3.84.28 \(\int \frac {2 e^{2 x} \log ^2(x)+4 e^x \log ^4(x)+2 \log ^6(x)+e^{\frac {3 x+15 \log (x)}{\log (x)}} (-3 e^x x^2+3 e^x x^2 \log (x)+(-3 x^2+e^x (2 x-x^2)) \log ^2(x)+(-2 x+3 x^2) \log ^3(x)+2 x \log ^4(x))}{2 e^{2 x} \log ^2(x)+4 e^x \log ^4(x)+2 \log ^6(x)} \, dx\)

Optimal. Leaf size=31 \[ x+\frac {e^{3 \left (5+\frac {x}{\log (x)}\right )} x^2}{2 \left (e^x+\log ^2(x)\right )} \]

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Rubi [B]  time = 4.84, antiderivative size = 72, normalized size of antiderivative = 2.32, number of steps used = 5, number of rules used = 4, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6741, 12, 6742, 2288} \begin {gather*} x+\frac {x e^{\frac {3 x}{\log (x)}+15} \left (e^x x-x \log ^3(x)+x \log ^2(x)-e^x x \log (x)\right )}{2 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x) \left (e^x+\log ^2(x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(2*x)*Log[x]^2 + 4*E^x*Log[x]^4 + 2*Log[x]^6 + E^((3*x + 15*Log[x])/Log[x])*(-3*E^x*x^2 + 3*E^x*x^2*L
og[x] + (-3*x^2 + E^x*(2*x - x^2))*Log[x]^2 + (-2*x + 3*x^2)*Log[x]^3 + 2*x*Log[x]^4))/(2*E^(2*x)*Log[x]^2 + 4
*E^x*Log[x]^4 + 2*Log[x]^6),x]

[Out]

x + (E^(15 + (3*x)/Log[x])*x*(E^x*x - E^x*x*Log[x] + x*Log[x]^2 - x*Log[x]^3))/(2*(Log[x]^(-2) - Log[x]^(-1))*
Log[x]^2*(E^x + Log[x]^2)^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 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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

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 \frac {2 e^{2 x} \log ^2(x)+4 e^x \log ^4(x)+2 \log ^6(x)+e^{\frac {3 x+15 \log (x)}{\log (x)}} \left (-3 e^x x^2+3 e^x x^2 \log (x)+\left (-3 x^2+e^x \left (2 x-x^2\right )\right ) \log ^2(x)+\left (-2 x+3 x^2\right ) \log ^3(x)+2 x \log ^4(x)\right )}{2 \log ^2(x) \left (e^x+\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {2 e^{2 x} \log ^2(x)+4 e^x \log ^4(x)+2 \log ^6(x)+e^{\frac {3 x+15 \log (x)}{\log (x)}} \left (-3 e^x x^2+3 e^x x^2 \log (x)+\left (-3 x^2+e^x \left (2 x-x^2\right )\right ) \log ^2(x)+\left (-2 x+3 x^2\right ) \log ^3(x)+2 x \log ^4(x)\right )}{\log ^2(x) \left (e^x+\log ^2(x)\right )^2} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {e^{15+\frac {3 x}{\log (x)}} x \left (-3 e^x x+3 e^x x \log (x)+2 e^x \log ^2(x)-3 x \log ^2(x)-e^x x \log ^2(x)-2 \log ^3(x)+3 x \log ^3(x)+2 \log ^4(x)\right )}{\log ^2(x) \left (e^x+\log ^2(x)\right )^2}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {e^{15+\frac {3 x}{\log (x)}} x \left (-3 e^x x+3 e^x x \log (x)+2 e^x \log ^2(x)-3 x \log ^2(x)-e^x x \log ^2(x)-2 \log ^3(x)+3 x \log ^3(x)+2 \log ^4(x)\right )}{\log ^2(x) \left (e^x+\log ^2(x)\right )^2} \, dx\\ &=x+\frac {e^{15+\frac {3 x}{\log (x)}} x \left (e^x x-e^x x \log (x)+x \log ^2(x)-x \log ^3(x)\right )}{2 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x) \left (e^x+\log ^2(x)\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 33, normalized size = 1.06 \begin {gather*} \frac {1}{2} \left (2 x+\frac {e^{15+\frac {3 x}{\log (x)}} x^2}{e^x+\log ^2(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(2*x)*Log[x]^2 + 4*E^x*Log[x]^4 + 2*Log[x]^6 + E^((3*x + 15*Log[x])/Log[x])*(-3*E^x*x^2 + 3*E^x
*x^2*Log[x] + (-3*x^2 + E^x*(2*x - x^2))*Log[x]^2 + (-2*x + 3*x^2)*Log[x]^3 + 2*x*Log[x]^4))/(2*E^(2*x)*Log[x]
^2 + 4*E^x*Log[x]^4 + 2*Log[x]^6),x]

[Out]

(2*x + (E^(15 + (3*x)/Log[x])*x^2)/(E^x + Log[x]^2))/2

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fricas [A]  time = 0.79, size = 41, normalized size = 1.32 \begin {gather*} \frac {x^{2} e^{\left (\frac {3 \, {\left (x + 5 \, \log \relax (x)\right )}}{\log \relax (x)}\right )} + 2 \, x \log \relax (x)^{2} + 2 \, x e^{x}}{2 \, {\left (\log \relax (x)^{2} + e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)^4+(3*x^2-2*x)*log(x)^3+((-x^2+2*x)*exp(x)-3*x^2)*log(x)^2+3*x^2*exp(x)*log(x)-3*exp(x)*
x^2)*exp((15*log(x)+3*x)/log(x))+2*log(x)^6+4*exp(x)*log(x)^4+2*exp(x)^2*log(x)^2)/(2*log(x)^6+4*exp(x)*log(x)
^4+2*exp(x)^2*log(x)^2),x, algorithm="fricas")

[Out]

1/2*(x^2*e^(3*(x + 5*log(x))/log(x)) + 2*x*log(x)^2 + 2*x*e^x)/(log(x)^2 + e^x)

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giac [B]  time = 0.17, size = 119, normalized size = 3.84 \begin {gather*} \frac {2 \, x^{3} e^{x} \log \relax (x)^{2} + x^{4} e^{\left (x + \frac {3 \, {\left (x + 5 \, \log \relax (x)\right )}}{\log \relax (x)}\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 2 \, x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} e^{\left (\frac {3 \, {\left (x + 5 \, \log \relax (x)\right )}}{\log \relax (x)}\right )} - 4 \, x e^{x} \log \relax (x) + 8 \, x \log \relax (x)^{2} + 8 \, x e^{x}}{2 \, {\left (x^{2} e^{x} \log \relax (x)^{2} + x^{2} e^{\left (2 \, x\right )} + 4 \, \log \relax (x)^{2} + 4 \, e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)^4+(3*x^2-2*x)*log(x)^3+((-x^2+2*x)*exp(x)-3*x^2)*log(x)^2+3*x^2*exp(x)*log(x)-3*exp(x)*
x^2)*exp((15*log(x)+3*x)/log(x))+2*log(x)^6+4*exp(x)*log(x)^4+2*exp(x)^2*log(x)^2)/(2*log(x)^6+4*exp(x)*log(x)
^4+2*exp(x)^2*log(x)^2),x, algorithm="giac")

[Out]

1/2*(2*x^3*e^x*log(x)^2 + x^4*e^(x + 3*(x + 5*log(x))/log(x)) + 2*x^3*e^(2*x) + 2*x^2*e^(2*x) + 4*x^2*e^(3*(x
+ 5*log(x))/log(x)) - 4*x*e^x*log(x) + 8*x*log(x)^2 + 8*x*e^x)/(x^2*e^x*log(x)^2 + x^2*e^(2*x) + 4*log(x)^2 +
4*e^x)

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maple [A]  time = 0.05, size = 30, normalized size = 0.97

method	result	size
risch	\(x +\frac {x^{2} {\mathrm e}^{\frac {15 \ln \relax (x )+3 x}{\ln \relax (x )}}}{2 \ln \relax (x )^{2}+2 \,{\mathrm e}^{x}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*ln(x)^4+(3*x^2-2*x)*ln(x)^3+((-x^2+2*x)*exp(x)-3*x^2)*ln(x)^2+3*x^2*exp(x)*ln(x)-3*exp(x)*x^2)*exp((
15*ln(x)+3*x)/ln(x))+2*ln(x)^6+4*exp(x)*ln(x)^4+2*exp(x)^2*ln(x)^2)/(2*ln(x)^6+4*exp(x)*ln(x)^4+2*exp(x)^2*ln(
x)^2),x,method=_RETURNVERBOSE)

[Out]

x+1/2*x^2/(ln(x)^2+exp(x))*exp(3*(5*ln(x)+x)/ln(x))

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maxima [A]  time = 0.47, size = 38, normalized size = 1.23 \begin {gather*} \frac {x^{2} e^{\left (\frac {3 \, x}{\log \relax (x)} + 15\right )} + 2 \, x \log \relax (x)^{2} + 2 \, x e^{x}}{2 \, {\left (\log \relax (x)^{2} + e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)^4+(3*x^2-2*x)*log(x)^3+((-x^2+2*x)*exp(x)-3*x^2)*log(x)^2+3*x^2*exp(x)*log(x)-3*exp(x)*
x^2)*exp((15*log(x)+3*x)/log(x))+2*log(x)^6+4*exp(x)*log(x)^4+2*exp(x)^2*log(x)^2)/(2*log(x)^6+4*exp(x)*log(x)
^4+2*exp(x)^2*log(x)^2),x, algorithm="maxima")

[Out]

1/2*(x^2*e^(3*x/log(x) + 15) + 2*x*log(x)^2 + 2*x*e^x)/(log(x)^2 + e^x)

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mupad [B]  time = 5.53, size = 29, normalized size = 0.94 \begin {gather*} x+\frac {x^2\,{\mathrm {e}}^{15}\,{\mathrm {e}}^{\frac {3\,x}{\ln \relax (x)}}}{2\,\left ({\ln \relax (x)}^2+{\mathrm {e}}^x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(x)^6 + exp((3*x + 15*log(x))/log(x))*(2*x*log(x)^4 - 3*x^2*exp(x) - log(x)^3*(2*x - 3*x^2) + log(x)
^2*(exp(x)*(2*x - x^2) - 3*x^2) + 3*x^2*exp(x)*log(x)) + 4*exp(x)*log(x)^4 + 2*exp(2*x)*log(x)^2)/(2*log(x)^6
+ 4*exp(x)*log(x)^4 + 2*exp(2*x)*log(x)^2),x)

[Out]

x + (x^2*exp(15)*exp((3*x)/log(x)))/(2*(exp(x) + log(x)^2))

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sympy [A]  time = 0.44, size = 29, normalized size = 0.94 \begin {gather*} \frac {x^{2} e^{\frac {3 x + 15 \log {\relax (x )}}{\log {\relax (x )}}}}{2 e^{x} + 2 \log {\relax (x )}^{2}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*ln(x)**4+(3*x**2-2*x)*ln(x)**3+((-x**2+2*x)*exp(x)-3*x**2)*ln(x)**2+3*x**2*exp(x)*ln(x)-3*exp(
x)*x**2)*exp((15*ln(x)+3*x)/ln(x))+2*ln(x)**6+4*exp(x)*ln(x)**4+2*exp(x)**2*ln(x)**2)/(2*ln(x)**6+4*exp(x)*ln(
x)**4+2*exp(x)**2*ln(x)**2),x)

[Out]

x**2*exp((3*x + 15*log(x))/log(x))/(2*exp(x) + 2*log(x)**2) + x

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