3.58.41 \(\int \frac {32 x^3-6 x^4+e (-2 x-32 x^2+12 x^3)+2 x^2 \log (e^2 x)+(e^2 (2+32 x-12 x^2)+e (-32 x^2+6 x^3)-2 e x \log (e^2 x)) \log (16 x-3 x^2+\log (e^2 x))}{16 x^2-3 x^3+x \log (e^2 x)} \, dx\)

Optimal. Leaf size=26 \[ 1+\left (-x+e \log \left (-x (-16+3 x)+\log \left (e^2 x\right )\right )\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 20, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 3, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 12, 6686} \begin {gather*} \left (x-e \log \left (-3 x^2+16 x+\log (x)+2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*x^3 - 6*x^4 + E*(-2*x - 32*x^2 + 12*x^3) + 2*x^2*Log[E^2*x] + (E^2*(2 + 32*x - 12*x^2) + E*(-32*x^2 +
6*x^3) - 2*E*x*Log[E^2*x])*Log[16*x - 3*x^2 + Log[E^2*x]])/(16*x^2 - 3*x^3 + x*Log[E^2*x]),x]

[Out]

(x - E*Log[2 + 16*x - 3*x^2 + Log[x]])^2

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 {2 \left (x \left (2+16 x-3 x^2\right )+e \left (-1-16 x+6 x^2\right )+x \log (x)\right ) \left (x-e \log \left (2+16 x-3 x^2+\log (x)\right )\right )}{x \left (2+16 x-3 x^2+\log (x)\right )} \, dx\\ &=2 \int \frac {\left (x \left (2+16 x-3 x^2\right )+e \left (-1-16 x+6 x^2\right )+x \log (x)\right ) \left (x-e \log \left (2+16 x-3 x^2+\log (x)\right )\right )}{x \left (2+16 x-3 x^2+\log (x)\right )} \, dx\\ &=\left (x-e \log \left (2+16 x-3 x^2+\log (x)\right )\right )^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 21, normalized size = 0.81 \begin {gather*} \left (-x+e \log \left (2+16 x-3 x^2+\log (x)\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x^3 - 6*x^4 + E*(-2*x - 32*x^2 + 12*x^3) + 2*x^2*Log[E^2*x] + (E^2*(2 + 32*x - 12*x^2) + E*(-32*
x^2 + 6*x^3) - 2*E*x*Log[E^2*x])*Log[16*x - 3*x^2 + Log[E^2*x]])/(16*x^2 - 3*x^3 + x*Log[E^2*x]),x]

[Out]

(-x + E*Log[2 + 16*x - 3*x^2 + Log[x]])^2

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 44, normalized size = 1.69 \begin {gather*} -2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \left (x e^{2}\right )\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \left (x e^{2}\right )\right )^{2} + x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*exp(1))*log(log(exp(2)*x)-3*x^2
+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32*x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorith
m="fricas")

[Out]

-2*x*e*log(-3*x^2 + 16*x + log(x*e^2)) + e^2*log(-3*x^2 + 16*x + log(x*e^2))^2 + x^2

________________________________________________________________________________________

giac [A]  time = 0.18, size = 40, normalized size = 1.54 \begin {gather*} -2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \relax (x) + 2\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \relax (x) + 2\right )^{2} + x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*exp(1))*log(log(exp(2)*x)-3*x^2
+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32*x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorith
m="giac")

[Out]

-2*x*e*log(-3*x^2 + 16*x + log(x) + 2) + e^2*log(-3*x^2 + 16*x + log(x) + 2)^2 + x^2

________________________________________________________________________________________

maple [A]  time = 0.09, size = 45, normalized size = 1.73

method	result	size
risch	\({\mathrm e}^{2} \ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right )^{2}-2 \,{\mathrm e} \ln \left (\ln \left ({\mathrm e}^{2} x \right )-3 x^{2}+16 x \right ) x +x^{2}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(1)*ln(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*exp(1))*ln(ln(exp(2)*x)-3*x^2+16*x)+2*
x^2*ln(exp(2)*x)+(12*x^3-32*x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*ln(exp(2)*x)-3*x^3+16*x^2),x,method=_RETURNVERBOS
E)

[Out]

exp(2)*ln(ln(exp(2)*x)-3*x^2+16*x)^2-2*exp(1)*ln(ln(exp(2)*x)-3*x^2+16*x)*x+x^2

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 40, normalized size = 1.54 \begin {gather*} -2 \, x e \log \left (-3 \, x^{2} + 16 \, x + \log \relax (x) + 2\right ) + e^{2} \log \left (-3 \, x^{2} + 16 \, x + \log \relax (x) + 2\right )^{2} + x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)*log(exp(2)*x)+(-12*x^2+32*x+2)*exp(1)^2+(6*x^3-32*x^2)*exp(1))*log(log(exp(2)*x)-3*x^2
+16*x)+2*x^2*log(exp(2)*x)+(12*x^3-32*x^2-2*x)*exp(1)-6*x^4+32*x^3)/(x*log(exp(2)*x)-3*x^3+16*x^2),x, algorith
m="maxima")

[Out]

-2*x*e*log(-3*x^2 + 16*x + log(x) + 2) + e^2*log(-3*x^2 + 16*x + log(x) + 2)^2 + x^2

________________________________________________________________________________________

mupad [B]  time = 4.00, size = 21, normalized size = 0.81 \begin {gather*} {\left (x-\mathrm {e}\,\ln \left (16\,x+\ln \relax (x)-3\,x^2+2\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)*(2*x + 32*x^2 - 12*x^3) - 32*x^3 + 6*x^4 - 2*x^2*log(x*exp(2)) + log(16*x + log(x*exp(2)) - 3*x^2
)*(exp(1)*(32*x^2 - 6*x^3) - exp(2)*(32*x - 12*x^2 + 2) + 2*x*exp(1)*log(x*exp(2))))/(x*log(x*exp(2)) + 16*x^2
- 3*x^3),x)

[Out]

(x - exp(1)*log(16*x + log(x) - 3*x^2 + 2))^2

________________________________________________________________________________________

sympy [B]  time = 0.49, size = 48, normalized size = 1.85 \begin {gather*} x^{2} - 2 e x \log {\left (- 3 x^{2} + 16 x + \log {\left (x e^{2} \right )} \right )} + e^{2} \log {\left (- 3 x^{2} + 16 x + \log {\left (x e^{2} \right )} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)*ln(exp(2)*x)+(-12*x**2+32*x+2)*exp(1)**2+(6*x**3-32*x**2)*exp(1))*ln(ln(exp(2)*x)-3*x*
*2+16*x)+2*x**2*ln(exp(2)*x)+(12*x**3-32*x**2-2*x)*exp(1)-6*x**4+32*x**3)/(x*ln(exp(2)*x)-3*x**3+16*x**2),x)

[Out]

x**2 - 2*E*x*log(-3*x**2 + 16*x + log(x*exp(2))) + exp(2)*log(-3*x**2 + 16*x + log(x*exp(2)))**2

________________________________________________________________________________________