3.81.94 \(\int \frac {e^5+(-12 e^5-3 x) \log (4)+3 \log (4) \log (x)+((3-6 x) \log (4)+3 \log (4) \log (x)) \log (2 x)}{e^5} \, dx\)

Optimal. Leaf size=28 \[ x-x \log (4) \left (3+3 \left (3-\frac {(-x+\log (x)) \log (2 x)}{e^5}\right )\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 2.14, number of steps used = 12, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 2295, 6688, 6742, 2304, 2361} \begin {gather*} -\frac {3 x^2 \log (4) \log (2 x)}{e^5}+\frac {3 x^2 \log (4)}{2 e^5}+x+\frac {3 x \log (4) \log (x) \log (2 x)}{e^5}-\frac {3 \left (x+4 e^5\right )^2 \log (4)}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5 + (-12*E^5 - 3*x)*Log[4] + 3*Log[4]*Log[x] + ((3 - 6*x)*Log[4] + 3*Log[4]*Log[x])*Log[2*x])/E^5,x]

[Out]

x + (3*x^2*Log[4])/(2*E^5) - (3*(4*E^5 + x)^2*Log[4])/(2*E^5) - (3*x^2*Log[4]*Log[2*x])/E^5 + (3*x*Log[4]*Log[
x]*Log[2*x])/E^5

Rule 12

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, 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 {\int \left (e^5+\left (-12 e^5-3 x\right ) \log (4)+3 \log (4) \log (x)+((3-6 x) \log (4)+3 \log (4) \log (x)) \log (2 x)\right ) \, dx}{e^5}\\ &=x-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {\int ((3-6 x) \log (4)+3 \log (4) \log (x)) \log (2 x) \, dx}{e^5}+\frac {(3 \log (4)) \int \log (x) \, dx}{e^5}\\ &=x-\frac {3 x \log (4)}{e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}+\frac {\int 3 \log (4) (1-2 x+\log (x)) \log (2 x) \, dx}{e^5}\\ &=x-\frac {3 x \log (4)}{e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}+\frac {(3 \log (4)) \int (1-2 x+\log (x)) \log (2 x) \, dx}{e^5}\\ &=x-\frac {3 x \log (4)}{e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}+\frac {(3 \log (4)) \int (\log (2 x)-2 x \log (2 x)+\log (x) \log (2 x)) \, dx}{e^5}\\ &=x-\frac {3 x \log (4)}{e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}+\frac {(3 \log (4)) \int \log (2 x) \, dx}{e^5}+\frac {(3 \log (4)) \int \log (x) \log (2 x) \, dx}{e^5}-\frac {(6 \log (4)) \int x \log (2 x) \, dx}{e^5}\\ &=x-\frac {6 x \log (4)}{e^5}+\frac {3 x^2 \log (4)}{2 e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}-\frac {3 x^2 \log (4) \log (2 x)}{e^5}+\frac {3 x \log (4) \log (x) \log (2 x)}{e^5}-\frac {(3 \log (4)) \int (-1+\log (x)) \, dx}{e^5}\\ &=x-\frac {3 x \log (4)}{e^5}+\frac {3 x^2 \log (4)}{2 e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}+\frac {3 x \log (4) \log (x)}{e^5}-\frac {3 x^2 \log (4) \log (2 x)}{e^5}+\frac {3 x \log (4) \log (x) \log (2 x)}{e^5}-\frac {(3 \log (4)) \int \log (x) \, dx}{e^5}\\ &=x+\frac {3 x^2 \log (4)}{2 e^5}-\frac {3 \left (4 e^5+x\right )^2 \log (4)}{2 e^5}-\frac {3 x^2 \log (4) \log (2 x)}{e^5}+\frac {3 x \log (4) \log (x) \log (2 x)}{e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.07 \begin {gather*} \frac {x \left (e^5 (1-12 \log (4))+3 \log (4) (-x+\log (x)) \log (2 x)\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5 + (-12*E^5 - 3*x)*Log[4] + 3*Log[4]*Log[x] + ((3 - 6*x)*Log[4] + 3*Log[4]*Log[x])*Log[2*x])/E^5
,x]

[Out]

(x*(E^5*(1 - 12*Log[4]) + 3*Log[4]*(-x + Log[x])*Log[2*x]))/E^5

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fricas [B]  time = 0.56, size = 53, normalized size = 1.89 \begin {gather*} -{\left (6 \, x^{2} \log \relax (2)^{2} - 6 \, x \log \relax (2) \log \relax (x)^{2} + 24 \, x e^{5} \log \relax (2) - x e^{5} + 6 \, {\left (x^{2} \log \relax (2) - x \log \relax (2)^{2}\right )} \log \relax (x)\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(2)*log(x)+2*(-6*x+3)*log(2))*log(2*x)+6*log(2)*log(x)+2*(-12*exp(5)-3*x)*log(2)+exp(5))/exp(
5),x, algorithm="fricas")

[Out]

-(6*x^2*log(2)^2 - 6*x*log(2)*log(x)^2 + 24*x*e^5*log(2) - x*e^5 + 6*(x^2*log(2) - x*log(2)^2)*log(x))*e^(-5)

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giac [B]  time = 0.15, size = 90, normalized size = 3.21 \begin {gather*} -{\left (6 \, x^{2} \log \relax (2)^{2} + 6 \, x^{2} \log \relax (2) \log \relax (x) - 6 \, x \log \relax (2)^{2} \log \relax (x) - 6 \, x \log \relax (2) \log \relax (x)^{2} - 3 \, x^{2} \log \relax (2) + 6 \, x \log \relax (2) \log \relax (x) - x e^{5} + 3 \, {\left (x^{2} + 8 \, x e^{5}\right )} \log \relax (2) - 6 \, {\left (x \log \relax (x) - x\right )} \log \relax (2) - 6 \, x \log \relax (2)\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(2)*log(x)+2*(-6*x+3)*log(2))*log(2*x)+6*log(2)*log(x)+2*(-12*exp(5)-3*x)*log(2)+exp(5))/exp(
5),x, algorithm="giac")

[Out]

-(6*x^2*log(2)^2 + 6*x^2*log(2)*log(x) - 6*x*log(2)^2*log(x) - 6*x*log(2)*log(x)^2 - 3*x^2*log(2) + 6*x*log(2)
*log(x) - x*e^5 + 3*(x^2 + 8*x*e^5)*log(2) - 6*(x*log(x) - x)*log(2) - 6*x*log(2))*e^(-5)

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maple [A]  time = 0.09, size = 40, normalized size = 1.43

method	result	size
norman	\(\left (-24 \ln \relax (2)+1\right ) x -6 x^{2} \ln \relax (2) {\mathrm e}^{-5} \ln \left (2 x \right )+6 \,{\mathrm e}^{-5} \ln \relax (2) x \ln \relax (x ) \ln \left (2 x \right )\)	\(40\)
risch	\(6 \,{\mathrm e}^{-5} \ln \relax (2) \ln \relax (x )^{2} x +3 \,{\mathrm e}^{-5} \ln \relax (2) \left (2 x \ln \relax (2)-2 x^{2}\right ) \ln \relax (x )-6 \,{\mathrm e}^{-5} x^{2} \ln \relax (2)^{2}-24 x \ln \relax (2)+x\)	\(49\)
default	\({\mathrm e}^{-5} \left (6 \ln \relax (x ) \ln \relax (2)^{2} x +6 \ln \relax (2) \ln \relax (x )^{2} x -6 x^{2} \ln \relax (2)^{2}-6 x^{2} \ln \relax (2) \ln \relax (x )-24 x \,{\mathrm e}^{5} \ln \relax (2)+x \,{\mathrm e}^{5}\right )\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*ln(2)*ln(x)+2*(-6*x+3)*ln(2))*ln(2*x)+6*ln(2)*ln(x)+2*(-12*exp(5)-3*x)*ln(2)+exp(5))/exp(5),x,method=_
RETURNVERBOSE)

[Out]

(-24*ln(2)+1)*x-6*x^2*ln(2)/exp(5)*ln(2*x)+6/exp(5)*ln(2)*x*ln(x)*ln(2*x)

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maxima [B]  time = 0.62, size = 81, normalized size = 2.89 \begin {gather*} {\left (3 \, x^{2} \log \relax (2) - 6 \, x \log \relax (2) \log \relax (x) + x e^{5} - 3 \, {\left (x^{2} + 8 \, x e^{5}\right )} \log \relax (2) + 6 \, {\left (x \log \relax (x) - x\right )} \log \relax (2) + 6 \, x \log \relax (2) - 6 \, {\left ({\left (x^{2} - x\right )} \log \relax (2) - {\left (x \log \relax (x) - x\right )} \log \relax (2)\right )} \log \left (2 \, x\right )\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(2)*log(x)+2*(-6*x+3)*log(2))*log(2*x)+6*log(2)*log(x)+2*(-12*exp(5)-3*x)*log(2)+exp(5))/exp(
5),x, algorithm="maxima")

[Out]

(3*x^2*log(2) - 6*x*log(2)*log(x) + x*e^5 - 3*(x^2 + 8*x*e^5)*log(2) + 6*(x*log(x) - x)*log(2) + 6*x*log(2) -
6*((x^2 - x)*log(2) - (x*log(x) - x)*log(2))*log(2*x))*e^(-5)

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mupad [B]  time = 5.79, size = 43, normalized size = 1.54 \begin {gather*} x\,{\mathrm {e}}^{-5}\,\left ({\mathrm {e}}^5+6\,\ln \relax (2)\,{\ln \relax (x)}^2+6\,{\ln \relax (2)}^2\,\ln \relax (x)-24\,{\mathrm {e}}^5\,\ln \relax (2)-6\,x\,{\ln \relax (2)}^2-6\,x\,\ln \relax (2)\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-5)*(exp(5) + 6*log(2)*log(x) - log(2*x)*(2*log(2)*(6*x - 3) - 6*log(2)*log(x)) - 2*log(2)*(3*x + 12*e
xp(5))),x)

[Out]

x*exp(-5)*(exp(5) + 6*log(2)*log(x)^2 + 6*log(2)^2*log(x) - 24*exp(5)*log(2) - 6*x*log(2)^2 - 6*x*log(2)*log(x
))

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sympy [B]  time = 0.32, size = 60, normalized size = 2.14 \begin {gather*} - \frac {6 x^{2} \log {\relax (2 )}^{2}}{e^{5}} + \frac {6 x \log {\relax (2 )} \log {\relax (x )}^{2}}{e^{5}} + x \left (1 - 24 \log {\relax (2 )}\right ) + \frac {\left (- 6 x^{2} \log {\relax (2 )} + 6 x \log {\relax (2 )}^{2}\right ) \log {\relax (x )}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*ln(2)*ln(x)+2*(-6*x+3)*ln(2))*ln(2*x)+6*ln(2)*ln(x)+2*(-12*exp(5)-3*x)*ln(2)+exp(5))/exp(5),x)

[Out]

-6*x**2*exp(-5)*log(2)**2 + 6*x*exp(-5)*log(2)*log(x)**2 + x*(1 - 24*log(2)) + (-6*x**2*log(2) + 6*x*log(2)**2
)*exp(-5)*log(x)

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