3.49.57 \(\int (64 x-32 x \log (2)+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))) \, dx\)

Optimal. Leaf size=21 \[ 16 \left (-e^x+x\right )^2 (2-\log (2)-\log (15)) \]

________________________________________________________________________________________

Rubi [B]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 2.62, number of steps used = 8, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 2194, 2187, 2176} \begin {gather*} 16 x^2 (2-\log (30))-32 e^x (x (2-\log (30))+2-\log (30))+32 e^x (2-\log (30))+16 e^{2 x} (2-\log (30)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[64*x - 32*x*Log[2] + E^(2*x)*(64 - 32*Log[2] - 32*Log[15]) - 32*x*Log[15] + E^x*(-64 - 64*x + (32 + 32*x)*
Log[2] + (32 + 32*x)*Log[15]),x]

[Out]

32*E^x*(2 - Log[30]) + 16*E^(2*x)*(2 - Log[30]) + 16*x^2*(2 - Log[30]) - 32*E^x*(2 + x*(2 - Log[30]) - Log[30]
)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 \left (x (64-32 \log (2))+e^{2 x} (64-32 \log (2)-32 \log (15))-32 x \log (15)+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx\\ &=\int \left (e^{2 x} (64-32 \log (2)-32 \log (15))+x (64-32 \log (2)-32 \log (15))+e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15))\right ) \, dx\\ &=16 x^2 (2-\log (30))+(32 (2-\log (30))) \int e^{2 x} \, dx+\int e^x (-64-64 x+(32+32 x) \log (2)+(32+32 x) \log (15)) \, dx\\ &=16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))+\int e^x (-64-64 x+(32+32 x) (\log (2)+\log (15))) \, dx\\ &=16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))+\int e^x (-32 (2-\log (30))-32 x (2-\log (30))) \, dx\\ &=16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))-32 e^x (2+x (2-\log (30))-\log (30))+(32 (2-\log (30))) \int e^x \, dx\\ &=32 e^x (2-\log (30))+16 e^{2 x} (2-\log (30))+16 x^2 (2-\log (30))-32 e^x (2+x (2-\log (30))-\log (30))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 15, normalized size = 0.71 \begin {gather*} -16 \left (e^x-x\right )^2 (-2+\log (30)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[64*x - 32*x*Log[2] + E^(2*x)*(64 - 32*Log[2] - 32*Log[15]) - 32*x*Log[15] + E^x*(-64 - 64*x + (32 +
32*x)*Log[2] + (32 + 32*x)*Log[15]),x]

[Out]

-16*(E^x - x)^2*(-2 + Log[30])

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 48, normalized size = 2.29 \begin {gather*} -16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \relax (2) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \relax (2) - 2\right )} e^{\left (2 \, x\right )} + 32 \, {\left (x \log \left (15\right ) + x \log \relax (2) - 2 \, x\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(15)-32*log(2)+64)*exp(x)^2+((32*x+32)*log(15)+(32*x+32)*log(2)-64*x-64)*exp(x)-32*x*log(15)
-32*x*log(2)+64*x,x, algorithm="fricas")

[Out]

-16*x^2*log(15) - 16*x^2*log(2) + 32*x^2 - 16*(log(15) + log(2) - 2)*e^(2*x) + 32*(x*log(15) + x*log(2) - 2*x)
*e^x

________________________________________________________________________________________

giac [B]  time = 0.13, size = 48, normalized size = 2.29 \begin {gather*} -16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \relax (2) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \relax (2) - 2\right )} e^{\left (2 \, x\right )} + 32 \, {\left (x \log \left (15\right ) + x \log \relax (2) - 2 \, x\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(15)-32*log(2)+64)*exp(x)^2+((32*x+32)*log(15)+(32*x+32)*log(2)-64*x-64)*exp(x)-32*x*log(15)
-32*x*log(2)+64*x,x, algorithm="giac")

[Out]

-16*x^2*log(15) - 16*x^2*log(2) + 32*x^2 - 16*(log(15) + log(2) - 2)*e^(2*x) + 32*(x*log(15) + x*log(2) - 2*x)
*e^x

________________________________________________________________________________________

maple [A]  time = 0.04, size = 45, normalized size = 2.14

method	result	size
norman	\(\left (-16 \ln \left (15\right )-16 \ln \relax (2)+32\right ) x^{2}+\left (-16 \ln \left (15\right )-16 \ln \relax (2)+32\right ) {\mathrm e}^{2 x}+\left (32 \ln \left (15\right )+32 \ln \relax (2)-64\right ) x \,{\mathrm e}^{x}\)	\(45\)
default	\(-16 \ln \relax (2) {\mathrm e}^{2 x}-16 \,{\mathrm e}^{2 x} \ln \left (15\right )+32 \,{\mathrm e}^{2 x}-64 \,{\mathrm e}^{x} x +32 x \ln \relax (2) {\mathrm e}^{x}+32 \,{\mathrm e}^{x} \ln \left (15\right ) x +32 x^{2}-16 x^{2} \ln \relax (2)-16 x^{2} \ln \left (15\right )\)	\(62\)
risch	\(-16 \ln \relax (2) {\mathrm e}^{2 x}-16 \,{\mathrm e}^{2 x} \ln \relax (5)-16 \,{\mathrm e}^{2 x} \ln \relax (3)+32 \,{\mathrm e}^{2 x}+32 \left (\ln \relax (2)+\ln \relax (5)+\ln \relax (3)-2\right ) x \,{\mathrm e}^{x}-16 x^{2} \ln \relax (5)-16 x^{2} \ln \relax (3)-16 x^{2} \ln \relax (2)+32 x^{2}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(15)-32*ln(2)+64)*exp(x)^2+((32*x+32)*ln(15)+(32*x+32)*ln(2)-64*x-64)*exp(x)-32*x*ln(15)-32*x*ln(2)
+64*x,x,method=_RETURNVERBOSE)

[Out]

(-16*ln(15)-16*ln(2)+32)*x^2+(-16*ln(15)-16*ln(2)+32)*exp(x)^2+(32*ln(15)+32*ln(2)-64)*x*exp(x)

________________________________________________________________________________________

maxima [B]  time = 0.45, size = 45, normalized size = 2.14 \begin {gather*} 32 \, x {\left (\log \relax (5) + \log \relax (3) + \log \relax (2) - 2\right )} e^{x} - 16 \, x^{2} \log \left (15\right ) - 16 \, x^{2} \log \relax (2) + 32 \, x^{2} - 16 \, {\left (\log \left (15\right ) + \log \relax (2) - 2\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(15)-32*log(2)+64)*exp(x)^2+((32*x+32)*log(15)+(32*x+32)*log(2)-64*x-64)*exp(x)-32*x*log(15)
-32*x*log(2)+64*x,x, algorithm="maxima")

[Out]

32*x*(log(5) + log(3) + log(2) - 2)*e^x - 16*x^2*log(15) - 16*x^2*log(2) + 32*x^2 - 16*(log(15) + log(2) - 2)*
e^(2*x)

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 16, normalized size = 0.76 \begin {gather*} -{\left (x-{\mathrm {e}}^x\right )}^2\,\left (16\,\ln \left (30\right )-32\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(64*x - 32*x*log(2) - 32*x*log(15) - exp(x)*(64*x - log(2)*(32*x + 32) - log(15)*(32*x + 32) + 64) - exp(2*
x)*(32*log(2) + 32*log(15) - 64),x)

[Out]

-(x - exp(x))^2*(16*log(30) - 32)

________________________________________________________________________________________

sympy [B]  time = 0.16, size = 51, normalized size = 2.43 \begin {gather*} x^{2} \left (- 16 \log {\left (15 \right )} - 16 \log {\relax (2 )} + 32\right ) + \left (- 64 x + 32 x \log {\relax (2 )} + 32 x \log {\left (15 \right )}\right ) e^{x} + \left (- 16 \log {\left (15 \right )} - 16 \log {\relax (2 )} + 32\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(15)-32*ln(2)+64)*exp(x)**2+((32*x+32)*ln(15)+(32*x+32)*ln(2)-64*x-64)*exp(x)-32*x*ln(15)-32*
x*ln(2)+64*x,x)

[Out]

x**2*(-16*log(15) - 16*log(2) + 32) + (-64*x + 32*x*log(2) + 32*x*log(15))*exp(x) + (-16*log(15) - 16*log(2) +
32)*exp(2*x)

________________________________________________________________________________________