 7)*e^(2*x)*log(x) - e^2*log(x) + ((2*log(2) - 7)*e^(2*x) - e^2)*log(log(x)))*e^(-x)
- (2*log(2) - 7)*integrate(e^x/x, x) + 2*Ei(x)*log(2) + 6*e*log(x) + 6*e*log(log(x)) - 7*Ei(x) + integrate(e^
(-x + 2)/x, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (2\,\ln \relax (2)+6\,{\mathrm {e}}^{1-x}-{\mathrm {e}}^{2-2\,x}-7\right )+{\mathrm {e}}^x\,\ln \relax (x)\,\left (2\,\ln \relax (2)+6\,{\mathrm {e}}^{1-x}-{\mathrm {e}}^{2-2\,x}-7\right )+\ln \left (x\,\ln \relax (x)\right )\,{\mathrm {e}}^x\,\ln \relax (x)\,\left (2\,x\,\ln \relax (2)-7\,x+x\,{\mathrm {e}}^{2-2\,x}\right )}{x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(2*log(2) + 6*exp(1 - x) - exp(2 - 2*x) - 7) + exp(x)*log(x)*(2*log(2) + 6*exp(1 - x) - exp(2 - 2*
x) - 7) + log(x*log(x))*exp(x)*log(x)*(2*x*log(2) - 7*x + x*exp(2 - 2*x)))/(x*log(x)),x)

[Out]

int((exp(x)*(2*log(2) + 6*exp(1 - x) - exp(2 - 2*x) - 7) + exp(x)*log(x)*(2*log(2) + 6*exp(1 - x) - exp(2 - 2*
x) - 7) + log(x*log(x))*exp(x)*log(x)*(2*x*log(2) - 7*x + x*exp(2 - 2*x)))/(x*log(x)), x)

________________________________________________________________________________________

sympy [B]  time = 20.57, size = 54, normalized size = 2.08 \begin {gather*} \left (- 7 \log {\left (x \log {\relax (x )} \right )} + 2 \log {\relax (2 )} \log {\left (x \log {\relax (x )} \right )}\right ) e^{x} + 6 e \log {\relax (x )} + 6 e \log {\left (\log {\relax (x )} \right )} - e^{2} e^{- x} \log {\left (x \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(-x+1)**2+2*x*ln(2)-7*x)*exp(x)*ln(x)*ln(x*ln(x))+(-exp(-x+1)**2+6*exp(-x+1)+2*ln(2)-7)*exp(x
)*ln(x)+(-exp(-x+1)**2+6*exp(-x+1)+2*ln(2)-7)*exp(x))/x/ln(x),x)

[Out]

(-7*log(x*log(x)) + 2*log(2)*log(x*log(x)))*exp(x) + 6*E*log(x) + 6*E*log(log(x)) - exp(2)*exp(-x)*log(x*log(x
))

________________________________________________________________________________________