3.20.43 \(\int (8 e^{5+8 x}-2 \log (\frac {3}{2 x})+(2-2 \log (\frac {3}{2 x})) \log (x)) \, dx\)

Optimal. Leaf size=21 \[ e^{5+8 x}-2 x \log \left (\frac {3}{2 x}\right ) \log (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2194, 2295, 2361, 12} \begin {gather*} e^{8 x+5}-2 x \log \left (\frac {3}{2 x}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8*E^(5 + 8*x) - 2*Log[3/(2*x)] + (2 - 2*Log[3/(2*x)])*Log[x],x]

[Out]

E^(5 + 8*x) - 2*x*Log[3/(2*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 2295

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

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (2 \int \log \left (\frac {3}{2 x}\right ) \, dx\right )+8 \int e^{5+8 x} \, dx+\int \left (2-2 \log \left (\frac {3}{2 x}\right )\right ) \log (x) \, dx\\ &=e^{5+8 x}-2 x-2 x \log \left (\frac {3}{2 x}\right )-2 x \log \left (\frac {3}{2 x}\right ) \log (x)-\int -2 \log \left (\frac {3}{2 x}\right ) \, dx\\ &=e^{5+8 x}-2 x-2 x \log \left (\frac {3}{2 x}\right )-2 x \log \left (\frac {3}{2 x}\right ) \log (x)+2 \int \log \left (\frac {3}{2 x}\right ) \, dx\\ &=e^{5+8 x}-2 x \log \left (\frac {3}{2 x}\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \begin {gather*} e^{5+8 x}-2 x \log \left (\frac {3}{2 x}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8*E^(5 + 8*x) - 2*Log[3/(2*x)] + (2 - 2*Log[3/(2*x)])*Log[x],x]

[Out]

E^(5 + 8*x) - 2*x*Log[3/(2*x)]*Log[x]

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fricas [A]  time = 0.54, size = 29, normalized size = 1.38 \begin {gather*} -2 \, x \log \left (\frac {3}{2}\right ) \log \left (\frac {3}{2 \, x}\right ) + 2 \, x \log \left (\frac {3}{2 \, x}\right )^{2} + e^{\left (8 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(3/2/x)+2)*log(x)-2*log(3/2/x)+8*exp(8*x+5),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.19, size = 52, normalized size = 2.48 \begin {gather*} -2 \, x \log \relax (3) \log \relax (x) + 2 \, x \log \relax (2) \log \relax (x) + 2 \, x \log \relax (x)^{2} + 2 \, x \log \relax (3) - 2 \, x \log \relax (2) - 2 \, x \log \relax (x) - 2 \, x \log \left (\frac {3}{2 \, x}\right ) + e^{\left (8 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(3/2/x)+2)*log(x)-2*log(3/2/x)+8*exp(8*x+5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 19, normalized size = 0.90

method	result	size
norman	\({\mathrm e}^{8 x +5}-2 x \ln \left (\frac {3}{2 x}\right ) \ln \relax (x )\)	\(19\)
risch	\(2 x \ln \relax (x )^{2}-\left (2+2 \ln \relax (3)-2 \ln \relax (2)\right ) x \ln \relax (x )+2 x \ln \relax (3)-2 x \ln \relax (2)-2 x \ln \left (\frac {3}{2 x}\right )+{\mathrm e}^{8 x +5}\)	\(49\)
default	\(2 x \ln \relax (2) \ln \relax (x )-2 x \ln \relax (2)-2 x \ln \relax (3) \ln \relax (x )+2 x \ln \relax (3)+2 x \ln \left (\frac {1}{x}\right )-2 \ln \relax (x ) \ln \left (\frac {1}{x}\right ) x +{\mathrm e}^{8 x +5}-2 x \ln \left (\frac {3}{2 x}\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(3/2/x)+2)*ln(x)-2*ln(3/2/x)+8*exp(8*x+5),x,method=_RETURNVERBOSE)

[Out]

exp(8*x+5)-2*x*ln(3/2/x)*ln(x)

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maxima [A]  time = 0.54, size = 18, normalized size = 0.86 \begin {gather*} -2 \, x \log \relax (x) \log \left (\frac {3}{2 \, x}\right ) + e^{\left (8 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(3/2/x)+2)*log(x)-2*log(3/2/x)+8*exp(8*x+5),x, algorithm="maxima")

[Out]

-2*x*log(x)*log(3/2/x) + e^(8*x + 5)

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mupad [B]  time = 1.18, size = 30, normalized size = 1.43 \begin {gather*} {\mathrm {e}}^{8\,x+5}+2\,x\,\ln \relax (2)\,\ln \relax (x)-2\,x\,\ln \relax (3)\,\ln \relax (x)-2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*exp(8*x + 5) - 2*log(3/(2*x)) - log(x)*(2*log(3/(2*x)) - 2),x)

[Out]

exp(8*x + 5) + 2*x*log(2)*log(x) - 2*x*log(3)*log(x) - 2*x*log(1/x)*log(x)

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sympy [A]  time = 0.29, size = 31, normalized size = 1.48 \begin {gather*} 2 x \log {\relax (x )}^{2} + \left (- 2 x \log {\relax (3 )} + 2 x \log {\relax (2 )}\right ) \log {\relax (x )} + e^{8 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(3/2/x)+2)*ln(x)-2*ln(3/2/x)+8*exp(8*x+5),x)

[Out]

2*x*log(x)**2 + (-2*x*log(3) + 2*x*log(2))*log(x) + exp(8*x + 5)

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