3.58.7 \(\int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} (e^x (64-64 x) \log (4)+(64 x^2-16 e^4 x^2) \log (4))}{16 e^{2 x}+400 x^2-160 x^3+16 x^4+e^8 x^4+e^x (160 x-32 x^2+8 e^4 x^2)+e^4 (40 x^3-8 x^4)} \, dx\)

Optimal. Leaf size=30 \[ 4^{\frac {4}{\frac {e^4 x}{4}+\frac {e^x+(5-x) x}{x}}} \]

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Rubi [A]  time = 0.83, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6, 6688, 12, 6706} \begin {gather*} 4^{\frac {16 x}{e^4 x^2+4 (5-x) x+4 e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4^((16*x)/(4*E^x + 20*x - 4*x^2 + E^4*x^2))*(E^x*(64 - 64*x)*Log[4] + (64*x^2 - 16*E^4*x^2)*Log[4]))/(16*
E^(2*x) + 400*x^2 - 160*x^3 + 16*x^4 + E^8*x^4 + E^x*(160*x - 32*x^2 + 8*E^4*x^2) + E^4*(40*x^3 - 8*x^4)),x]

[Out]

4^((16*x)/(4*E^x + 4*(5 - x)*x + E^4*x^2))

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \left (e^x (64-64 x) \log (4)+\left (64 x^2-16 e^4 x^2\right ) \log (4)\right )}{16 e^{2 x}+400 x^2-160 x^3+\left (16+e^8\right ) x^4+e^x \left (160 x-32 x^2+8 e^4 x^2\right )+e^4 \left (40 x^3-8 x^4\right )} \, dx\\ &=\int \frac {4^{2+\frac {16 x}{4 e^x-4 (-5+x) x+e^4 x^2}} \left (-4 e^x (-1+x)+4 \left (1-\frac {e^4}{4}\right ) x^2\right ) \log (4)}{\left (4 e^x-4 (-5+x) x+e^4 x^2\right )^2} \, dx\\ &=\log (4) \int \frac {4^{2+\frac {16 x}{4 e^x-4 (-5+x) x+e^4 x^2}} \left (-4 e^x (-1+x)+4 \left (1-\frac {e^4}{4}\right ) x^2\right )}{\left (4 e^x-4 (-5+x) x+e^4 x^2\right )^2} \, dx\\ &=4^{\frac {16 x}{4 e^x+4 (5-x) x+e^4 x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.90, size = 28, normalized size = 0.93 \begin {gather*} 4^{\frac {16 x}{4 e^x+20 x-4 x^2+e^4 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4^((16*x)/(4*E^x + 20*x - 4*x^2 + E^4*x^2))*(E^x*(64 - 64*x)*Log[4] + (64*x^2 - 16*E^4*x^2)*Log[4])
)/(16*E^(2*x) + 400*x^2 - 160*x^3 + 16*x^4 + E^8*x^4 + E^x*(160*x - 32*x^2 + 8*E^4*x^2) + E^4*(40*x^3 - 8*x^4)
),x]

[Out]

4^((16*x)/(4*E^x + 20*x - 4*x^2 + E^4*x^2))

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fricas [A]  time = 0.57, size = 26, normalized size = 0.87 \begin {gather*} 2^{\frac {32 \, x}{x^{2} e^{4} - 4 \, x^{2} + 20 \, x + 4 \, e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-64*x+64)*log(2)*exp(x)+2*(-16*x^2*exp(4)+64*x^2)*log(2))*exp(32*x*log(2)/(4*exp(x)+x^2*exp(4)-4
*x^2+20*x))/(16*exp(x)^2+(8*x^2*exp(4)-32*x^2+160*x)*exp(x)+x^4*exp(4)^2+(-8*x^4+40*x^3)*exp(4)+16*x^4-160*x^3
+400*x^2),x, algorithm="fricas")

[Out]

2^(32*x/(x^2*e^4 - 4*x^2 + 20*x + 4*e^x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {32 \, {\left (4 \, {\left (x - 1\right )} e^{x} \log \relax (2) + {\left (x^{2} e^{4} - 4 \, x^{2}\right )} \log \relax (2)\right )} 2^{\frac {32 \, x}{x^{2} e^{4} - 4 \, x^{2} + 20 \, x + 4 \, e^{x}}}}{x^{4} e^{8} + 16 \, x^{4} - 160 \, x^{3} + 400 \, x^{2} - 8 \, {\left (x^{4} - 5 \, x^{3}\right )} e^{4} + 8 \, {\left (x^{2} e^{4} - 4 \, x^{2} + 20 \, x\right )} e^{x} + 16 \, e^{\left (2 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-64*x+64)*log(2)*exp(x)+2*(-16*x^2*exp(4)+64*x^2)*log(2))*exp(32*x*log(2)/(4*exp(x)+x^2*exp(4)-4
*x^2+20*x))/(16*exp(x)^2+(8*x^2*exp(4)-32*x^2+160*x)*exp(x)+x^4*exp(4)^2+(-8*x^4+40*x^3)*exp(4)+16*x^4-160*x^3
+400*x^2),x, algorithm="giac")

[Out]

integrate(-32*(4*(x - 1)*e^x*log(2) + (x^2*e^4 - 4*x^2)*log(2))*2^(32*x/(x^2*e^4 - 4*x^2 + 20*x + 4*e^x))/(x^4
*e^8 + 16*x^4 - 160*x^3 + 400*x^2 - 8*(x^4 - 5*x^3)*e^4 + 8*(x^2*e^4 - 4*x^2 + 20*x)*e^x + 16*e^(2*x)), x)

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maple [A]  time = 0.55, size = 28, normalized size = 0.93

method	result	size
risch	\({\mathrm e}^{\frac {32 x \ln \relax (2)}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}\)	\(28\)
norman	\(\frac {\left ({\mathrm e}^{4}-4\right ) x^{2} {\mathrm e}^{\frac {32 x \ln \relax (2)}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}+20 x \,{\mathrm e}^{\frac {32 x \ln \relax (2)}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}+4 \,{\mathrm e}^{x} {\mathrm e}^{\frac {32 x \ln \relax (2)}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}}}{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{4}-4 x^{2}+20 x}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-64*x+64)*ln(2)*exp(x)+2*(-16*x^2*exp(4)+64*x^2)*ln(2))*exp(32*x*ln(2)/(4*exp(x)+x^2*exp(4)-4*x^2+20*x
))/(16*exp(x)^2+(8*x^2*exp(4)-32*x^2+160*x)*exp(x)+x^4*exp(4)^2+(-8*x^4+40*x^3)*exp(4)+16*x^4-160*x^3+400*x^2)
,x,method=_RETURNVERBOSE)

[Out]

exp(32*x*ln(2)/(4*exp(x)+x^2*exp(4)-4*x^2+20*x))

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maxima [A]  time = 0.64, size = 23, normalized size = 0.77 \begin {gather*} 2^{\frac {32 \, x}{x^{2} {\left (e^{4} - 4\right )} + 20 \, x + 4 \, e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-64*x+64)*log(2)*exp(x)+2*(-16*x^2*exp(4)+64*x^2)*log(2))*exp(32*x*log(2)/(4*exp(x)+x^2*exp(4)-4
*x^2+20*x))/(16*exp(x)^2+(8*x^2*exp(4)-32*x^2+160*x)*exp(x)+x^4*exp(4)^2+(-8*x^4+40*x^3)*exp(4)+16*x^4-160*x^3
+400*x^2),x, algorithm="maxima")

[Out]

2^(32*x/(x^2*(e^4 - 4) + 20*x + 4*e^x))

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mupad [B]  time = 4.71, size = 27, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{\frac {32\,x\,\ln \relax (2)}{20\,x+4\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^4-4\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((32*x*log(2))/(20*x + 4*exp(x) + x^2*exp(4) - 4*x^2))*(2*log(2)*(16*x^2*exp(4) - 64*x^2) + 2*exp(x)*
log(2)*(64*x - 64)))/(16*exp(2*x) + exp(x)*(160*x + 8*x^2*exp(4) - 32*x^2) + exp(4)*(40*x^3 - 8*x^4) + x^4*exp
(8) + 400*x^2 - 160*x^3 + 16*x^4),x)

[Out]

exp((32*x*log(2))/(20*x + 4*exp(x) + x^2*exp(4) - 4*x^2))

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sympy [A]  time = 0.52, size = 27, normalized size = 0.90 \begin {gather*} e^{\frac {32 x \log {\relax (2 )}}{- 4 x^{2} + x^{2} e^{4} + 20 x + 4 e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-64*x+64)*ln(2)*exp(x)+2*(-16*x**2*exp(4)+64*x**2)*ln(2))*exp(32*x*ln(2)/(4*exp(x)+x**2*exp(4)-4
*x**2+20*x))/(16*exp(x)**2+(8*x**2*exp(4)-32*x**2+160*x)*exp(x)+x**4*exp(4)**2+(-8*x**4+40*x**3)*exp(4)+16*x**
4-160*x**3+400*x**2),x)

[Out]

exp(32*x*log(2)/(-4*x**2 + x**2*exp(4) + 20*x + 4*exp(x)))

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