3.39.7 \(\int \frac {16 e^{4+x} x+e^4 (8 x-56 x^2+64 x^3-8 x^5)+e^4 (32 x-32 x^2-8 x^3+8 x^4) \log (4)+e^4 (4 x^2-2 x^3) \log ^2(4)+(16 e^{4+x}+e^4 (8-56 x+64 x^2-8 x^4)+e^4 (32-32 x-8 x^2+8 x^3) \log (4)+e^4 (4 x-2 x^2) \log ^2(4)) \log (4 e^x+4 x-16 x^2+16 x^3-4 x^4+(16 x-16 x^2+4 x^3) \log (4)+(-4+4 x-x^2) \log ^2(4))}{4 e^x+4 x-16 x^2+16 x^3-4 x^4+(16 x-16 x^2+4 x^3) \log (4)+(-4+4 x-x^2) \log ^2(4)} \, dx\)

Optimal. Leaf size=35 \[ e^4 \left (x+\log \left (4 \left (e^x+x-(2-x)^2 \left (x-\frac {\log (4)}{2}\right )^2\right )\right )\right )^2 \]

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Rubi [A]  time = 1.17, antiderivative size = 58, normalized size of antiderivative = 1.66, number of steps used = 3, number of rules used = 3, integrand size = 258, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 12, 6686} \begin {gather*} e^4 \left (\log \left (-4 x^4+16 x^3-16 x^2+4 x+4 e^x-(2-x)^2 \log ^2(4)+4 (2-x)^2 x \log (4)\right )+x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*E^(4 + x)*x + E^4*(8*x - 56*x^2 + 64*x^3 - 8*x^5) + E^4*(32*x - 32*x^2 - 8*x^3 + 8*x^4)*Log[4] + E^4*(
4*x^2 - 2*x^3)*Log[4]^2 + (16*E^(4 + x) + E^4*(8 - 56*x + 64*x^2 - 8*x^4) + E^4*(32 - 32*x - 8*x^2 + 8*x^3)*Lo
g[4] + E^4*(4*x - 2*x^2)*Log[4]^2)*Log[4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*Log[4]
+ (-4 + 4*x - x^2)*Log[4]^2])/(4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*Log[4] + (-4 +
4*x - x^2)*Log[4]^2),x]

[Out]

E^4*(x + Log[4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + 4*(2 - x)^2*x*Log[4] - (2 - x)^2*Log[4]^2])^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 e^4 \left (8 e^x-4 x^4+4 x^3 \log (4)+2 x \left (-14-8 \log (4)+\log ^2(4)\right )+4 (1+\log (256))-x^2 \left (-32+\log ^2(4)+\log (256)\right )\right ) \left (x+\log \left (4 e^x+4 x-16 x^2+16 x^3-4 x^4+4 (-2+x)^2 x \log (4)-(-2+x)^2 \log ^2(4)\right )\right )}{4 e^x+4 x-16 x^2+16 x^3-4 x^4+4 (-2+x)^2 x \log (4)-(-2+x)^2 \log ^2(4)} \, dx\\ &=\left (2 e^4\right ) \int \frac {\left (8 e^x-4 x^4+4 x^3 \log (4)+2 x \left (-14-8 \log (4)+\log ^2(4)\right )+4 (1+\log (256))-x^2 \left (-32+\log ^2(4)+\log (256)\right )\right ) \left (x+\log \left (4 e^x+4 x-16 x^2+16 x^3-4 x^4+4 (-2+x)^2 x \log (4)-(-2+x)^2 \log ^2(4)\right )\right )}{4 e^x+4 x-16 x^2+16 x^3-4 x^4+4 (-2+x)^2 x \log (4)-(-2+x)^2 \log ^2(4)} \, dx\\ &=e^4 \left (x+\log \left (4 e^x+4 x-16 x^2+16 x^3-4 x^4+4 (2-x)^2 x \log (4)-(2-x)^2 \log ^2(4)\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 4.93, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {16 e^{4+x} x+e^4 \left (8 x-56 x^2+64 x^3-8 x^5\right )+e^4 \left (32 x-32 x^2-8 x^3+8 x^4\right ) \log (4)+e^4 \left (4 x^2-2 x^3\right ) \log ^2(4)+\left (16 e^{4+x}+e^4 \left (8-56 x+64 x^2-8 x^4\right )+e^4 \left (32-32 x-8 x^2+8 x^3\right ) \log (4)+e^4 \left (4 x-2 x^2\right ) \log ^2(4)\right ) \log \left (4 e^x+4 x-16 x^2+16 x^3-4 x^4+\left (16 x-16 x^2+4 x^3\right ) \log (4)+\left (-4+4 x-x^2\right ) \log ^2(4)\right )}{4 e^x+4 x-16 x^2+16 x^3-4 x^4+\left (16 x-16 x^2+4 x^3\right ) \log (4)+\left (-4+4 x-x^2\right ) \log ^2(4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(16*E^(4 + x)*x + E^4*(8*x - 56*x^2 + 64*x^3 - 8*x^5) + E^4*(32*x - 32*x^2 - 8*x^3 + 8*x^4)*Log[4] +
E^4*(4*x^2 - 2*x^3)*Log[4]^2 + (16*E^(4 + x) + E^4*(8 - 56*x + 64*x^2 - 8*x^4) + E^4*(32 - 32*x - 8*x^2 + 8*x
^3)*Log[4] + E^4*(4*x - 2*x^2)*Log[4]^2)*Log[4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*L
og[4] + (-4 + 4*x - x^2)*Log[4]^2])/(4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*Log[4] +
(-4 + 4*x - x^2)*Log[4]^2),x]

[Out]

Integrate[(16*E^(4 + x)*x + E^4*(8*x - 56*x^2 + 64*x^3 - 8*x^5) + E^4*(32*x - 32*x^2 - 8*x^3 + 8*x^4)*Log[4] +
E^4*(4*x^2 - 2*x^3)*Log[4]^2 + (16*E^(4 + x) + E^4*(8 - 56*x + 64*x^2 - 8*x^4) + E^4*(32 - 32*x - 8*x^2 + 8*x
^3)*Log[4] + E^4*(4*x - 2*x^2)*Log[4]^2)*Log[4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*L
og[4] + (-4 + 4*x - x^2)*Log[4]^2])/(4*E^x + 4*x - 16*x^2 + 16*x^3 - 4*x^4 + (16*x - 16*x^2 + 4*x^3)*Log[4] +
(-4 + 4*x - x^2)*Log[4]^2), x]

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fricas [B]  time = 0.87, size = 147, normalized size = 4.20 \begin {gather*} x^{2} e^{4} + 2 \, x e^{4} \log \left (-4 \, {\left ({\left (x^{2} - 4 \, x + 4\right )} e^{4} \log \relax (2)^{2} - 2 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{4} \log \relax (2) + {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2} - x\right )} e^{4} - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )}\right ) + e^{4} \log \left (-4 \, {\left ({\left (x^{2} - 4 \, x + 4\right )} e^{4} \log \relax (2)^{2} - 2 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{4} \log \relax (2) + {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2} - x\right )} e^{4} - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(4)*exp(x)+4*(-2*x^2+4*x)*exp(4)*log(2)^2+2*(8*x^3-8*x^2-32*x+32)*exp(4)*log(2)+(-8*x^4+64*x
^2-56*x+8)*exp(4))*log(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x)+
16*x*exp(4)*exp(x)+4*(-2*x^3+4*x^2)*exp(4)*log(2)^2+2*(8*x^4-8*x^3-32*x^2+32*x)*exp(4)*log(2)+(-8*x^5+64*x^3-5
6*x^2+8*x)*exp(4))/(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x),x,
algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{4} \log \relax (2)^{2} - 2 \, {\left (x^{4} - x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{4} \log \relax (2) + {\left (x^{5} - 8 \, x^{3} + 7 \, x^{2} - x\right )} e^{4} - 2 \, x e^{\left (x + 4\right )} + {\left ({\left (x^{2} - 2 \, x\right )} e^{4} \log \relax (2)^{2} - 2 \, {\left (x^{3} - x^{2} - 4 \, x + 4\right )} e^{4} \log \relax (2) + {\left (x^{4} - 8 \, x^{2} + 7 \, x - 1\right )} e^{4} - 2 \, e^{\left (x + 4\right )}\right )} \log \left (-4 \, x^{4} + 16 \, x^{3} - 4 \, {\left (x^{2} - 4 \, x + 4\right )} \log \relax (2)^{2} - 16 \, x^{2} + 8 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \relax (2) + 4 \, x + 4 \, e^{x}\right )\right )}}{x^{4} - 4 \, x^{3} + {\left (x^{2} - 4 \, x + 4\right )} \log \relax (2)^{2} + 4 \, x^{2} - 2 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \relax (2) - x - e^{x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(4)*exp(x)+4*(-2*x^2+4*x)*exp(4)*log(2)^2+2*(8*x^3-8*x^2-32*x+32)*exp(4)*log(2)+(-8*x^4+64*x
^2-56*x+8)*exp(4))*log(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x)+
16*x*exp(4)*exp(x)+4*(-2*x^3+4*x^2)*exp(4)*log(2)^2+2*(8*x^4-8*x^3-32*x^2+32*x)*exp(4)*log(2)+(-8*x^5+64*x^3-5
6*x^2+8*x)*exp(4))/(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x),x,
algorithm="giac")

[Out]

integrate(2*((x^3 - 2*x^2)*e^4*log(2)^2 - 2*(x^4 - x^3 - 4*x^2 + 4*x)*e^4*log(2) + (x^5 - 8*x^3 + 7*x^2 - x)*e
^4 - 2*x*e^(x + 4) + ((x^2 - 2*x)*e^4*log(2)^2 - 2*(x^3 - x^2 - 4*x + 4)*e^4*log(2) + (x^4 - 8*x^2 + 7*x - 1)*
e^4 - 2*e^(x + 4))*log(-4*x^4 + 16*x^3 - 4*(x^2 - 4*x + 4)*log(2)^2 - 16*x^2 + 8*(x^3 - 4*x^2 + 4*x)*log(2) +
4*x + 4*e^x))/(x^4 - 4*x^3 + (x^2 - 4*x + 4)*log(2)^2 + 4*x^2 - 2*(x^3 - 4*x^2 + 4*x)*log(2) - x - e^x), x)

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maple [B]  time = 0.06, size = 134, normalized size = 3.83

method	result	size
risch	\(x^{2} {\mathrm e}^{4}+2 x \,{\mathrm e}^{4} \ln \left (4 \,{\mathrm e}^{x}+4 \left (-x^{2}+4 x -4\right ) \ln \relax (2)^{2}+2 \left (4 x^{3}-16 x^{2}+16 x \right ) \ln \relax (2)-4 x^{4}+16 x^{3}-16 x^{2}+4 x \right )+{\mathrm e}^{4} \ln \left (4 \,{\mathrm e}^{x}+4 \left (-x^{2}+4 x -4\right ) \ln \relax (2)^{2}+2 \left (4 x^{3}-16 x^{2}+16 x \right ) \ln \relax (2)-4 x^{4}+16 x^{3}-16 x^{2}+4 x \right )^{2}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*exp(4)*exp(x)+4*(-2*x^2+4*x)*exp(4)*ln(2)^2+2*(8*x^3-8*x^2-32*x+32)*exp(4)*ln(2)+(-8*x^4+64*x^2-56*x+
8)*exp(4))*ln(4*exp(x)+4*(-x^2+4*x-4)*ln(2)^2+2*(4*x^3-16*x^2+16*x)*ln(2)-4*x^4+16*x^3-16*x^2+4*x)+16*x*exp(4)
*exp(x)+4*(-2*x^3+4*x^2)*exp(4)*ln(2)^2+2*(8*x^4-8*x^3-32*x^2+32*x)*exp(4)*ln(2)+(-8*x^5+64*x^3-56*x^2+8*x)*ex
p(4))/(4*exp(x)+4*(-x^2+4*x-4)*ln(2)^2+2*(4*x^3-16*x^2+16*x)*ln(2)-4*x^4+16*x^3-16*x^2+4*x),x,method=_RETURNVE
RBOSE)

[Out]

x^2*exp(4)+2*x*exp(4)*ln(4*exp(x)+4*(-x^2+4*x-4)*ln(2)^2+2*(4*x^3-16*x^2+16*x)*ln(2)-4*x^4+16*x^3-16*x^2+4*x)+
exp(4)*ln(4*exp(x)+4*(-x^2+4*x-4)*ln(2)^2+2*(4*x^3-16*x^2+16*x)*ln(2)-4*x^4+16*x^3-16*x^2+4*x)^2

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maxima [B]  time = 0.61, size = 138, normalized size = 3.94 \begin {gather*} x^{2} e^{4} + 4 \, x e^{4} \log \relax (2) + e^{4} \log \left (-x^{4} + 2 \, x^{3} {\left (\log \relax (2) + 2\right )} - {\left (\log \relax (2)^{2} + 8 \, \log \relax (2) + 4\right )} x^{2} + {\left (4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) + 1\right )} x - 4 \, \log \relax (2)^{2} + e^{x}\right )^{2} + 2 \, {\left (x e^{4} + 2 \, e^{4} \log \relax (2)\right )} \log \left (-x^{4} + 2 \, x^{3} {\left (\log \relax (2) + 2\right )} - {\left (\log \relax (2)^{2} + 8 \, \log \relax (2) + 4\right )} x^{2} + {\left (4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) + 1\right )} x - 4 \, \log \relax (2)^{2} + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(4)*exp(x)+4*(-2*x^2+4*x)*exp(4)*log(2)^2+2*(8*x^3-8*x^2-32*x+32)*exp(4)*log(2)+(-8*x^4+64*x
^2-56*x+8)*exp(4))*log(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x)+
16*x*exp(4)*exp(x)+4*(-2*x^3+4*x^2)*exp(4)*log(2)^2+2*(8*x^4-8*x^3-32*x^2+32*x)*exp(4)*log(2)+(-8*x^5+64*x^3-5
6*x^2+8*x)*exp(4))/(4*exp(x)+4*(-x^2+4*x-4)*log(2)^2+2*(4*x^3-16*x^2+16*x)*log(2)-4*x^4+16*x^3-16*x^2+4*x),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.10, size = 63, normalized size = 1.80 \begin {gather*} {\mathrm {e}}^4\,{\left (x+\ln \left (4\,x+4\,{\mathrm {e}}^x-4\,{\ln \relax (2)}^2\,\left (x^2-4\,x+4\right )+2\,\ln \relax (2)\,\left (4\,x^3-16\,x^2+16\,x\right )-16\,x^2+16\,x^3-4\,x^4\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*(8*x - 56*x^2 + 64*x^3 - 8*x^5) - log(4*x + 4*exp(x) - 4*log(2)^2*(x^2 - 4*x + 4) + 2*log(2)*(16*x
- 16*x^2 + 4*x^3) - 16*x^2 + 16*x^3 - 4*x^4)*(exp(4)*(56*x - 64*x^2 + 8*x^4 - 8) - 16*exp(4)*exp(x) + 2*exp(4
)*log(2)*(32*x + 8*x^2 - 8*x^3 - 32) - 4*exp(4)*log(2)^2*(4*x - 2*x^2)) + 2*exp(4)*log(2)*(32*x - 32*x^2 - 8*x
^3 + 8*x^4) + 16*x*exp(4)*exp(x) + 4*exp(4)*log(2)^2*(4*x^2 - 2*x^3))/(4*x + 4*exp(x) - 4*log(2)^2*(x^2 - 4*x
+ 4) + 2*log(2)*(16*x - 16*x^2 + 4*x^3) - 16*x^2 + 16*x^3 - 4*x^4),x)

[Out]

exp(4)*(x + log(4*x + 4*exp(x) - 4*log(2)^2*(x^2 - 4*x + 4) + 2*log(2)*(16*x - 16*x^2 + 4*x^3) - 16*x^2 + 16*x
^3 - 4*x^4))^2

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sympy [B]  time = 1.01, size = 133, normalized size = 3.80 \begin {gather*} x^{2} e^{4} + 2 x e^{4} \log {\left (- 4 x^{4} + 16 x^{3} - 16 x^{2} + 4 x + \left (- 4 x^{2} + 16 x - 16\right ) \log {\relax (2 )}^{2} + \left (8 x^{3} - 32 x^{2} + 32 x\right ) \log {\relax (2 )} + 4 e^{x} \right )} + e^{4} \log {\left (- 4 x^{4} + 16 x^{3} - 16 x^{2} + 4 x + \left (- 4 x^{2} + 16 x - 16\right ) \log {\relax (2 )}^{2} + \left (8 x^{3} - 32 x^{2} + 32 x\right ) \log {\relax (2 )} + 4 e^{x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(4)*exp(x)+4*(-2*x**2+4*x)*exp(4)*ln(2)**2+2*(8*x**3-8*x**2-32*x+32)*exp(4)*ln(2)+(-8*x**4+6
4*x**2-56*x+8)*exp(4))*ln(4*exp(x)+4*(-x**2+4*x-4)*ln(2)**2+2*(4*x**3-16*x**2+16*x)*ln(2)-4*x**4+16*x**3-16*x*
*2+4*x)+16*x*exp(4)*exp(x)+4*(-2*x**3+4*x**2)*exp(4)*ln(2)**2+2*(8*x**4-8*x**3-32*x**2+32*x)*exp(4)*ln(2)+(-8*
x**5+64*x**3-56*x**2+8*x)*exp(4))/(4*exp(x)+4*(-x**2+4*x-4)*ln(2)**2+2*(4*x**3-16*x**2+16*x)*ln(2)-4*x**4+16*x
**3-16*x**2+4*x),x)

[Out]

x**2*exp(4) + 2*x*exp(4)*log(-4*x**4 + 16*x**3 - 16*x**2 + 4*x + (-4*x**2 + 16*x - 16)*log(2)**2 + (8*x**3 - 3
2*x**2 + 32*x)*log(2) + 4*exp(x)) + exp(4)*log(-4*x**4 + 16*x**3 - 16*x**2 + 4*x + (-4*x**2 + 16*x - 16)*log(2
)**2 + (8*x**3 - 32*x**2 + 32*x)*log(2) + 4*exp(x))**2

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