3.18.43 \(\int e^{x+\log ^2(5)} (-28-10 x+(6+6 x) \log (4)) \, dx\)

Optimal. Leaf size=25 \[ 1+2 e^{x+\log ^2(5)} x \left (-5+3 \left (-\frac {3}{x}+\log (4)\right )\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.64, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2187, 2176, 2194} \begin {gather*} 2 (5-\log (64)) e^{x+\log ^2(5)}-2 e^{x+\log ^2(5)} (x (5-\log (64))+14-3 \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(x + Log[5]^2)*(-28 - 10*x + (6 + 6*x)*Log[4]),x]

[Out]

-2*E^(x + Log[5]^2)*(14 - 3*Log[4] + x*(5 - Log[64])) + 2*E^(x + Log[5]^2)*(5 - Log[64])

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+\log ^2(5)} (-2 (14-3 \log (4))-2 x (5-\log (64))) \, dx\\ &=-2 e^{x+\log ^2(5)} (14-3 \log (4)+x (5-\log (64)))+(2 (5-\log (64))) \int e^{x+\log ^2(5)} \, dx\\ &=-2 e^{x+\log ^2(5)} (14-3 \log (4)+x (5-\log (64)))+2 e^{x+\log ^2(5)} (5-\log (64))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 0.72 \begin {gather*} 2 e^{x+\log ^2(5)} (-9+x (-5+\log (64))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(x + Log[5]^2)*(-28 - 10*x + (6 + 6*x)*Log[4]),x]

[Out]

2*E^(x + Log[5]^2)*(-9 + x*(-5 + Log[64]))

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fricas [A]  time = 0.62, size = 19, normalized size = 0.76 \begin {gather*} 2 \, {\left (6 \, x \log \relax (2) - 5 \, x - 9\right )} e^{\left (\log \relax (5)^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(6*x+6)*log(2)-10*x-28)*exp(log(5)^2+x),x, algorithm="fricas")

[Out]

2*(6*x*log(2) - 5*x - 9)*e^(log(5)^2 + x)

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giac [A]  time = 0.25, size = 19, normalized size = 0.76 \begin {gather*} 2 \, {\left (6 \, x \log \relax (2) - 5 \, x - 9\right )} e^{\left (\log \relax (5)^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(6*x+6)*log(2)-10*x-28)*exp(log(5)^2+x),x, algorithm="giac")

[Out]

2*(6*x*log(2) - 5*x - 9)*e^(log(5)^2 + x)

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maple [A]  time = 0.05, size = 19, normalized size = 0.76

method	result	size
risch	\(\left (12 x \ln \relax (2)-10 x -18\right ) {\mathrm e}^{\ln \relax (5)^{2}+x}\)	\(19\)
gosper	\(2 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \left (6 x \ln \relax (2)-5 x -9\right )\)	\(20\)
norman	\(\left (12 \ln \relax (2)-10\right ) x \,{\mathrm e}^{\ln \relax (5)^{2}+x}-18 \,{\mathrm e}^{\ln \relax (5)^{2}+x}\)	\(26\)
meijerg	\(\left (12 \ln \relax (2)-10\right ) {\mathrm e}^{\ln \relax (5)^{2}} \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )-12 \ln \relax (2) {\mathrm e}^{\ln \relax (5)^{2}} \left (1-{\mathrm e}^{x}\right )+28 \,{\mathrm e}^{\ln \relax (5)^{2}} \left (1-{\mathrm e}^{x}\right )\)	\(53\)
derivativedivides	\(-10 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \left (\ln \relax (5)^{2}+x \right )-18 \,{\mathrm e}^{\ln \relax (5)^{2}+x}+10 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (5)^{2}+12 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (2) \left (\ln \relax (5)^{2}+x \right )-12 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (2) \ln \relax (5)^{2}\)	\(71\)
default	\(-10 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \left (\ln \relax (5)^{2}+x \right )-18 \,{\mathrm e}^{\ln \relax (5)^{2}+x}+10 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (5)^{2}+12 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (2) \left (\ln \relax (5)^{2}+x \right )-12 \,{\mathrm e}^{\ln \relax (5)^{2}+x} \ln \relax (2) \ln \relax (5)^{2}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(6*x+6)*ln(2)-10*x-28)*exp(ln(5)^2+x),x,method=_RETURNVERBOSE)

[Out]

(12*x*ln(2)-10*x-18)*exp(ln(5)^2+x)

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maxima [B]  time = 0.48, size = 61, normalized size = 2.44 \begin {gather*} 12 \, {\left (x e^{\left (\log \relax (5)^{2}\right )} - e^{\left (\log \relax (5)^{2}\right )}\right )} e^{x} \log \relax (2) - 10 \, {\left (x e^{\left (\log \relax (5)^{2}\right )} - e^{\left (\log \relax (5)^{2}\right )}\right )} e^{x} + 12 \, e^{\left (\log \relax (5)^{2} + x\right )} \log \relax (2) - 28 \, e^{\left (\log \relax (5)^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(6*x+6)*log(2)-10*x-28)*exp(log(5)^2+x),x, algorithm="maxima")

[Out]

12*(x*e^(log(5)^2) - e^(log(5)^2))*e^x*log(2) - 10*(x*e^(log(5)^2) - e^(log(5)^2))*e^x + 12*e^(log(5)^2 + x)*l
og(2) - 28*e^(log(5)^2 + x)

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mupad [B]  time = 1.08, size = 19, normalized size = 0.76 \begin {gather*} -{\mathrm {e}}^{x+{\ln \relax (5)}^2}\,\left (10\,x-12\,x\,\ln \relax (2)+18\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x + log(5)^2)*(10*x - 2*log(2)*(6*x + 6) + 28),x)

[Out]

-exp(x + log(5)^2)*(10*x - 12*x*log(2) + 18)

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sympy [A]  time = 0.11, size = 19, normalized size = 0.76 \begin {gather*} \left (- 10 x + 12 x \log {\relax (2 )} - 18\right ) e^{x + \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(6*x+6)*ln(2)-10*x-28)*exp(ln(5)**2+x),x)

[Out]

(-10*x + 12*x*log(2) - 18)*exp(x + log(5)**2)

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