3.103.77 \(\int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log (-12-4 e+4 e^5-4 e^x)}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+(3 x+e x-e^5 x+e^x x) \log ^2(-12-4 e+4 e^5-4 e^x)+(3 x+e x-e^5 x+e^x x) \log (x)} \, dx\)

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

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Rubi [A]  time = 0.60, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, integrand size = 141, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6, 6688, 6684} \begin {gather*} \log \left (2 x+\log ^2\left (-4 \left (e^x+3+e-e^5\right )\right )+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + E^5*(-1 - 2*x) + 6*x + E*(1 + 2*x) + E^x*(1 + 2*x) + 2*E^x*x*Log[-12 - 4*E + 4*E^5 - 4*E^x])/(6*x^2 +
2*E*x^2 - 2*E^5*x^2 + 2*E^x*x^2 + (3*x + E*x - E^5*x + E^x*x)*Log[-12 - 4*E + 4*E^5 - 4*E^x]^2 + (3*x + E*x -
E^5*x + E^x*x)*Log[x]),x]

[Out]

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

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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{-2 e^5 x^2+2 e^x x^2+(6+2 e) x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx\\ &=\int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{2 e^x x^2+\left (6+2 e-2 e^5\right ) x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx\\ &=\int \frac {\left (3+e-e^5+e^x\right ) (1+2 x)+2 e^x x \log \left (-4 \left (3+e-e^5+e^x\right )\right )}{\left (e^x+3 \left (1-\frac {1}{3} e \left (-1+e^4\right )\right )\right ) x \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right )} \, dx\\ &=\log \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.28, size = 23, normalized size = 0.92 \begin {gather*} \log \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + E^5*(-1 - 2*x) + 6*x + E*(1 + 2*x) + E^x*(1 + 2*x) + 2*E^x*x*Log[-12 - 4*E + 4*E^5 - 4*E^x])/(6
*x^2 + 2*E*x^2 - 2*E^5*x^2 + 2*E^x*x^2 + (3*x + E*x - E^5*x + E^x*x)*Log[-12 - 4*E + 4*E^5 - 4*E^x]^2 + (3*x +
E*x - E^5*x + E^x*x)*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.59, size = 24, normalized size = 0.96 \begin {gather*} \log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(2*x+1)*exp(x)+(-2*x-1)*exp(5)+(2*x+1)*exp(1)+6*x+3)
/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(
x)+2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm="fricas")

[Out]

log(log(4*e^5 - 4*e - 4*e^x - 12)^2 + 2*x + log(x))

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giac [A]  time = 0.49, size = 24, normalized size = 0.96 \begin {gather*} \log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(2*x+1)*exp(x)+(-2*x-1)*exp(5)+(2*x+1)*exp(1)+6*x+3)
/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(
x)+2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm="giac")

[Out]

log(log(4*e^5 - 4*e - 4*e^x - 12)^2 + 2*x + log(x))

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maple [A]  time = 0.06, size = 25, normalized size = 1.00

method	result	size
risch	\(\ln \left (\ln \relax (x )+\ln \left (-4 \,{\mathrm e}^{x}+4 \,{\mathrm e}^{5}-4 \,{\mathrm e}-12\right )^{2}+2 x \right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(2*x+1)*exp(x)+(-2*x-1)*exp(5)+(2*x+1)*exp(1)+6*x+3)/((exp(
x)*x-x*exp(5)+x*exp(1)+3*x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(x)+2*exp(
x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)+ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+2*x)

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maxima [C]  time = 0.55, size = 57, normalized size = 2.28 \begin {gather*} \log \left (-\pi ^{2} + 4 i \, \pi \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, {\left (-i \, \pi - 2 \, \log \relax (2)\right )} \log \left (-e^{5} + e + e^{x} + 3\right ) + \log \left (-e^{5} + e + e^{x} + 3\right )^{2} + 2 \, x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(2*x+1)*exp(x)+(-2*x-1)*exp(5)+(2*x+1)*exp(1)+6*x+3)
/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(
x)+2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm="maxima")

[Out]

log(-pi^2 + 4*I*pi*log(2) + 4*log(2)^2 - 2*(-I*pi - 2*log(2))*log(-e^5 + e + e^x + 3) + log(-e^5 + e + e^x + 3
)^2 + 2*x + log(x))

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mupad [B]  time = 7.53, size = 24, normalized size = 0.96 \begin {gather*} \ln \left ({\ln \left (4\,{\mathrm {e}}^5-4\,\mathrm {e}-4\,{\mathrm {e}}^x-12\right )}^2+2\,x+\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + exp(x)*(2*x + 1) + exp(1)*(2*x + 1) - exp(5)*(2*x + 1) + 2*x*exp(x)*log(4*exp(5) - 4*exp(1) - 4*exp
(x) - 12) + 3)/(2*x^2*exp(x) + 2*x^2*exp(1) - 2*x^2*exp(5) + log(x)*(3*x + x*exp(1) - x*exp(5) + x*exp(x)) + l
og(4*exp(5) - 4*exp(1) - 4*exp(x) - 12)^2*(3*x + x*exp(1) - x*exp(5) + x*exp(x)) + 6*x^2),x)

[Out]

log(2*x + log(x) + log(4*exp(5) - 4*exp(1) - 4*exp(x) - 12)^2)

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sympy [A]  time = 3.41, size = 27, normalized size = 1.08 \begin {gather*} \log {\left (2 x + \log {\relax (x )} + \log {\left (- 4 e^{x} - 12 - 4 e + 4 e^{5} \right )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(2*x+1)*exp(x)+(-2*x-1)*exp(5)+(2*x+1)*exp(1)+6*x+3)/
((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)**2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(x)
+2*exp(x)*x**2-2*x**2*exp(5)+2*x**2*exp(1)+6*x**2),x)

[Out]

log(2*x + log(x) + log(-4*exp(x) - 12 - 4*E + 4*exp(5))**2)

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