3.9.4 \(\int \frac {25+e^2 (1-5 x)-125 x+e (-10+50 x)+(-50 x+20 e x-2 e^2 x) \log (4 x)}{9 x^3-6 x^2 \log (x)+x \log ^2(x)+(12 x^3-4 x^2 \log (x)) \log (4 x)+4 x^3 \log ^2(4 x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {(5-e)^2}{3 x-\log (x)+2 x \log (4 x)} \]

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Rubi [A]  time = 0.21, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {(5-e)^2}{\log (x)-x (2 \log (4 x)+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + E^2*(1 - 5*x) - 125*x + E*(-10 + 50*x) + (-50*x + 20*E*x - 2*E^2*x)*Log[4*x])/(9*x^3 - 6*x^2*Log[x]
+ x*Log[x]^2 + (12*x^3 - 4*x^2*Log[x])*Log[4*x] + 4*x^3*Log[4*x]^2),x]

[Out]

-((5 - E)^2/(Log[x] - x*(3 + 2*Log[4*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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.92 \begin {gather*} -\frac {(-5+e)^2}{\log (x)-x (3+2 \log (4 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + E^2*(1 - 5*x) - 125*x + E*(-10 + 50*x) + (-50*x + 20*E*x - 2*E^2*x)*Log[4*x])/(9*x^3 - 6*x^2*L
og[x] + x*Log[x]^2 + (12*x^3 - 4*x^2*Log[x])*Log[4*x] + 4*x^3*Log[4*x]^2),x]

[Out]

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

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fricas [A]  time = 0.71, size = 28, normalized size = 1.12 \begin {gather*} \frac {e^{2} - 10 \, e + 25}{4 \, x \log \relax (2) + {\left (2 \, x - 1\right )} \log \relax (x) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2+20*x*exp(1)-50*x)*log(4*x)+(-5*x+1)*exp(1)^2+(50*x-10)*exp(1)-125*x+25)/(4*x^3*log(4
*x)^2+(-4*x^2*log(x)+12*x^3)*log(4*x)+x*log(x)^2-6*x^2*log(x)+9*x^3),x, algorithm="fricas")

[Out]

(e^2 - 10*e + 25)/(4*x*log(2) + (2*x - 1)*log(x) + 3*x)

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giac [A]  time = 0.81, size = 29, normalized size = 1.16 \begin {gather*} \frac {e^{2} - 10 \, e + 25}{4 \, x \log \relax (2) + 2 \, x \log \relax (x) + 3 \, x - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2+20*x*exp(1)-50*x)*log(4*x)+(-5*x+1)*exp(1)^2+(50*x-10)*exp(1)-125*x+25)/(4*x^3*log(4
*x)^2+(-4*x^2*log(x)+12*x^3)*log(4*x)+x*log(x)^2-6*x^2*log(x)+9*x^3),x, algorithm="giac")

[Out]

(e^2 - 10*e + 25)/(4*x*log(2) + 2*x*log(x) + 3*x - log(x))

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maple [C]  time = 0.14, size = 87, normalized size = 3.48

method	result	size
risch	\(\frac {i {\mathrm e}^{2}}{4 i x \ln \relax (2)+2 i x \ln \relax (x )+3 i x -i \ln \relax (x )}-\frac {10 i {\mathrm e}}{4 i x \ln \relax (2)+2 i x \ln \relax (x )+3 i x -i \ln \relax (x )}+\frac {25 i}{4 i x \ln \relax (2)+2 i x \ln \relax (x )+3 i x -i \ln \relax (x )}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(1)^2+20*x*exp(1)-50*x)*ln(4*x)+(-5*x+1)*exp(1)^2+(50*x-10)*exp(1)-125*x+25)/(4*x^3*ln(4*x)^2+(-
4*x^2*ln(x)+12*x^3)*ln(4*x)+x*ln(x)^2-6*x^2*ln(x)+9*x^3),x,method=_RETURNVERBOSE)

[Out]

I/(4*I*x*ln(2)+2*I*x*ln(x)+3*I*x-I*ln(x))*exp(2)-10*I/(4*I*x*ln(2)+2*I*x*ln(x)+3*I*x-I*ln(x))*exp(1)+25*I/(4*I
*x*ln(2)+2*I*x*ln(x)+3*I*x-I*ln(x))

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maxima [A]  time = 0.68, size = 28, normalized size = 1.12 \begin {gather*} \frac {e^{2} - 10 \, e + 25}{x {\left (4 \, \log \relax (2) + 3\right )} + {\left (2 \, x - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2+20*x*exp(1)-50*x)*log(4*x)+(-5*x+1)*exp(1)^2+(50*x-10)*exp(1)-125*x+25)/(4*x^3*log(4
*x)^2+(-4*x^2*log(x)+12*x^3)*log(4*x)+x*log(x)^2-6*x^2*log(x)+9*x^3),x, algorithm="maxima")

[Out]

(e^2 - 10*e + 25)/(x*(4*log(2) + 3) + (2*x - 1)*log(x))

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mupad [B]  time = 1.25, size = 24, normalized size = 0.96 \begin {gather*} -\frac {{\left (\mathrm {e}-5\right )}^2}{\ln \relax (x)-x\,\left (2\,\ln \left (4\,x\right )+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(125*x + log(4*x)*(50*x - 20*x*exp(1) + 2*x*exp(2)) + exp(2)*(5*x - 1) - exp(1)*(50*x - 10) - 25)/(x*log(
x)^2 - 6*x^2*log(x) - log(4*x)*(4*x^2*log(x) - 12*x^3) + 9*x^3 + 4*x^3*log(4*x)^2),x)

[Out]

-(exp(1) - 5)^2/(log(x) - x*(2*log(4*x) + 3))

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sympy [A]  time = 0.41, size = 27, normalized size = 1.08 \begin {gather*} \frac {- 10 e + e^{2} + 25}{4 x \log {\relax (2 )} + 3 x + \left (2 x - 1\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)**2+20*x*exp(1)-50*x)*ln(4*x)+(-5*x+1)*exp(1)**2+(50*x-10)*exp(1)-125*x+25)/(4*x**3*ln(
4*x)**2+(-4*x**2*ln(x)+12*x**3)*ln(4*x)+x*ln(x)**2-6*x**2*ln(x)+9*x**3),x)

[Out]

(-10*E + exp(2) + 25)/(4*x*log(2) + 3*x + (2*x - 1)*log(x))

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