3.77.98 \(\int \frac {e^x x+(1-2 x+e^x (-4 x-x^2)) \log (x)+(1+e^x (3+x)) \log ^2(x)+(e^x (1-2 x-x^2) \log (x)+e^x (1+x) \log ^2(x)) \log (\frac {x^2}{5 \log (x)})}{\log (x)} \, dx\)

Optimal. Leaf size=32 \[ x (x-\log (x)) \left (-1-e^x-e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \]

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Rubi [B]  time = 1.46, antiderivative size = 70, normalized size of antiderivative = 2.19, number of steps used = 4, number of rules used = 3, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6742, 2295, 2288} \begin {gather*} -x^2-\frac {e^x \left (-x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x^2 \log (x)+x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x \log ^2(x)\right )}{\log (x)}+x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*x + (1 - 2*x + E^x*(-4*x - x^2))*Log[x] + (1 + E^x*(3 + x))*Log[x]^2 + (E^x*(1 - 2*x - x^2)*Log[x] +
E^x*(1 + x)*Log[x]^2)*Log[x^2/(5*Log[x])])/Log[x],x]

[Out]

-x^2 + x*Log[x] - (E^x*(x^2*Log[x] - x*Log[x]^2 + x^2*Log[x]*Log[x^2/(5*Log[x])] - x*Log[x]^2*Log[x^2/(5*Log[x
])]))/Log[x]

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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 (1-2 x+\log (x)+\frac {e^x \left (x-4 x \log (x)-x^2 \log (x)+3 \log ^2(x)+x \log ^2(x)+\log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-2 x \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)}\right ) \, dx\\ &=x-x^2+\int \log (x) \, dx+\int \frac {e^x \left (x-4 x \log (x)-x^2 \log (x)+3 \log ^2(x)+x \log ^2(x)+\log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-2 x \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)} \, dx\\ &=-x^2+x \log (x)-\frac {e^x \left (x^2 \log (x)-x \log ^2(x)+x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )\right )}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 30, normalized size = 0.94 \begin {gather*} -x (x-\log (x)) \left (1+e^x+e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x + (1 - 2*x + E^x*(-4*x - x^2))*Log[x] + (1 + E^x*(3 + x))*Log[x]^2 + (E^x*(1 - 2*x - x^2)*Log
[x] + E^x*(1 + x)*Log[x]^2)*Log[x^2/(5*Log[x])])/Log[x],x]

[Out]

-(x*(x - Log[x])*(1 + E^x + E^x*Log[x^2/(5*Log[x])]))

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fricas [A]  time = 0.54, size = 48, normalized size = 1.50 \begin {gather*} -x^{2} e^{x} - x^{2} + {\left (x e^{x} + x\right )} \log \relax (x) - {\left (x^{2} e^{x} - x e^{x} \log \relax (x)\right )} \log \left (\frac {x^{2}}{5 \, \log \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/log(x))+((3+x)*exp(x)+1)*log(x)^2+((
-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)*x)/log(x),x, algorithm="fricas")

[Out]

-x^2*e^x - x^2 + (x*e^x + x)*log(x) - (x^2*e^x - x*e^x*log(x))*log(1/5*x^2/log(x))

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giac [B]  time = 0.17, size = 64, normalized size = 2.00 \begin {gather*} -2 \, x^{2} e^{x} \log \relax (x) + 2 \, x e^{x} \log \relax (x)^{2} + x^{2} e^{x} \log \left (5 \, \log \relax (x)\right ) - x e^{x} \log \relax (x) \log \left (5 \, \log \relax (x)\right ) - x^{2} e^{x} + x e^{x} \log \relax (x) - x^{2} + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/log(x))+((3+x)*exp(x)+1)*log(x)^2+((
-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)*x)/log(x),x, algorithm="giac")

[Out]

-2*x^2*e^x*log(x) + 2*x*e^x*log(x)^2 + x^2*e^x*log(5*log(x)) - x*e^x*log(x)*log(5*log(x)) - x^2*e^x + x*e^x*lo
g(x) - x^2 + x*log(x)

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maple [C]  time = 0.23, size = 437, normalized size = 13.66

method	result	size
risch	\(x^{2} \ln \relax (5) {\mathrm e}^{x}+2 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+x \,{\mathrm e}^{x} \ln \relax (x )-2 x^{2} {\mathrm e}^{x} \ln \relax (x )-x^{2}-{\mathrm e}^{x} x^{2}+x \ln \relax (x )-x \ln \relax (5) {\mathrm e}^{x} \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+\left ({\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x} \ln \relax (x )\right ) \ln \left (\ln \relax (x )\right )+\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{3} {\mathrm e}^{x} \ln \relax (x )}{2}+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x} \ln \relax (x )+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x} \ln \relax (x )}{2}-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \ln \relax (x )}{2}-i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2}}{\ln \relax (x )}\right )^{3} {\mathrm e}^{x}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}\)	\(437\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(x)*ln(x)^2+(-x^2-2*x+1)*exp(x)*ln(x))*ln(1/5*x^2/ln(x))+((3+x)*exp(x)+1)*ln(x)^2+((-x^2-4*x)*e
xp(x)+1-2*x)*ln(x)+exp(x)*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

x^2*ln(5)*exp(x)+2*x*exp(x)*ln(x)^2+x*exp(x)*ln(x)-2*x^2*exp(x)*ln(x)-x^2-exp(x)*x^2+x*ln(x)-x*ln(5)*exp(x)*ln
(x)-1/2*I*Pi*x*csgn(I*x^2/ln(x))*csgn(I*x^2)*csgn(I/ln(x))*exp(x)*ln(x)+1/2*I*Pi*x*csgn(I*x^2/ln(x))^2*csgn(I*
x^2)*exp(x)*ln(x)-1/2*I*Pi*x^2*csgn(I*x^2/ln(x))^2*csgn(I/ln(x))*exp(x)+1/2*I*Pi*x^2*csgn(I*x^2/ln(x))*csgn(I*
x^2)*csgn(I/ln(x))*exp(x)-1/2*I*Pi*x^2*csgn(I*x^2/ln(x))^2*csgn(I*x^2)*exp(x)+(exp(x)*x^2-x*exp(x)*ln(x))*ln(l
n(x))+1/2*I*Pi*x*csgn(I*x^2/ln(x))^2*csgn(I/ln(x))*exp(x)*ln(x)-1/2*I*Pi*x*csgn(I*x^2/ln(x))^3*exp(x)*ln(x)+I*
Pi*x*csgn(I*x)*csgn(I*x^2)^2*exp(x)*ln(x)+1/2*I*Pi*x^2*csgn(I*x^2)^3*exp(x)-1/2*I*Pi*x*csgn(I*x^2)^3*exp(x)*ln
(x)-1/2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)*exp(x)*ln(x)-I*Pi*x^2*csgn(I*x)*csgn(I*x^2)^2*exp(x)+1/2*I*Pi*x^2*csgn(
I*x^2/ln(x))^3*exp(x)+1/2*I*Pi*x^2*csgn(I*x)^2*csgn(I*x^2)*exp(x)

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maxima [B]  time = 0.48, size = 81, normalized size = 2.53 \begin {gather*} {\left (x^{2} - x \log \relax (x)\right )} e^{x} \log \left (\log \relax (x)\right ) - x^{2} + {\left (x^{2} \log \relax (5) + 2 \, x \log \relax (x)^{2} - {\left (2 \, x^{2} + x {\left (\log \relax (5) - 1\right )}\right )} \log \relax (x) + 2 \, x - 2\right )} e^{x} - {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 4 \, {\left (x - 1\right )} e^{x} + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/log(x))+((3+x)*exp(x)+1)*log(x)^2+((
-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)*x)/log(x),x, algorithm="maxima")

[Out]

(x^2 - x*log(x))*e^x*log(log(x)) - x^2 + (x^2*log(5) + 2*x*log(x)^2 - (2*x^2 + x*(log(5) - 1))*log(x) + 2*x -
2)*e^x - (x^2 - 2*x + 2)*e^x - 4*(x - 1)*e^x + x*log(x)

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mupad [B]  time = 5.97, size = 57, normalized size = 1.78 \begin {gather*} \ln \left (\frac {x^2}{5\,\ln \relax (x)}\right )\,\left (\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x}-{\mathrm {e}}^x+x\,{\mathrm {e}}^x\,\ln \relax (x)\right )-x^2\,{\mathrm {e}}^x+\ln \relax (x)\,\left (x+x\,{\mathrm {e}}^x\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(exp(x)*(x + 3) + 1) - log(x^2/(5*log(x)))*(exp(x)*log(x)*(2*x + x^2 - 1) - exp(x)*log(x)^2*(x +
1)) + x*exp(x) - log(x)*(2*x + exp(x)*(4*x + x^2) - 1))/log(x),x)

[Out]

log(x^2/(5*log(x)))*((exp(x)*(x - x^3))/x - exp(x) + x*exp(x)*log(x)) - x^2*exp(x) + log(x)*(x + x*exp(x)) - x
^2

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sympy [A]  time = 6.62, size = 48, normalized size = 1.50 \begin {gather*} - x^{2} + x \log {\relax (x )} + \left (- x^{2} \log {\left (\frac {x^{2}}{5 \log {\relax (x )}} \right )} - x^{2} + x \log {\relax (x )} \log {\left (\frac {x^{2}}{5 \log {\relax (x )}} \right )} + x \log {\relax (x )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)*ln(x)**2+(-x**2-2*x+1)*exp(x)*ln(x))*ln(1/5*x**2/ln(x))+((3+x)*exp(x)+1)*ln(x)**2+((-
x**2-4*x)*exp(x)+1-2*x)*ln(x)+exp(x)*x)/ln(x),x)

[Out]

-x**2 + x*log(x) + (-x**2*log(x**2/(5*log(x))) - x**2 + x*log(x)*log(x**2/(5*log(x))) + x*log(x))*exp(x)

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