3.26.63 \(\int (1+e^x (1+x)+4^{12+4 x} e^x (-1+e+(-4+4 e) \log (4))) \, dx\)

Optimal. Leaf size=29 \[ x+e^x x \left (2-\frac {4^{4 (3+x)} (1-e)+x}{x}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2176, 2194, 2287, 12} \begin {gather*} x-(1-e) 256^{x+3} e^x-e^x+e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + E^x*(1 + x) + 4^(12 + 4*x)*E^x*(-1 + E + (-4 + 4*E)*Log[4]),x]

[Out]

-E^x - 256^(3 + x)*(1 - E)*E^x + x + E^x*(1 + 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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x-((1-e) (1+\log (256))) \int 4^{12+4 x} e^x \, dx+\int e^x (1+x) \, dx\\ &=x+e^x (1+x)-((1-e) (1+\log (256))) \int 16777216 e^{x (1+\log (256))} \, dx-\int e^x \, dx\\ &=-e^x+x+e^x (1+x)-(16777216 (1-e) (1+\log (256))) \int e^{x (1+\log (256))} \, dx\\ &=-e^x-256^{3+x} (1-e) e^x+x+e^x (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.69 \begin {gather*} 16777216 (-1+e) e^{x (1+\log (256))}+x+e^x x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + E^x*(1 + x) + 4^(12 + 4*x)*E^x*(-1 + E + (-4 + 4*E)*Log[4]),x]

[Out]

16777216*(-1 + E)*E^(x*(1 + Log[256])) + x + E^x*x

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fricas [A]  time = 0.65, size = 20, normalized size = 0.69 \begin {gather*} 2^{8 \, x + 24} {\left (e - 1\right )} e^{x} + x e^{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*exp(1)-4)*log(2)+exp(1)-1)*exp(x)*exp(2*(4*x+12)*log(2))+(x+1)*exp(x)+1,x, algorithm="fricas")

[Out]

2^(8*x + 24)*(e - 1)*e^x + x*e^x + x

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giac [A]  time = 0.19, size = 39, normalized size = 1.34 \begin {gather*} x e^{x} + x + \frac {{\left (8 \, {\left (e - 1\right )} \log \relax (2) + e - 1\right )} e^{\left (8 \, x \log \relax (2) + x + 24 \, \log \relax (2)\right )}}{8 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*exp(1)-4)*log(2)+exp(1)-1)*exp(x)*exp(2*(4*x+12)*log(2))+(x+1)*exp(x)+1,x, algorithm="giac")

[Out]

x*e^x + x + (8*(e - 1)*log(2) + e - 1)*e^(8*x*log(2) + x + 24*log(2))/(8*log(2) + 1)

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maple [A]  time = 0.04, size = 24, normalized size = 0.83

method	result	size
norman	\(x +{\mathrm e}^{x} x +\left ({\mathrm e}-1\right ) {\mathrm e}^{x} {\mathrm e}^{2 \left (4 x +12\right ) \ln \relax (2)}\)	\(24\)
risch	\(2^{8 x +24} {\mathrm e}^{x +1}-{\mathrm e}^{x} 2^{8 x +24}+{\mathrm e}^{x} x +x\)	\(30\)
default	\({\mathrm e}^{x} x +x +\frac {8 \,{\mathrm e}^{x +2 \left (4 x +12\right ) \ln \relax (2)} {\mathrm e} \ln \relax (2)}{8 \ln \relax (2)+1}+\frac {{\mathrm e}^{x +2 \left (4 x +12\right ) \ln \relax (2)} {\mathrm e}}{8 \ln \relax (2)+1}-\frac {8 \,{\mathrm e}^{x +2 \left (4 x +12\right ) \ln \relax (2)} \ln \relax (2)}{8 \ln \relax (2)+1}-\frac {{\mathrm e}^{x +2 \left (4 x +12\right ) \ln \relax (2)}}{8 \ln \relax (2)+1}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(4*exp(1)-4)*ln(2)+exp(1)-1)*exp(x)*exp(2*(4*x+12)*ln(2))+(x+1)*exp(x)+1,x,method=_RETURNVERBOSE)

[Out]

x+exp(x)*x+(exp(1)-1)*exp(x)*exp(2*(4*x+12)*ln(2))

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maxima [A]  time = 0.56, size = 40, normalized size = 1.38 \begin {gather*} {\left (x - 1\right )} e^{x} + x + \frac {16777216 \, {\left (8 \, {\left (e - 1\right )} \log \relax (2) + e - 1\right )} e^{\left (8 \, x \log \relax (2) + x\right )}}{8 \, \log \relax (2) + 1} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*exp(1)-4)*log(2)+exp(1)-1)*exp(x)*exp(2*(4*x+12)*log(2))+(x+1)*exp(x)+1,x, algorithm="maxima")

[Out]

(x - 1)*e^x + x + 16777216*(8*(e - 1)*log(2) + e - 1)*e^(8*x*log(2) + x)/(8*log(2) + 1) + e^x

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mupad [B]  time = 0.12, size = 35, normalized size = 1.21 \begin {gather*} x+x\,{\mathrm {e}}^x+\frac {16777216\,2^{8\,x}\,{\mathrm {e}}^x\,\left (\mathrm {e}-8\,\ln \relax (2)+8\,\mathrm {e}\,\ln \relax (2)-1\right )}{\ln \left (256\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(x + 1) + exp(2*log(2)*(4*x + 12))*exp(x)*(exp(1) + 2*log(2)*(4*exp(1) - 4) - 1) + 1,x)

[Out]

x + x*exp(x) + (16777216*2^(8*x)*exp(x)*(exp(1) - 8*log(2) + 8*exp(1)*log(2) - 1))/(log(256) + 1)

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sympy [A]  time = 0.38, size = 41, normalized size = 1.41 \begin {gather*} x e^{x} + x + \frac {16777216 \left (-1 + e + \left (-8 + 8 e\right ) \log {\relax (2 )}\right ) e^{x} e^{8 x \log {\relax (2 )}}}{1 + 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*exp(1)-4)*ln(2)+exp(1)-1)*exp(x)*exp(2*(4*x+12)*ln(2))+(x+1)*exp(x)+1,x)

[Out]

x*exp(x) + x + 16777216*(-1 + E + (-8 + 8*E)*log(2))*exp(x)*exp(8*x*log(2))/(1 + 8*log(2))

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