3.94.38 \(\int \frac {4 e^{2 x}+(e^x (-x+x^2)+e^{2 x} (-10 x+20 x^2)) \log (x)+e^{2 x} (-8+8 x) \log (x) \log (\log (x))}{15 x^3 \log (x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^x \left (2+4 e^x \left (5+\frac {2 \log (\log (x))}{x}\right )\right )}{30 x} \]

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Rubi [A]  time = 1.08, antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 6, number of rules used = 5, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 6688, 6742, 2197, 2288} \begin {gather*} \frac {2 e^{2 x} \left (5 x^2 \log (x)+2 x \log (x) \log (\log (x))\right )}{15 x^3 \log (x)}+\frac {e^x}{15 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(2*x) + (E^x*(-x + x^2) + E^(2*x)*(-10*x + 20*x^2))*Log[x] + E^(2*x)*(-8 + 8*x)*Log[x]*Log[Log[x]])/(
15*x^3*Log[x]),x]

[Out]

E^x/(15*x) + (2*E^(2*x)*(5*x^2*Log[x] + 2*x*Log[x]*Log[Log[x]]))/(15*x^3*Log[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\frac {1}{15} \int \frac {4 e^{2 x}+\left (e^x \left (-x+x^2\right )+e^{2 x} \left (-10 x+20 x^2\right )\right ) \log (x)+e^{2 x} (-8+8 x) \log (x) \log (\log (x))}{x^3 \log (x)} \, dx\\ &=\frac {1}{15} \int \frac {e^x \left (4 e^x+\log (x) \left (x \left (-1+x+10 e^x (-1+2 x)\right )+8 e^x (-1+x) \log (\log (x))\right )\right )}{x^3 \log (x)} \, dx\\ &=\frac {1}{15} \int \left (\frac {e^x (-1+x)}{x^2}+\frac {2 e^{2 x} \left (2-5 x \log (x)+10 x^2 \log (x)-4 \log (x) \log (\log (x))+4 x \log (x) \log (\log (x))\right )}{x^3 \log (x)}\right ) \, dx\\ &=\frac {1}{15} \int \frac {e^x (-1+x)}{x^2} \, dx+\frac {2}{15} \int \frac {e^{2 x} \left (2-5 x \log (x)+10 x^2 \log (x)-4 \log (x) \log (\log (x))+4 x \log (x) \log (\log (x))\right )}{x^3 \log (x)} \, dx\\ &=\frac {e^x}{15 x}+\frac {2 e^{2 x} \left (5 x^2 \log (x)+2 x \log (x) \log (\log (x))\right )}{15 x^3 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 26, normalized size = 0.96 \begin {gather*} \frac {e^x \left (x+10 e^x x+4 e^x \log (\log (x))\right )}{15 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(2*x) + (E^x*(-x + x^2) + E^(2*x)*(-10*x + 20*x^2))*Log[x] + E^(2*x)*(-8 + 8*x)*Log[x]*Log[Log[
x]])/(15*x^3*Log[x]),x]

[Out]

(E^x*(x + 10*E^x*x + 4*E^x*Log[Log[x]]))/(15*x^2)

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fricas [A]  time = 0.64, size = 26, normalized size = 0.96 \begin {gather*} \frac {10 \, x e^{\left (2 \, x\right )} + x e^{x} + 4 \, e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((8*x-8)*exp(x)^2*log(x)*log(log(x))+((20*x^2-10*x)*exp(x)^2+(x^2-x)*exp(x))*log(x)+4*exp(x)^2)
/x^3/log(x),x, algorithm="fricas")

[Out]

1/15*(10*x*e^(2*x) + x*e^x + 4*e^(2*x)*log(log(x)))/x^2

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giac [A]  time = 0.21, size = 35, normalized size = 1.30 \begin {gather*} \frac {4 \, x e^{\left (2 \, x\right )} \log \relax (x) + 10 \, x e^{\left (2 \, x\right )} + x e^{x} + 4 \, e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((8*x-8)*exp(x)^2*log(x)*log(log(x))+((20*x^2-10*x)*exp(x)^2+(x^2-x)*exp(x))*log(x)+4*exp(x)^2)
/x^3/log(x),x, algorithm="giac")

[Out]

1/15*(4*x*e^(2*x)*log(x) + 10*x*e^(2*x) + x*e^x + 4*e^(2*x)*log(log(x)))/x^2

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maple [A]  time = 0.04, size = 27, normalized size = 1.00

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{2 x} \ln \left (\ln \relax (x )\right )}{15 x^{2}}+\frac {{\mathrm e}^{x} \left (10 \,{\mathrm e}^{x}+1\right )}{15 x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*((8*x-8)*exp(x)^2*ln(x)*ln(ln(x))+((20*x^2-10*x)*exp(x)^2+(x^2-x)*exp(x))*ln(x)+4*exp(x)^2)/x^3/ln(x)
,x,method=_RETURNVERBOSE)

[Out]

4/15/x^2*exp(2*x)*ln(ln(x))+1/15*exp(x)*(10*exp(x)+1)/x

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maxima [C]  time = 0.42, size = 37, normalized size = 1.37 \begin {gather*} \frac {4 \, e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )}{15 \, x^{2}} + \frac {4}{3} \, {\rm Ei}\left (2 \, x\right ) + \frac {1}{15} \, {\rm Ei}\relax (x) - \frac {1}{15} \, \Gamma \left (-1, -x\right ) - \frac {4}{3} \, \Gamma \left (-1, -2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((8*x-8)*exp(x)^2*log(x)*log(log(x))+((20*x^2-10*x)*exp(x)^2+(x^2-x)*exp(x))*log(x)+4*exp(x)^2)
/x^3/log(x),x, algorithm="maxima")

[Out]

4/15*e^(2*x)*log(log(x))/x^2 + 4/3*Ei(2*x) + 1/15*Ei(x) - 1/15*gamma(-1, -x) - 4/3*gamma(-1, -2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\frac {4\,{\mathrm {e}}^{2\,x}}{15}-\frac {\ln \relax (x)\,\left ({\mathrm {e}}^{2\,x}\,\left (10\,x-20\,x^2\right )+{\mathrm {e}}^x\,\left (x-x^2\right )\right )}{15}+\frac {\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)\,\left (8\,x-8\right )}{15}}{x^3\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(2*x))/15 - (log(x)*(exp(2*x)*(10*x - 20*x^2) + exp(x)*(x - x^2)))/15 + (log(log(x))*exp(2*x)*log(x
)*(8*x - 8))/15)/(x^3*log(x)),x)

[Out]

int(((4*exp(2*x))/15 - (log(x)*(exp(2*x)*(10*x - 20*x^2) + exp(x)*(x - x^2)))/15 + (log(log(x))*exp(2*x)*log(x
)*(8*x - 8))/15)/(x^3*log(x)), x)

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sympy [A]  time = 0.48, size = 31, normalized size = 1.15 \begin {gather*} \frac {15 x^{2} e^{x} + \left (150 x^{2} + 60 x \log {\left (\log {\relax (x )} \right )}\right ) e^{2 x}}{225 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((8*x-8)*exp(x)**2*ln(x)*ln(ln(x))+((20*x**2-10*x)*exp(x)**2+(x**2-x)*exp(x))*ln(x)+4*exp(x)**2
)/x**3/ln(x),x)

[Out]

(15*x**2*exp(x) + (150*x**2 + 60*x*log(log(x)))*exp(2*x))/(225*x**3)

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