3.17.53 \(\int \frac {e^{\frac {256 e^x x}{\log ^4(2)}} (e^x (256 x-4096 x^2-4352 x^3)+(1-34 x) \log ^4(2))}{5 \log ^4(2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} e^{\frac {256 e^x x}{\log ^4(2)}} \left (x-17 x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 2288} \begin {gather*} \frac {\left (-17 x^3-16 x^2+x\right ) e^{x+\frac {256 e^x x}{\log ^4(2)}}}{5 \left (e^x x+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((256*E^x*x)/Log[2]^4)*(E^x*(256*x - 4096*x^2 - 4352*x^3) + (1 - 34*x)*Log[2]^4))/(5*Log[2]^4),x]

[Out]

(E^(x + (256*E^x*x)/Log[2]^4)*(x - 16*x^2 - 17*x^3))/(5*(E^x + E^x*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 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} &=\frac {\int e^{\frac {256 e^x x}{\log ^4(2)}} \left (e^x \left (256 x-4096 x^2-4352 x^3\right )+(1-34 x) \log ^4(2)\right ) \, dx}{5 \log ^4(2)}\\ &=\frac {e^{x+\frac {256 e^x x}{\log ^4(2)}} \left (x-16 x^2-17 x^3\right )}{5 \left (e^x+e^x x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} -\frac {1}{5} e^{\frac {256 e^x x}{\log ^4(2)}} x (-1+17 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((256*E^x*x)/Log[2]^4)*(E^x*(256*x - 4096*x^2 - 4352*x^3) + (1 - 34*x)*Log[2]^4))/(5*Log[2]^4),x]

[Out]

-1/5*(E^((256*E^x*x)/Log[2]^4)*x*(-1 + 17*x))

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fricas [A]  time = 0.72, size = 21, normalized size = 0.91 \begin {gather*} -\frac {1}{5} \, {\left (17 \, x^{2} - x\right )} e^{\left (\frac {256 \, x e^{x}}{\log \relax (2)^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="fricas")

[Out]

-1/5*(17*x^2 - x)*e^(256*x*e^x/log(2)^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (34 \, x - 1\right )} \log \relax (2)^{4} + 256 \, {\left (17 \, x^{3} + 16 \, x^{2} - x\right )} e^{x}\right )} e^{\left (\frac {256 \, x e^{x}}{\log \relax (2)^{4}}\right )}}{5 \, \log \relax (2)^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="giac")

[Out]

integrate(-1/5*((34*x - 1)*log(2)^4 + 256*(17*x^3 + 16*x^2 - x)*e^x)*e^(256*x*e^x/log(2)^4)/log(2)^4, x)

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maple [A]  time = 0.08, size = 19, normalized size = 0.83

method	result	size
risch	\(-\frac {\left (17 x -1\right ) x \,{\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \relax (2)^{4}}}}{5}\)	\(19\)
norman	\(\frac {\frac {x \ln \relax (2)^{3} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \relax (2)^{4}}}}{5}-\frac {17 x^{2} \ln \relax (2)^{3} {\mathrm e}^{\frac {256 x \,{\mathrm e}^{x}}{\ln \relax (2)^{4}}}}{5}}{\ln \relax (2)^{3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*ln(2)^4)*exp(256*x*exp(x)/ln(2)^4)/ln(2)^4,x,method=_RETU
RNVERBOSE)

[Out]

-1/5*(17*x-1)*x*exp(256*x*exp(x)/ln(2)^4)

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maxima [A]  time = 1.04, size = 33, normalized size = 1.43 \begin {gather*} -\frac {{\left (17 \, x^{2} \log \relax (2)^{4} - x \log \relax (2)^{4}\right )} e^{\left (\frac {256 \, x e^{x}}{\log \relax (2)^{4}}\right )}}{5 \, \log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="maxima")

[Out]

-1/5*(17*x^2*log(2)^4 - x*log(2)^4)*e^(256*x*e^x/log(2)^4)/log(2)^4

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mupad [B]  time = 0.11, size = 18, normalized size = 0.78 \begin {gather*} -\frac {x\,{\mathrm {e}}^{\frac {256\,x\,{\mathrm {e}}^x}{{\ln \relax (2)}^4}}\,\left (17\,x-1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((256*x*exp(x))/log(2)^4)*(log(2)^4*(34*x - 1) + exp(x)*(4096*x^2 - 256*x + 4352*x^3)))/(5*log(2)^4),
x)

[Out]

-(x*exp((256*x*exp(x))/log(2)^4)*(17*x - 1))/5

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sympy [A]  time = 0.22, size = 20, normalized size = 0.87 \begin {gather*} \frac {\left (- 17 x^{2} + x\right ) e^{\frac {256 x e^{x}}{\log {\relax (2 )}^{4}}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x**3-4096*x**2+256*x)*exp(x)+(-34*x+1)*ln(2)**4)*exp(256*x*exp(x)/ln(2)**4)/ln(2)**4,x)

[Out]

(-17*x**2 + x)*exp(256*x*exp(x)/log(2)**4)/5

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