3.13.91 \(\int \frac {1+4 e^{2 x} x+e^x (4+x)+(4+4 e^x x) \log (x)}{18 x} \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{9} \left (-\frac {1}{4}-e^x-\log (x)\right )^2 \]

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Rubi [B]  time = 0.07, antiderivative size = 39, normalized size of antiderivative = 2.05, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2194, 2301, 2288} \begin {gather*} \frac {e^{2 x}}{9}+\frac {1}{144} (4 \log (x)+1)^2+\frac {e^x (x+4 x \log (x))}{18 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 4*E^(2*x)*x + E^x*(4 + x) + (4 + 4*E^x*x)*Log[x])/(18*x),x]

[Out]

E^(2*x)/9 + (1 + 4*Log[x])^2/144 + (E^x*(x + 4*x*Log[x]))/(18*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{18} \int \frac {1+4 e^{2 x} x+e^x (4+x)+\left (4+4 e^x x\right ) \log (x)}{x} \, dx\\ &=\frac {1}{18} \int \left (4 e^{2 x}+\frac {1+4 \log (x)}{x}+\frac {e^x (4+x+4 x \log (x))}{x}\right ) \, dx\\ &=\frac {1}{18} \int \frac {1+4 \log (x)}{x} \, dx+\frac {1}{18} \int \frac {e^x (4+x+4 x \log (x))}{x} \, dx+\frac {2}{9} \int e^{2 x} \, dx\\ &=\frac {e^{2 x}}{9}+\frac {1}{144} (1+4 \log (x))^2+\frac {e^x (x+4 x \log (x))}{18 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{144} \left (1+4 e^x+4 \log (x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*E^(2*x)*x + E^x*(4 + x) + (4 + 4*E^x*x)*Log[x])/(18*x),x]

[Out]

(1 + 4*E^x + 4*Log[x])^2/144

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fricas [A]  time = 0.77, size = 27, normalized size = 1.42 \begin {gather*} \frac {1}{18} \, {\left (4 \, e^{x} + 1\right )} \log \relax (x) + \frac {1}{9} \, \log \relax (x)^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} + \frac {1}{18} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*((4*exp(x)*x+4)*log(x)+4*x*exp(x)^2+(4+x)*exp(x)+1)/x,x, algorithm="fricas")

[Out]

1/18*(4*e^x + 1)*log(x) + 1/9*log(x)^2 + 1/9*e^(2*x) + 1/18*e^x

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giac [A]  time = 0.28, size = 27, normalized size = 1.42 \begin {gather*} \frac {2}{9} \, e^{x} \log \relax (x) + \frac {1}{9} \, \log \relax (x)^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} + \frac {1}{18} \, e^{x} + \frac {1}{18} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*((4*exp(x)*x+4)*log(x)+4*x*exp(x)^2+(4+x)*exp(x)+1)/x,x, algorithm="giac")

[Out]

2/9*e^x*log(x) + 1/9*log(x)^2 + 1/9*e^(2*x) + 1/18*e^x + 1/18*log(x)

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maple [A]  time = 0.04, size = 28, normalized size = 1.47

method	result	size
default	\(\frac {\ln \relax (x )}{18}+\frac {2 \,{\mathrm e}^{x} \ln \relax (x )}{9}+\frac {{\mathrm e}^{x}}{18}+\frac {{\mathrm e}^{2 x}}{9}+\frac {\ln \relax (x )^{2}}{9}\)	\(28\)
norman	\(\frac {\ln \relax (x )}{18}+\frac {2 \,{\mathrm e}^{x} \ln \relax (x )}{9}+\frac {{\mathrm e}^{x}}{18}+\frac {{\mathrm e}^{2 x}}{9}+\frac {\ln \relax (x )^{2}}{9}\)	\(28\)
risch	\(\frac {\ln \relax (x )}{18}+\frac {2 \,{\mathrm e}^{x} \ln \relax (x )}{9}+\frac {{\mathrm e}^{x}}{18}+\frac {{\mathrm e}^{2 x}}{9}+\frac {\ln \relax (x )^{2}}{9}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*((4*exp(x)*x+4)*ln(x)+4*x*exp(x)^2+(4+x)*exp(x)+1)/x,x,method=_RETURNVERBOSE)

[Out]

1/18*ln(x)+2/9*exp(x)*ln(x)+1/18*exp(x)+1/9*exp(x)^2+1/9*ln(x)^2

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maxima [A]  time = 0.69, size = 27, normalized size = 1.42 \begin {gather*} \frac {2}{9} \, e^{x} \log \relax (x) + \frac {1}{9} \, \log \relax (x)^{2} + \frac {1}{9} \, e^{\left (2 \, x\right )} + \frac {1}{18} \, e^{x} + \frac {1}{18} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*((4*exp(x)*x+4)*log(x)+4*x*exp(x)^2+(4+x)*exp(x)+1)/x,x, algorithm="maxima")

[Out]

2/9*e^x*log(x) + 1/9*log(x)^2 + 1/9*e^(2*x) + 1/18*e^x + 1/18*log(x)

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mupad [B]  time = 0.98, size = 27, normalized size = 1.42 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{9}+\frac {{\mathrm {e}}^x}{18}+\frac {\ln \relax (x)}{18}+\frac {2\,{\mathrm {e}}^x\,\ln \relax (x)}{9}+\frac {{\ln \relax (x)}^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(2*x))/9 + (exp(x)*(x + 4))/18 + (log(x)*(4*x*exp(x) + 4))/18 + 1/18)/x,x)

[Out]

exp(2*x)/9 + exp(x)/18 + log(x)/18 + (2*exp(x)*log(x))/9 + log(x)^2/9

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sympy [A]  time = 0.34, size = 29, normalized size = 1.53 \begin {gather*} \frac {\left (36 \log {\relax (x )} + 9\right ) e^{x}}{162} + \frac {e^{2 x}}{9} + \frac {\log {\relax (x )}^{2}}{9} + \frac {\log {\relax (x )}}{18} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*((4*exp(x)*x+4)*ln(x)+4*x*exp(x)**2+(4+x)*exp(x)+1)/x,x)

[Out]

(36*log(x) + 9)*exp(x)/162 + exp(2*x)/9 + log(x)**2/9 + log(x)/18

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