3.57.87 \(\int e^x (-5 x+(10-5 x) \log (2)) \, dx\)

Optimal. Leaf size=16 \[ e^x (5-5 (x+(-3+x) \log (2))) \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.69, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2187, 2176, 2194} \begin {gather*} 5 e^x (1+\log (2))-e^x (5 x (1+\log (2))-\log (1024)) \end {gather*}

Antiderivative was successfully verified.

[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 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 2187

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 16, normalized size = 1.00 \begin {gather*} -5 e^x (-1+x+x \log (2)-\log (8)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5*E^x*(-1 + x + x*Log[2] - Log[8])

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fricas [A]  time = 1.03, size = 13, normalized size = 0.81 \begin {gather*} -5 \, {\left ({\left (x - 3\right )} \log \relax (2) + x - 1\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*((x - 3)*log(2) + x - 1)*e^x

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giac [A]  time = 0.19, size = 15, normalized size = 0.94 \begin {gather*} -5 \, {\left (x \log \relax (2) + x - 3 \, \log \relax (2) - 1\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 16, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.36, size = 23, normalized size = 1.44 \begin {gather*} -5 \, {\left (x - 1\right )} e^{x} \log \relax (2) - 5 \, {\left (x - 1\right )} e^{x} + 10 \, e^{x} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 19, normalized size = 1.19 \begin {gather*} {\mathrm {e}}^x\,\left (15\,\ln \relax (2)+5\right )-x\,{\mathrm {e}}^x\,\left (\ln \left (32\right )+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.09, size = 19, normalized size = 1.19 \begin {gather*} \left (- 5 x - 5 x \log {\relax (2 )} + 5 + 15 \log {\relax (2 )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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