3.31.80 \(\int \frac {e^{\log ^2(\frac {1}{4} e^{-x} (4 e^{2 x}+4 x+e^x (19+4 x)))} (8+8 e^x+8 e^{2 x}-8 x) \log (\frac {1}{4} e^{-x} (4 e^{2 x}+4 x+e^x (19+4 x)))}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx\)

Optimal. Leaf size=20 \[ e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \]

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Rubi [A]  time = 8.07, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 12, 6742, 6706} \begin {gather*} e^{\log ^2\left (e^{-x} x+x+e^x+\frac {19}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)]^2*(8 + 8*E^x + 8*E^(2*x) - 8*x)*Log[(4*E^(2*x) + 4*x +
E^x*(19 + 4*x))/(4*E^x)])/(4*E^(2*x) + 4*x + E^x*(19 + 4*x)),x]

[Out]

E^Log[19/4 + E^x + x + x/E^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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

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 {8 e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \left (1+e^x+e^{2 x}-x\right ) \log \left (\frac {19}{4}+e^x+x+e^{-x} x\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx\\ &=8 \int \frac {e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \left (1+e^x+e^{2 x}-x\right ) \log \left (\frac {19}{4}+e^x+x+e^{-x} x\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx\\ &=e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.86, size = 20, normalized size = 1.00 \begin {gather*} e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)]^2*(8 + 8*E^x + 8*E^(2*x) - 8*x)*Log[(4*E^(2*x) +
4*x + E^x*(19 + 4*x))/(4*E^x)])/(4*E^(2*x) + 4*x + E^x*(19 + 4*x)),x]

[Out]

E^Log[19/4 + E^x + x + x/E^x]^2

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fricas [A]  time = 0.58, size = 28, normalized size = 1.40 \begin {gather*} e^{\left (\log \left (\frac {1}{4} \, {\left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2
+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="fricas")

[Out]

e^(log(1/4*((4*x + 19)*e^x + 4*x + 4*e^(2*x))*e^(-x))^2)

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giac [B]  time = 0.32, size = 85, normalized size = 4.25 \begin {gather*} e^{\left (x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, x \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right ) - 4 \, \log \relax (2) \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right ) + \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2
+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="giac")

[Out]

e^(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*x*log(4*x*e^x + 4*x + 4*e^(2*x) + 19*e^x) - 4*log(2)*log(4*x*e^x + 4*x +
4*e^(2*x) + 19*e^x) + log(4*x*e^x + 4*x + 4*e^(2*x) + 19*e^x)^2)

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maple [C]  time = 0.20, size = 198, normalized size = 9.90

method	result	size
risch	\({\mathrm e}^{\frac {\left (i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{3}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{2} \mathrm {csgn}\left (i \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )+2 \ln \left ({\mathrm e}^{x}\right )-2 \ln \left (x \left ({\mathrm e}^{x}+1\right )+{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{2}}{4}}\)	\(198\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(x)^2+8*exp(x)-8*x+8)*ln(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))*exp(ln(1/4*(4*exp(x)^2+(4*x+19
)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2+(4*x+19)*exp(x)+4*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/4*(I*Pi*csgn(I*exp(-x)*(x*(exp(x)+1)+exp(2*x)+19/4*exp(x)))^3-I*Pi*csgn(I*exp(-x)*(x*(exp(x)+1)+exp(2*x)
+19/4*exp(x)))^2*csgn(I*exp(-x))-I*Pi*csgn(I*exp(-x)*(x*(exp(x)+1)+exp(2*x)+19/4*exp(x)))^2*csgn(I*(x*(exp(x)+
1)+exp(2*x)+19/4*exp(x)))+I*Pi*csgn(I*exp(-x)*(x*(exp(x)+1)+exp(2*x)+19/4*exp(x)))*csgn(I*exp(-x))*csgn(I*(x*(
exp(x)+1)+exp(2*x)+19/4*exp(x)))+2*ln(exp(x))-2*ln(x*(exp(x)+1)+exp(2*x)+19/4*exp(x)))^2)

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maxima [B]  time = 0.68, size = 82, normalized size = 4.10 \begin {gather*} e^{\left (x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, x \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right ) - 4 \, \log \relax (2) \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right ) + \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2
+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="maxima")

[Out]

e^(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*x*log((4*x + 19)*e^x + 4*x + 4*e^(2*x)) - 4*log(2)*log((4*x + 19)*e^x + 4
*x + 4*e^(2*x)) + log((4*x + 19)*e^x + 4*x + 4*e^(2*x))^2)

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mupad [B]  time = 2.01, size = 15, normalized size = 0.75 \begin {gather*} {\mathrm {e}}^{{\ln \left (x+{\mathrm {e}}^x+x\,{\mathrm {e}}^{-x}+\frac {19}{4}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(exp(-x)*(x + exp(2*x) + (exp(x)*(4*x + 19))/4))^2)*log(exp(-x)*(x + exp(2*x) + (exp(x)*(4*x + 19)
)/4))*(8*exp(2*x) - 8*x + 8*exp(x) + 8))/(4*x + 4*exp(2*x) + exp(x)*(4*x + 19)),x)

[Out]

exp(log(x + exp(x) + x*exp(-x) + 19/4)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)**2+8*exp(x)-8*x+8)*ln(1/4*(4*exp(x)**2+(4*x+19)*exp(x)+4*x)/exp(x))*exp(ln(1/4*(4*exp(x)**
2+(4*x+19)*exp(x)+4*x)/exp(x))**2)/(4*exp(x)**2+(4*x+19)*exp(x)+4*x),x)

[Out]

Timed out

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