3.12.62 \(\int (-968 x+e^{4-e^3+2 x} (1+2 x)-88 x \log (2)-2 x \log ^2(2)) \, dx\)

Optimal. Leaf size=28 \[ 5-x \left (-e^{4-e^3+2 x}+x (22+\log (2))^2\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 2176, 2194} \begin {gather*} x^2 \left (-(22+\log (2))^2\right )-\frac {1}{2} e^{2 x-e^3+4}+\frac {1}{2} e^{2 x-e^3+4} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-968*x + E^(4 - E^3 + 2*x)*(1 + 2*x) - 88*x*Log[2] - 2*x*Log[2]^2,x]

[Out]

-1/2*E^(4 - E^3 + 2*x) + (E^(4 - E^3 + 2*x)*(1 + 2*x))/2 - x^2*(22 + Log[2])^2

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 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 (e^{4-e^3+2 x} (1+2 x)+x (-968-88 \log (2))-2 x \log ^2(2)\right ) \, dx\\ &=\int \left (e^{4-e^3+2 x} (1+2 x)+x \left (-968-88 \log (2)-2 \log ^2(2)\right )\right ) \, dx\\ &=-x^2 (22+\log (2))^2+\int e^{4-e^3+2 x} (1+2 x) \, dx\\ &=\frac {1}{2} e^{4-e^3+2 x} (1+2 x)-x^2 (22+\log (2))^2-\int e^{4-e^3+2 x} \, dx\\ &=-\frac {1}{2} e^{4-e^3+2 x}+\frac {1}{2} e^{4-e^3+2 x} (1+2 x)-x^2 (22+\log (2))^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.93 \begin {gather*} e^{4-e^3+2 x} x-x^2 (22+\log (2))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-968*x + E^(4 - E^3 + 2*x)*(1 + 2*x) - 88*x*Log[2] - 2*x*Log[2]^2,x]

[Out]

E^(4 - E^3 + 2*x)*x - x^2*(22 + Log[2])^2

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fricas [A]  time = 0.61, size = 34, normalized size = 1.21 \begin {gather*} -x^{2} \log \relax (2)^{2} - 44 \, x^{2} \log \relax (2) - 484 \, x^{2} + x e^{\left (2 \, x - e^{3} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(-exp(3)+2*x+4)-2*x*log(2)^2-88*x*log(2)-968*x,x, algorithm="fricas")

[Out]

-x^2*log(2)^2 - 44*x^2*log(2) - 484*x^2 + x*e^(2*x - e^3 + 4)

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giac [A]  time = 0.36, size = 34, normalized size = 1.21 \begin {gather*} -x^{2} \log \relax (2)^{2} - 44 \, x^{2} \log \relax (2) - 484 \, x^{2} + x e^{\left (2 \, x - e^{3} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(-exp(3)+2*x+4)-2*x*log(2)^2-88*x*log(2)-968*x,x, algorithm="giac")

[Out]

-x^2*log(2)^2 - 44*x^2*log(2) - 484*x^2 + x*e^(2*x - e^3 + 4)

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maple [A]  time = 0.04, size = 30, normalized size = 1.07

method	result	size
norman	\(\left (-\ln \relax (2)^{2}-44 \ln \relax (2)-484\right ) x^{2}+{\mathrm e}^{-{\mathrm e}^{3}+2 x +4} x\)	\(30\)
risch	\({\mathrm e}^{-{\mathrm e}^{3}+2 x +4} x -x^{2} \ln \relax (2)^{2}-44 x^{2} \ln \relax (2)-484 x^{2}\)	\(35\)
default	\(\frac {{\mathrm e}^{-{\mathrm e}^{3}+2 x +4} \left (-{\mathrm e}^{3}+2 x +4\right )}{2}-2 \,{\mathrm e}^{-{\mathrm e}^{3}+2 x +4}+\frac {{\mathrm e}^{-{\mathrm e}^{3}+2 x +4} {\mathrm e}^{3}}{2}-484 x^{2}-44 x^{2} \ln \relax (2)-x^{2} \ln \relax (2)^{2}\)	\(70\)
derivativedivides	\(-121 \left (-{\mathrm e}^{3}+2 x +4\right )^{2}-968 \,{\mathrm e}^{3}+1936 x +3872+\frac {{\mathrm e}^{-{\mathrm e}^{3}+2 x +4} \left (-{\mathrm e}^{3}+2 x +4\right )}{2}-2 \,{\mathrm e}^{-{\mathrm e}^{3}+2 x +4}+\frac {{\mathrm e}^{-{\mathrm e}^{3}+2 x +4} {\mathrm e}^{3}}{2}-22 \ln \relax (2) {\mathrm e}^{3} \left (-{\mathrm e}^{3}+2 x +4\right )-11 \ln \relax (2) \left (-{\mathrm e}^{3}+2 x +4\right )^{2}+88 \ln \relax (2) \left (-{\mathrm e}^{3}+2 x +4\right )-\frac {\ln \relax (2)^{2} {\mathrm e}^{3} \left (-{\mathrm e}^{3}+2 x +4\right )}{2}-\frac {\ln \relax (2)^{2} \left (-{\mathrm e}^{3}+2 x +4\right )^{2}}{4}+2 \ln \relax (2)^{2} \left (-{\mathrm e}^{3}+2 x +4\right )-242 \,{\mathrm e}^{3} \left (-{\mathrm e}^{3}+2 x +4\right )\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x+1)*exp(-exp(3)+2*x+4)-2*x*ln(2)^2-88*x*ln(2)-968*x,x,method=_RETURNVERBOSE)

[Out]

(-ln(2)^2-44*ln(2)-484)*x^2+exp(-exp(3)+2*x+4)*x

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maxima [A]  time = 0.43, size = 34, normalized size = 1.21 \begin {gather*} -x^{2} \log \relax (2)^{2} - 44 \, x^{2} \log \relax (2) - 484 \, x^{2} + x e^{\left (2 \, x - e^{3} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(-exp(3)+2*x+4)-2*x*log(2)^2-88*x*log(2)-968*x,x, algorithm="maxima")

[Out]

-x^2*log(2)^2 - 44*x^2*log(2) - 484*x^2 + x*e^(2*x - e^3 + 4)

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mupad [B]  time = 0.09, size = 24, normalized size = 0.86 \begin {gather*} x\,{\mathrm {e}}^{2\,x-{\mathrm {e}}^3+4}-x^2\,{\left (\ln \relax (2)+22\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x - exp(3) + 4)*(2*x + 1) - 88*x*log(2) - 2*x*log(2)^2 - 968*x,x)

[Out]

x*exp(2*x - exp(3) + 4) - x^2*(log(2) + 22)^2

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sympy [A]  time = 0.10, size = 27, normalized size = 0.96 \begin {gather*} x^{2} \left (-484 - 44 \log {\relax (2 )} - \log {\relax (2 )}^{2}\right ) + x e^{2 x - e^{3} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(-exp(3)+2*x+4)-2*x*ln(2)**2-88*x*ln(2)-968*x,x)

[Out]

x**2*(-484 - 44*log(2) - log(2)**2) + x*exp(2*x - exp(3) + 4)

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