3.77.77 \(\int e^{1-\log ^2(-4 x) \log (x)} (2 x-4 e^4 x+2 e^8 x+(-x+2 e^4 x-e^8 x) \log ^2(-4 x)+(-2 x+4 e^4 x-2 e^8 x) \log (-4 x) \log (x)) \, dx\)

Optimal. Leaf size=25 \[ e^{1-\log ^2(-4 x) \log (x)} \left (x-e^4 x\right )^2 \]

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Rubi [B]  time = 0.24, antiderivative size = 72, normalized size of antiderivative = 2.88, number of steps used = 3, number of rules used = 2, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6, 2288} \begin {gather*} \frac {e x^{-\log ^2(-4 x)} \left (\left (1-e^4\right )^2 x \log ^2(-4 x)+2 \left (1-e^4\right )^2 x \log (x) \log (-4 x)\right )}{\frac {\log ^2(-4 x)}{x}+\frac {2 \log (x) \log (-4 x)}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(1 - Log[-4*x]^2*Log[x])*(2*x - 4*E^4*x + 2*E^8*x + (-x + 2*E^4*x - E^8*x)*Log[-4*x]^2 + (-2*x + 4*E^4*x
- 2*E^8*x)*Log[-4*x]*Log[x]),x]

[Out]

(E*((1 - E^4)^2*x*Log[-4*x]^2 + 2*(1 - E^4)^2*x*Log[-4*x]*Log[x]))/(x^Log[-4*x]^2*(Log[-4*x]^2/x + (2*Log[-4*x
]*Log[x])/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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{1-\log ^2(-4 x) \log (x)} \left (2 e^8 x+\left (2-4 e^4\right ) x+\left (-x+2 e^4 x-e^8 x\right ) \log ^2(-4 x)+\left (-2 x+4 e^4 x-2 e^8 x\right ) \log (-4 x) \log (x)\right ) \, dx\\ &=\int e^{1-\log ^2(-4 x) \log (x)} \left (\left (2-4 e^4+2 e^8\right ) x+\left (-x+2 e^4 x-e^8 x\right ) \log ^2(-4 x)+\left (-2 x+4 e^4 x-2 e^8 x\right ) \log (-4 x) \log (x)\right ) \, dx\\ &=\frac {e x^{-\log ^2(-4 x)} \left (\left (1-e^4\right )^2 x \log ^2(-4 x)+2 \left (1-e^4\right )^2 x \log (-4 x) \log (x)\right )}{\frac {\log ^2(-4 x)}{x}+\frac {2 \log (-4 x) \log (x)}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 21, normalized size = 0.84 \begin {gather*} e \left (-1+e^4\right )^2 x^{2-\log ^2(-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(1 - Log[-4*x]^2*Log[x])*(2*x - 4*E^4*x + 2*E^8*x + (-x + 2*E^4*x - E^8*x)*Log[-4*x]^2 + (-2*x + 4
*E^4*x - 2*E^8*x)*Log[-4*x]*Log[x]),x]

[Out]

E*(-1 + E^4)^2*x^(2 - Log[-4*x]^2)

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fricas [A]  time = 0.96, size = 48, normalized size = 1.92 \begin {gather*} {\left (x^{2} e^{8} - 2 \, x^{2} e^{4} + x^{2}\right )} \cos \left (\pi \log \left (-4 \, x\right )^{2}\right ) e^{\left (2 \, \log \relax (2) \log \left (-4 \, x\right )^{2} - \log \left (-4 \, x\right )^{3} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*e^8 - 2*x^2*e^4 + x^2)*cos(pi*log(-4*x)^2)*e^(2*log(2)*log(-4*x)^2 - log(-4*x)^3 + 1)

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giac [B]  time = 0.34, size = 675, normalized size = 27.00 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2*e^(-6*pi^2*log(2)*sgn(x) - 11/2*pi^2*log(abs(x))*sgn(x) + 6*pi^2*log(2) + 19/2*pi^2*log(abs(x)) - 4*log(
2)^2*log(abs(x)) - 4*log(2)*log(abs(x))^2 - log(abs(x))^3 + 9)*tan(9/4*pi^3*sgn(x) - pi*log(2)^2*sgn(x) - 2*pi
*log(2)*log(abs(x))*sgn(x) - 3/4*pi*log(abs(x))^2*sgn(x) - 9/4*pi^3 + pi*log(2)^2 + 6*pi*log(2)*log(abs(x)) +
11/4*pi*log(abs(x))^2)^2 - 2*x^2*e^(-6*pi^2*log(2)*sgn(x) - 11/2*pi^2*log(abs(x))*sgn(x) + 6*pi^2*log(2) + 19/
2*pi^2*log(abs(x)) - 4*log(2)^2*log(abs(x)) - 4*log(2)*log(abs(x))^2 - log(abs(x))^3 + 5)*tan(9/4*pi^3*sgn(x)
- pi*log(2)^2*sgn(x) - 2*pi*log(2)*log(abs(x))*sgn(x) - 3/4*pi*log(abs(x))^2*sgn(x) - 9/4*pi^3 + pi*log(2)^2 +
6*pi*log(2)*log(abs(x)) + 11/4*pi*log(abs(x))^2)^2 + x^2*e^(-6*pi^2*log(2)*sgn(x) - 11/2*pi^2*log(abs(x))*sgn
(x) + 6*pi^2*log(2) + 19/2*pi^2*log(abs(x)) - 4*log(2)^2*log(abs(x)) - 4*log(2)*log(abs(x))^2 - log(abs(x))^3
+ 1)*tan(9/4*pi^3*sgn(x) - pi*log(2)^2*sgn(x) - 2*pi*log(2)*log(abs(x))*sgn(x) - 3/4*pi*log(abs(x))^2*sgn(x) -
9/4*pi^3 + pi*log(2)^2 + 6*pi*log(2)*log(abs(x)) + 11/4*pi*log(abs(x))^2)^2 - x^2*e^(-6*pi^2*log(2)*sgn(x) -
11/2*pi^2*log(abs(x))*sgn(x) + 6*pi^2*log(2) + 19/2*pi^2*log(abs(x)) - 4*log(2)^2*log(abs(x)) - 4*log(2)*log(a
bs(x))^2 - log(abs(x))^3 + 9) + 2*x^2*e^(-6*pi^2*log(2)*sgn(x) - 11/2*pi^2*log(abs(x))*sgn(x) + 6*pi^2*log(2)
+ 19/2*pi^2*log(abs(x)) - 4*log(2)^2*log(abs(x)) - 4*log(2)*log(abs(x))^2 - log(abs(x))^3 + 5) - x^2*e^(-6*pi^
2*log(2)*sgn(x) - 11/2*pi^2*log(abs(x))*sgn(x) + 6*pi^2*log(2) + 19/2*pi^2*log(abs(x)) - 4*log(2)^2*log(abs(x)
) - 4*log(2)*log(abs(x))^2 - log(abs(x))^3 + 1))/(tan(9/4*pi^3*sgn(x) - pi*log(2)^2*sgn(x) - 2*pi*log(2)*log(a
bs(x))*sgn(x) - 3/4*pi*log(abs(x))^2*sgn(x) - 9/4*pi^3 + pi*log(2)^2 + 6*pi*log(2)*log(abs(x)) + 11/4*pi*log(a
bs(x))^2)^2 + 1)

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maple [C]  time = 0.14, size = 37, normalized size = 1.48

method	result	size
risch	\(x^{2} \left ({\mathrm e}^{8}-2 \,{\mathrm e}^{4}+1\right ) x^{-\left (i \pi \,\mathrm {csgn}\left (i x \right )+\ln \relax (x )+2 \ln \relax (2)\right )^{2}} {\mathrm e}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(4)^2+4*x*exp(4)-2*x)*ln(-4*x)*ln(x)+(-x*exp(4)^2+2*x*exp(4)-x)*ln(-4*x)^2+2*x*exp(4)^2-4*x*exp(
4)+2*x)*exp(-ln(-4*x)^2*ln(x)+1),x,method=_RETURNVERBOSE)

[Out]

x^2*(exp(8)-2*exp(4)+1)*x^(-(I*Pi*csgn(I*x)+ln(x)+2*ln(2))^2)*exp(1)

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maxima [C]  time = 0.56, size = 57, normalized size = 2.28 \begin {gather*} x^{2} {\left (e^{9} - 2 \, e^{5} + e\right )} e^{\left (\pi ^{2} \log \relax (x) - 4 i \, \pi \log \relax (2) \log \relax (x) - 4 \, \log \relax (2)^{2} \log \relax (x) - 2 i \, \pi \log \relax (x)^{2} - 4 \, \log \relax (2) \log \relax (x)^{2} - \log \relax (x)^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*(e^9 - 2*e^5 + e)*e^(pi^2*log(x) - 4*I*pi*log(2)*log(x) - 4*log(2)^2*log(x) - 2*I*pi*log(x)^2 - 4*log(2)*l
og(x)^2 - log(x)^3)

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mupad [B]  time = 5.53, size = 21, normalized size = 0.84 \begin {gather*} x^{2-{\ln \left (-4\,x\right )}^2}\,\mathrm {e}\,{\left ({\mathrm {e}}^4-1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(1 - log(-4*x)^2*log(x))*(4*x*exp(4) - 2*x - 2*x*exp(8) + log(-4*x)^2*(x - 2*x*exp(4) + x*exp(8)) + lo
g(-4*x)*log(x)*(2*x - 4*x*exp(4) + 2*x*exp(8))),x)

[Out]

x^(2 - log(-4*x)^2)*exp(1)*(exp(4) - 1)^2

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sympy [C]  time = 0.91, size = 36, normalized size = 1.44 \begin {gather*} \left (- 2 x^{2} e^{4} + x^{2} + x^{2} e^{8}\right ) e^{- \left (\log {\relax (x )} + \log {\relax (4 )} + i \pi \right )^{2} \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*x**2*exp(4) + x**2 + x**2*exp(8))*exp(-(log(x) + log(4) + I*pi)**2*log(x) + 1)

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