3.46.2 \(\int \frac {5 e^{4 x}+e^{4 x} (-5-20 x) \log (x)-11 \log ^2(x)+5 e^{e^x+x} \log ^2(x)}{5 \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ -1+e^{e^x}-\frac {11 x}{5}-\frac {e^{4 x} x}{\log (x)} \]

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Rubi [A]  time = 0.19, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {12, 6742, 2282, 2194, 2288} \begin {gather*} -\frac {11 x}{5}+e^{e^x}-\frac {e^{4 x} x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^(4*x) + E^(4*x)*(-5 - 20*x)*Log[x] - 11*Log[x]^2 + 5*E^(E^x + x)*Log[x]^2)/(5*Log[x]^2),x]

[Out]

E^E^x - (11*x)/5 - (E^(4*x)*x)/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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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 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}{5} \int \frac {5 e^{4 x}+e^{4 x} (-5-20 x) \log (x)-11 \log ^2(x)+5 e^{e^x+x} \log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (-11+5 e^{e^x+x}-\frac {5 e^{4 x} (-1+\log (x)+4 x \log (x))}{\log ^2(x)}\right ) \, dx\\ &=-\frac {11 x}{5}+\int e^{e^x+x} \, dx-\int \frac {e^{4 x} (-1+\log (x)+4 x \log (x))}{\log ^2(x)} \, dx\\ &=-\frac {11 x}{5}-\frac {e^{4 x} x}{\log (x)}+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}-\frac {11 x}{5}-\frac {e^{4 x} x}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 0.96 \begin {gather*} e^{e^x}-\frac {11 x}{5}-\frac {e^{4 x} x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(4*x) + E^(4*x)*(-5 - 20*x)*Log[x] - 11*Log[x]^2 + 5*E^(E^x + x)*Log[x]^2)/(5*Log[x]^2),x]

[Out]

E^E^x - (11*x)/5 - (E^(4*x)*x)/Log[x]

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fricas [A]  time = 0.51, size = 34, normalized size = 1.42 \begin {gather*} -\frac {{\left (11 \, x e^{x} \log \relax (x) + 5 \, x e^{\left (5 \, x\right )} - 5 \, e^{\left (x + e^{x}\right )} \log \relax (x)\right )} e^{\left (-x\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(x)*log(x)^2*exp(exp(x))-11*log(x)^2+(-20*x-5)*exp(4*x)*log(x)+5*exp(4*x))/log(x)^2,x, alg
orithm="fricas")

[Out]

-1/5*(11*x*e^x*log(x) + 5*x*e^(5*x) - 5*e^(x + e^x)*log(x))*e^(-x)/log(x)

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giac [A]  time = 0.12, size = 34, normalized size = 1.42 \begin {gather*} -\frac {{\left (11 \, x e^{x} \log \relax (x) + 5 \, x e^{\left (5 \, x\right )} - 5 \, e^{\left (x + e^{x}\right )} \log \relax (x)\right )} e^{\left (-x\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(x)*log(x)^2*exp(exp(x))-11*log(x)^2+(-20*x-5)*exp(4*x)*log(x)+5*exp(4*x))/log(x)^2,x, alg
orithm="giac")

[Out]

-1/5*(11*x*e^x*log(x) + 5*x*e^(5*x) - 5*e^(x + e^x)*log(x))*e^(-x)/log(x)

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maple [A]  time = 0.04, size = 19, normalized size = 0.79

method	result	size
default	\(-\frac {11 x}{5}-\frac {{\mathrm e}^{4 x} x}{\ln \relax (x )}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(19\)
risch	\(-\frac {11 x}{5}-\frac {{\mathrm e}^{4 x} x}{\ln \relax (x )}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(5*exp(x)*ln(x)^2*exp(exp(x))-11*ln(x)^2+(-20*x-5)*exp(4*x)*ln(x)+5*exp(4*x))/ln(x)^2,x,method=_RETURN
VERBOSE)

[Out]

-11/5*x-exp(4*x)*x/ln(x)+exp(exp(x))

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maxima [A]  time = 0.41, size = 18, normalized size = 0.75 \begin {gather*} -\frac {11}{5} \, x - \frac {x e^{\left (4 \, x\right )}}{\log \relax (x)} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(x)*log(x)^2*exp(exp(x))-11*log(x)^2+(-20*x-5)*exp(4*x)*log(x)+5*exp(4*x))/log(x)^2,x, alg
orithm="maxima")

[Out]

-11/5*x - x*e^(4*x)/log(x) + e^(e^x)

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mupad [B]  time = 3.32, size = 18, normalized size = 0.75 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}-\frac {11\,x}{5}-\frac {x\,{\mathrm {e}}^{4\,x}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x) - (11*log(x)^2)/5 - (exp(4*x)*log(x)*(20*x + 5))/5 + exp(exp(x))*exp(x)*log(x)^2)/log(x)^2,x)

[Out]

exp(exp(x)) - (11*x)/5 - (x*exp(4*x))/log(x)

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sympy [A]  time = 0.38, size = 19, normalized size = 0.79 \begin {gather*} - \frac {x e^{4 x}}{\log {\relax (x )}} - \frac {11 x}{5} + e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(x)*ln(x)**2*exp(exp(x))-11*ln(x)**2+(-20*x-5)*exp(4*x)*ln(x)+5*exp(4*x))/ln(x)**2,x)

[Out]

-x*exp(4*x)/log(x) - 11*x/5 + exp(exp(x))

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