\(\int e^x (-5 x+(10-5 x) \log (2)) \, dx\) [5687]

Optimal result

Integrand size = 16, antiderivative size = 16 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=e^x (5-5 (x+(-3+x) \log (2))) \]

[Out]

(5-5*ln(2)*(-3+x)-5*x)*exp(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2218, 2207, 2225} \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=5 e^x (1+\log (2))-e^x (5 x (1+\log (2))-\log (1024)) \]

[In]

Int[E^x*(-5*x + (10 - 5*x)*Log[2]),x]

[Out]

5*E^x*(1 + Log[2]) - E^x*(5*x*(1 + Log[2]) - Log[1024])

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 2218

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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]

Rubi steps \begin{align*} \text {integral}& = \int e^x (-5 x (1+\log (2))+\log (1024)) \, dx \\ & = -e^x (5 x (1+\log (2))-\log (1024))+(5 (1+\log (2))) \int e^x \, dx \\ & = 5 e^x (1+\log (2))-e^x (5 x (1+\log (2))-\log (1024)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=-5 e^x (-1+x+x \log (2)-\log (8)) \]

[In]

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

[Out]

-5*E^x*(-1 + x + x*Log[2] - Log[8])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
gosper	\(-5 \,{\mathrm e}^{x} \left (x \ln \left (2\right )-3 \ln \left (2\right )+x -1\right )\)	\(16\)
risch	\(\left (-5 x \ln \left (2\right )+15 \ln \left (2\right )-5 x +5\right ) {\mathrm e}^{x}\)	\(18\)
norman	\(\left (5+15 \ln \left (2\right )\right ) {\mathrm e}^{x}+\left (-5 \ln \left (2\right )-5\right ) x \,{\mathrm e}^{x}\)	\(21\)
parallelrisch	\(-5 x \ln \left (2\right ) {\mathrm e}^{x}+15 \,{\mathrm e}^{x} \ln \left (2\right )-5 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x}\)	\(24\)
default	\(-5 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x}+10 \,{\mathrm e}^{x} \ln \left (2\right )-5 \ln \left (2\right ) \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )\)	\(30\)
parts	\(-5 x \ln \left (2\right ) {\mathrm e}^{x}+10 \,{\mathrm e}^{x} \ln \left (2\right )-5 \,{\mathrm e}^{x} x -\left (-5 \ln \left (2\right )-5\right ) {\mathrm e}^{x}\)	\(30\)
meijerg	\(-\left (5+5 \ln \left (2\right )\right ) \left (1-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}\right )-10 \left (1-{\mathrm e}^{x}\right ) \ln \left (2\right )\)	\(31\)

[In]

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

[Out]

-5*exp(x)*(x*ln(2)-3*ln(2)+x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=-5 \, {\left ({\left (x - 3\right )} \log \left (2\right ) + x - 1\right )} e^{x} \]

[In]

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

[Out]

-5*((x - 3)*log(2) + x - 1)*e^x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=\left (- 5 x - 5 x \log {\left (2 \right )} + 5 + 15 \log {\left (2 \right )}\right ) e^{x} \]

[In]

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

[Out]

(-5*x - 5*x*log(2) + 5 + 15*log(2))*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=-5 \, {\left (x - 1\right )} e^{x} \log \left (2\right ) - 5 \, {\left (x - 1\right )} e^{x} + 10 \, e^{x} \log \left (2\right ) \]

[In]

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

[Out]

-5*(x - 1)*e^x*log(2) - 5*(x - 1)*e^x + 10*e^x*log(2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx=-5 \, {\left (x \log \left (2\right ) + x - 3 \, \log \left (2\right ) - 1\right )} e^{x} \]

[In]

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

[Out]

-5*(x*log(2) + x - 3*log(2) - 1)*e^x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int e^x (-5 x+(10-5 x) \log (2)) \, dx={\mathrm {e}}^x\,\left (15\,\ln \left (2\right )+5\right )-x\,{\mathrm {e}}^x\,\left (\ln \left (32\right )+5\right ) \]

[In]

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

[Out]

exp(x)*(15*log(2) + 5) - x*exp(x)*(log(32) + 5)