3.4.2 \(\int (-16-6 x+(-16-8 x) \log (4 e^x)-4 \log ^2(4 e^x)) \, dx\)

Optimal. Leaf size=20 \[ 1+5 x^2-4 x \left (2+\log \left (4 e^x\right )\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 2.15, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2551, 12, 32, 2157, 30} \begin {gather*} -3 x^2+\frac {4}{3} (x+2)^3-16 x-\frac {4}{3} \log ^3\left (4 e^x\right )-4 (x+2)^2 \log \left (4 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-16 - 6*x + (-16 - 8*x)*Log[4*E^x] - 4*Log[4*E^x]^2,x]

[Out]

-16*x - 3*x^2 + (4*(2 + x)^3)/3 - 4*(2 + x)^2*Log[4*E^x] - (4*Log[4*E^x]^3)/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-16 x-3 x^2-4 \int \log ^2\left (4 e^x\right ) \, dx+\int (-16-8 x) \log \left (4 e^x\right ) \, dx\\ &=-16 x-3 x^2-4 (2+x)^2 \log \left (4 e^x\right )+\frac {1}{16} \int 64 (2+x)^2 \, dx-4 \operatorname {Subst}\left (\int x^2 \, dx,x,\log \left (4 e^x\right )\right )\\ &=-16 x-3 x^2-4 (2+x)^2 \log \left (4 e^x\right )-\frac {4}{3} \log ^3\left (4 e^x\right )+4 \int (2+x)^2 \, dx\\ &=-16 x-3 x^2+\frac {4}{3} (2+x)^3-4 (2+x)^2 \log \left (4 e^x\right )-\frac {4}{3} \log ^3\left (4 e^x\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.01, size = 49, normalized size = 2.45 \begin {gather*} -16 x-3 x^2+\frac {4 x^3}{3}-4 x^2 \log \left (4 e^x\right )-8 \log ^2\left (4 e^x\right )-\frac {4}{3} \log ^3\left (4 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-16 - 6*x + (-16 - 8*x)*Log[4*E^x] - 4*Log[4*E^x]^2,x]

[Out]

-16*x - 3*x^2 + (4*x^3)/3 - 4*x^2*Log[4*E^x] - 8*Log[4*E^x]^2 - (4*Log[4*E^x]^3)/3

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 32, normalized size = 1.60 \begin {gather*} -4 \, x^{3} - 16 \, x \log \relax (2)^{2} - 11 \, x^{2} - 16 \, {\left (x^{2} + 2 \, x\right )} \log \relax (2) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*log(4*exp(x))^2+(-8*x-16)*log(4*exp(x))-6*x-16,x, algorithm="fricas")

[Out]

-4*x^3 - 16*x*log(2)^2 - 11*x^2 - 16*(x^2 + 2*x)*log(2) - 16*x

________________________________________________________________________________________

giac [A]  time = 0.27, size = 33, normalized size = 1.65 \begin {gather*} -4 \, x^{3} - 16 \, x^{2} \log \relax (2) - 16 \, x \log \relax (2)^{2} - 11 \, x^{2} - 32 \, x \log \relax (2) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*log(4*exp(x))^2+(-8*x-16)*log(4*exp(x))-6*x-16,x, algorithm="giac")

[Out]

-4*x^3 - 16*x^2*log(2) - 16*x*log(2)^2 - 11*x^2 - 32*x*log(2) - 16*x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 36, normalized size = 1.80

method	result	size
norman	\(-16 \ln \left (4 \,{\mathrm e}^{x}\right )-5 \ln \left (4 \,{\mathrm e}^{x}\right )^{2}-6 \ln \left (4 \,{\mathrm e}^{x}\right ) x -4 x \ln \left (4 \,{\mathrm e}^{x}\right )^{2}\)	\(36\)
default	\(-16 x -4 \ln \left (4 \,{\mathrm e}^{x}\right ) x^{2}-16 \ln \left (4 \,{\mathrm e}^{x}\right ) x +\frac {4 x^{3}}{3}+5 x^{2}-\frac {4 \ln \left (4 \,{\mathrm e}^{x}\right )^{3}}{3}\)	\(42\)
risch	\(-16 x \ln \relax (2)^{2}-16 \ln \relax (2) \ln \left ({\mathrm e}^{x}\right ) x -4 x \ln \left ({\mathrm e}^{x}\right )^{2}-32 x \ln \relax (2)-16 \ln \left ({\mathrm e}^{x}\right ) x +5 x^{2}-16 x\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4*ln(4*exp(x))^2+(-8*x-16)*ln(4*exp(x))-6*x-16,x,method=_RETURNVERBOSE)

[Out]

-16*ln(4*exp(x))-5*ln(4*exp(x))^2-6*ln(4*exp(x))*x-4*x*ln(4*exp(x))^2

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 37, normalized size = 1.85 \begin {gather*} \frac {4}{3} \, x^{3} - \frac {4}{3} \, \log \left (4 \, e^{x}\right )^{3} + 5 \, x^{2} - 4 \, {\left (x^{2} + 4 \, x\right )} \log \left (4 \, e^{x}\right ) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*log(4*exp(x))^2+(-8*x-16)*log(4*exp(x))-6*x-16,x, algorithm="maxima")

[Out]

4/3*x^3 - 4/3*log(4*e^x)^3 + 5*x^2 - 4*(x^2 + 4*x)*log(4*e^x) - 16*x

________________________________________________________________________________________

mupad [B]  time = 0.36, size = 32, normalized size = 1.60 \begin {gather*} -4\,x^3+\left (-8\,\ln \relax (4)-11\right )\,x^2+\left (-16\,\ln \relax (4)-4\,{\ln \relax (4)}^2-16\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- 6*x - 4*log(4*exp(x))^2 - log(4*exp(x))*(8*x + 16) - 16,x)

[Out]

- x*(16*log(4) + 4*log(4)^2 + 16) - x^2*(8*log(4) + 11) - 4*x^3

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 32, normalized size = 1.60 \begin {gather*} - 4 x^{3} + x^{2} \left (- 16 \log {\relax (2 )} - 11\right ) + x \left (- 32 \log {\relax (2 )} - 16 - 16 \log {\relax (2 )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*ln(4*exp(x))**2+(-8*x-16)*ln(4*exp(x))-6*x-16,x)

[Out]

-4*x**3 + x**2*(-16*log(2) - 11) + x*(-32*log(2) - 16 - 16*log(2)**2)

________________________________________________________________________________________