3.65.23 \(\int \frac {-4+5 x+(1-x) \log (4)+(-4 x+x^2+x \log (4)) \log (e^5 (-4 x+x^2)+e^5 x \log (4))}{-4 x+x^2+x \log (4)} \, dx\)

Optimal. Leaf size=20 \[ 2-2 x+\log (x)+x \log \left (e^5 x (-4+x+\log (4))\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 7, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6, 1593, 6742, 72, 2487, 31, 8} \begin {gather*} 3 x+x \log (-x (-x+4-\log (4)))+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*x + Log[x] + x*Log[-(x*(4 - x - Log[4]))]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2487

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + (Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p
*(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] - Dist[r*s*(p + q), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1
), x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && NeQ[p + q, 0] && IGtQ[s, 0] &&
LtQ[s, 4]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{x^2+x (-4+\log (4))} \, dx\\ &=\int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{x (-4+x+\log (4))} \, dx\\ &=\int \left (5+\frac {4-x (5-\log (4))-\log (4)}{x (4-x-\log (4))}+\log (x (-4+x+\log (4)))\right ) \, dx\\ &=5 x+\int \frac {4-x (5-\log (4))-\log (4)}{x (4-x-\log (4))} \, dx+\int \log (x (-4+x+\log (4))) \, dx\\ &=5 x+x \log (-x (4-x-\log (4)))-2 \int 1 \, dx+(-4+\log (4)) \int \frac {1}{-4+x+\log (4)} \, dx+\int \left (\frac {1}{x}+\frac {4-\log (4)}{-4+x+\log (4)}\right ) \, dx\\ &=3 x+\log (x)+x \log (-x (4-x-\log (4)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 16, normalized size = 0.80 \begin {gather*} 3 x+\log (x)+x \log (x (-4+x+\log (4))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*x + Log[x] + x*Log[x*(-4 + x + Log[4])]

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fricas [A]  time = 0.61, size = 27, normalized size = 1.35 \begin {gather*} x \log \left (2 \, x e^{5} \log \relax (2) + {\left (x^{2} - 4 \, x\right )} e^{5}\right ) - 2 \, x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2*(-x+1)*log(2)+5*x-4)/(2*x*log(2)+x^2
-4*x),x, algorithm="fricas")

[Out]

x*log(2*x*e^5*log(2) + (x^2 - 4*x)*e^5) - 2*x + log(x)

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giac [A]  time = 0.21, size = 21, normalized size = 1.05 \begin {gather*} x \log \left (x^{2} + 2 \, x \log \relax (2) - 4 \, x\right ) + 3 \, x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2*(-x+1)*log(2)+5*x-4)/(2*x*log(2)+x^2
-4*x),x, algorithm="giac")

[Out]

x*log(x^2 + 2*x*log(2) - 4*x) + 3*x + log(x)

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maple [A]  time = 0.48, size = 28, normalized size = 1.40

method	result	size
norman	\(x \ln \left (2 x \,{\mathrm e}^{5} \ln \relax (2)+\left (x^{2}-4 x \right ) {\mathrm e}^{5}\right )-2 x +\ln \relax (x )\)	\(28\)
risch	\(x \ln \left (2 x \,{\mathrm e}^{5} \ln \relax (2)+\left (x^{2}-4 x \right ) {\mathrm e}^{5}\right )-2 x +\ln \relax (x )\)	\(28\)
default	\(\ln \relax (x )+\ln \left (2 x \,{\mathrm e}^{5} \ln \relax (2)+x^{2} {\mathrm e}^{5}-4 x \,{\mathrm e}^{5}\right ) x -2 x\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*ln(2)+x^2-4*x)*ln(2*x*exp(5)*ln(2)+(x^2-4*x)*exp(5))+2*(1-x)*ln(2)+5*x-4)/(2*x*ln(2)+x^2-4*x),x,meth
od=_RETURNVERBOSE)

[Out]

x*ln(2*x*exp(5)*ln(2)+(x^2-4*x)*exp(5))-2*x+ln(x)

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maxima [B]  time = 0.52, size = 102, normalized size = 5.10 \begin {gather*} -{\left (\frac {\log \left (x + 2 \, \log \relax (2) - 4\right )}{\log \relax (2) - 2} - \frac {\log \relax (x)}{\log \relax (2) - 2}\right )} \log \relax (2) + {\left (x + 2 \, \log \relax (2) - 4\right )} \log \left (x + 2 \, \log \relax (2) - 4\right ) - 2 \, \log \relax (2) \log \left (x + 2 \, \log \relax (2) - 4\right ) + x \log \relax (x) + 3 \, x + \frac {2 \, \log \left (x + 2 \, \log \relax (2) - 4\right )}{\log \relax (2) - 2} - \frac {2 \, \log \relax (x)}{\log \relax (2) - 2} + 5 \, \log \left (x + 2 \, \log \relax (2) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2*(-x+1)*log(2)+5*x-4)/(2*x*log(2)+x^2
-4*x),x, algorithm="maxima")

[Out]

-(log(x + 2*log(2) - 4)/(log(2) - 2) - log(x)/(log(2) - 2))*log(2) + (x + 2*log(2) - 4)*log(x + 2*log(2) - 4)
- 2*log(2)*log(x + 2*log(2) - 4) + x*log(x) + 3*x + 2*log(x + 2*log(2) - 4)/(log(2) - 2) - 2*log(x)/(log(2) -
2) + 5*log(x + 2*log(2) - 4)

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mupad [B]  time = 4.41, size = 30, normalized size = 1.50 \begin {gather*} \ln \relax (x)-2\,x+x\,\ln \left (2\,x\,{\mathrm {e}}^5\,\ln \relax (2)-{\mathrm {e}}^5\,\left (4\,x-x^2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(2*x*exp(5)*log(2) - exp(5)*(4*x - x^2))*(2*x*log(2) - 4*x + x^2) - 2*log(2)*(x - 1) - 4)/(2*x*l
og(2) - 4*x + x^2),x)

[Out]

log(x) - 2*x + x*log(2*x*exp(5)*log(2) - exp(5)*(4*x - x^2))

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sympy [A]  time = 0.19, size = 29, normalized size = 1.45 \begin {gather*} x \log {\left (2 x e^{5} \log {\relax (2 )} + \left (x^{2} - 4 x\right ) e^{5} \right )} - 2 x + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(2*x*exp(5)*log(2) + (x**2 - 4*x)*exp(5)) - 2*x + log(x)

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