3.83.1 \(\int \frac {3-6 e^{-11 x} \log (5)+e^{-11 x} (-6+66 x) \log (5) \log (x)}{2 \log (5)} \, dx\)

Optimal. Leaf size=20 \[ x \left (\frac {3}{2 \log (5)}-3 e^{-11 x} \log (x)\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 37, normalized size of antiderivative = 1.85, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2194, 2176, 2554} \begin {gather*} \frac {3 x}{2 \log (5)}-\frac {3}{11} e^{-11 x} \log (x)+\frac {3}{11} e^{-11 x} (1-11 x) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - (6*Log[5])/E^(11*x) + ((-6 + 66*x)*Log[5]*Log[x])/E^(11*x))/(2*Log[5]),x]

[Out]

(3*x)/(2*Log[5]) - (3*Log[x])/(11*E^(11*x)) + (3*(1 - 11*x)*Log[x])/(11*E^(11*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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (3-6 e^{-11 x} \log (5)+e^{-11 x} (-6+66 x) \log (5) \log (x)\right ) \, dx}{2 \log (5)}\\ &=\frac {3 x}{2 \log (5)}+\frac {1}{2} \int e^{-11 x} (-6+66 x) \log (x) \, dx-3 \int e^{-11 x} \, dx\\ &=\frac {3 e^{-11 x}}{11}+\frac {3 x}{2 \log (5)}-\frac {3}{11} e^{-11 x} \log (x)+\frac {3}{11} e^{-11 x} (1-11 x) \log (x)-\frac {1}{2} \int -6 e^{-11 x} \, dx\\ &=\frac {3 e^{-11 x}}{11}+\frac {3 x}{2 \log (5)}-\frac {3}{11} e^{-11 x} \log (x)+\frac {3}{11} e^{-11 x} (1-11 x) \log (x)+3 \int e^{-11 x} \, dx\\ &=\frac {3 x}{2 \log (5)}-\frac {3}{11} e^{-11 x} \log (x)+\frac {3}{11} e^{-11 x} (1-11 x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 1.00 \begin {gather*} \frac {3 \left (x-2 e^{-11 x} x \log (5) \log (x)\right )}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - (6*Log[5])/E^(11*x) + ((-6 + 66*x)*Log[5]*Log[x])/E^(11*x))/(2*Log[5]),x]

[Out]

(3*(x - (2*x*Log[5]*Log[x])/E^(11*x)))/Log[25]

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fricas [A]  time = 0.71, size = 21, normalized size = 1.05 \begin {gather*} -\frac {3 \, {\left (2 \, x e^{\left (-11 \, x\right )} \log \relax (5) \log \relax (x) - x\right )}}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((66*x-6)*log(5)*exp(-11*x)*log(x)-6*log(5)*exp(-11*x)+3)/log(5),x, algorithm="fricas")

[Out]

-3/2*(2*x*e^(-11*x)*log(5)*log(x) - x)/log(5)

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giac [A]  time = 0.13, size = 36, normalized size = 1.80 \begin {gather*} -\frac {3 \, {\left (2 \, {\left (11 \, x e^{\left (-11 \, x\right )} \log \relax (x) + e^{\left (-11 \, x\right )}\right )} \log \relax (5) - 2 \, e^{\left (-11 \, x\right )} \log \relax (5) - 11 \, x\right )}}{22 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((66*x-6)*log(5)*exp(-11*x)*log(x)-6*log(5)*exp(-11*x)+3)/log(5),x, algorithm="giac")

[Out]

-3/22*(2*(11*x*e^(-11*x)*log(x) + e^(-11*x))*log(5) - 2*e^(-11*x)*log(5) - 11*x)/log(5)

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maple [A]  time = 0.06, size = 18, normalized size = 0.90

method	result	size
norman	\(\frac {3 x}{2 \ln \relax (5)}-3 x \,{\mathrm e}^{-11 x} \ln \relax (x )\)	\(18\)
risch	\(\frac {3 x}{2 \ln \relax (5)}-3 x \,{\mathrm e}^{-11 x} \ln \relax (x )\)	\(18\)
default	\(\frac {3 x -6 \ln \relax (5) {\mathrm e}^{-11 x} \ln \relax (x ) x}{2 \ln \relax (5)}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((66*x-6)*ln(5)*exp(-11*x)*ln(x)-6*ln(5)*exp(-11*x)+3)/ln(5),x,method=_RETURNVERBOSE)

[Out]

3/2*x/ln(5)-3*x*exp(-11*x)*ln(x)

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maxima [A]  time = 0.37, size = 21, normalized size = 1.05 \begin {gather*} -\frac {3 \, {\left (2 \, x e^{\left (-11 \, x\right )} \log \relax (5) \log \relax (x) - x\right )}}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((66*x-6)*log(5)*exp(-11*x)*log(x)-6*log(5)*exp(-11*x)+3)/log(5),x, algorithm="maxima")

[Out]

-3/2*(2*x*e^(-11*x)*log(5)*log(x) - x)/log(5)

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mupad [B]  time = 5.10, size = 17, normalized size = 0.85 \begin {gather*} \frac {3\,x}{2\,\ln \relax (5)}-3\,x\,{\mathrm {e}}^{-11\,x}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(-11*x)*log(5)*log(x)*(66*x - 6))/2 - 3*exp(-11*x)*log(5) + 3/2)/log(5),x)

[Out]

(3*x)/(2*log(5)) - 3*x*exp(-11*x)*log(x)

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sympy [A]  time = 0.30, size = 19, normalized size = 0.95 \begin {gather*} \frac {3 x}{2 \log {\relax (5 )}} - 3 x e^{- 11 x} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((66*x-6)*ln(5)*exp(-11*x)*ln(x)-6*ln(5)*exp(-11*x)+3)/ln(5),x)

[Out]

3*x/(2*log(5)) - 3*x*exp(-11*x)*log(x)

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