\(\int (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))) \, dx\) [7824]

Optimal result

Integrand size = 27, antiderivative size = 20 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=e^x \left (11-4^{2 \left (-\frac {5}{4}+x\right )}-x\right ) \]

[Out]

(11-x-exp(4*(x-5/4)*ln(2)))*exp(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2207, 2225, 2325, 12} \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=e^x (10-x)-2^{4 x-5} e^x+e^x \]

[In]

Int[E^x*(10 - x) + 2^(-5 + 4*x)*E^x*(-1 - 2*Log[4]),x]

[Out]

E^x - 2^(-5 + 4*x)*E^x + E^x*(10 - 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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2325

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{align*} \text {integral}& = (-1-\log (16)) \int 2^{-5+4 x} e^x \, dx+\int e^x (10-x) \, dx \\ & = e^x (10-x)+(-1-\log (16)) \int \frac {1}{32} e^{x (1+\log (16))} \, dx+\int e^x \, dx \\ & = e^x+e^x (10-x)+\frac {1}{32} (-1-\log (16)) \int e^{x (1+\log (16))} \, dx \\ & = e^x-2^{-5+4 x} e^x+e^x (10-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=-\frac {1}{32} e^x \left (-352+16^x+32 x\right ) \]

[In]

Integrate[E^x*(10 - x) + 2^(-5 + 4*x)*E^x*(-1 - 2*Log[4]),x]

[Out]

-1/32*(E^x*(-352 + 16^x + 32*x))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\left (2^{-5+4 x}+x -11\right ) {\mathrm e}^{x}\)	\(15\)
norman	\(-{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left (-5+4 x \right ) \ln \left (2\right )}+11 \,{\mathrm e}^{x}\)	\(24\)
parallelrisch	\(-{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left (-5+4 x \right ) \ln \left (2\right )}+11 \,{\mathrm e}^{x}\)	\(24\)
default	\(-{\mathrm e}^{x} x +11 \,{\mathrm e}^{x}+\frac {\left (-4 \ln \left (2\right )-1\right ) {\mathrm e}^{x +\left (-5+4 x \right ) \ln \left (2\right )}}{1+4 \ln \left (2\right )}\)	\(37\)
meijerg	\(-11+\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}+\frac {1-{\mathrm e}^{x \ln \left (2\right ) \left (4+\frac {1}{\ln \left (2\right )}\right )}}{32+\frac {8}{\ln \left (2\right )}}+\frac {1-{\mathrm e}^{x \ln \left (2\right ) \left (4+\frac {1}{\ln \left (2\right )}\right )}}{32 \ln \left (2\right ) \left (4+\frac {1}{\ln \left (2\right )}\right )}+10 \,{\mathrm e}^{x}\)	\(70\)

[In]

int((-4*ln(2)-1)*exp(x)*exp((-5+4*x)*ln(2))+(10-x)*exp(x),x,method=_RETURNVERBOSE)

[Out]

-(2^(-5+4*x)+x-11)*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=-2^{4 \, x - 5} e^{x} - {\left (x - 11\right )} e^{x} \]

[In]

integrate((-4*log(2)-1)*exp(x)*exp((-5+4*x)*log(2))+(10-x)*exp(x),x, algorithm="fricas")

[Out]

-2^(4*x - 5)*e^x - (x - 11)*e^x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (15) = 30\).

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=\left (11 - x\right ) e^{x} + \frac {\left (- 4 \log {\left (2 \right )} - 1\right ) e^{x} e^{4 x \log {\left (2 \right )}}}{32 + 128 \log {\left (2 \right )}} \]

[In]

integrate((-4*ln(2)-1)*exp(x)*exp((-5+4*x)*ln(2))+(10-x)*exp(x),x)

[Out]

(11 - x)*exp(x) + (-4*log(2) - 1)*exp(x)*exp(4*x*log(2))/(32 + 128*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=-{\left (x - 1\right )} e^{x} - \frac {1}{32} \, e^{\left (4 \, x \log \left (2\right ) + x\right )} + 10 \, e^{x} \]

[In]

integrate((-4*log(2)-1)*exp(x)*exp((-5+4*x)*log(2))+(10-x)*exp(x),x, algorithm="maxima")

[Out]

-(x - 1)*e^x - 1/32*e^(4*x*log(2) + x) + 10*e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=-{\left (x - 11\right )} e^{x} - e^{\left (4 \, x \log \left (2\right ) + x - 5 \, \log \left (2\right )\right )} \]

[In]

integrate((-4*log(2)-1)*exp(x)*exp((-5+4*x)*log(2))+(10-x)*exp(x),x, algorithm="giac")

[Out]

-(x - 11)*e^x - e^(4*x*log(2) + x - 5*log(2))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \left (e^x (10-x)+2^{-5+4 x} e^x (-1-2 \log (4))\right ) \, dx=-\frac {{\mathrm {e}}^x\,\left (32\,x+2^{4\,x}-352\right )}{32} \]

[In]

int(- exp(x)*(x - 10) - exp(log(2)*(4*x - 5))*exp(x)*(4*log(2) + 1),x)

[Out]

-(exp(x)*(32*x + 2^(4*x) - 352))/32