3.8.74 \(\int (8 x+6 x^2+8 x^3+e^{2 x} (-6 x^2-4 x^3)+6 x^2 \log (4)) \, dx\)

Optimal. Leaf size=21 \[ 2 x^2 \left (2+x+x \left (-e^{2 x}+x+\log (4)\right )\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 30, normalized size of antiderivative = 1.43, number of steps used = 12, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6, 1593, 2196, 2176, 2194} \begin {gather*} 2 x^4-2 e^{2 x} x^3+2 x^3 (1+\log (4))+4 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8*x + 6*x^2 + 8*x^3 + E^(2*x)*(-6*x^2 - 4*x^3) + 6*x^2*Log[4],x]

[Out]

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

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8 x+8 x^3+e^{2 x} \left (-6 x^2-4 x^3\right )+x^2 (6+6 \log (4))\right ) \, dx\\ &=4 x^2+2 x^4+2 x^3 (1+\log (4))+\int e^{2 x} \left (-6 x^2-4 x^3\right ) \, dx\\ &=4 x^2+2 x^4+2 x^3 (1+\log (4))+\int e^{2 x} (-6-4 x) x^2 \, dx\\ &=4 x^2+2 x^4+2 x^3 (1+\log (4))+\int \left (-6 e^{2 x} x^2-4 e^{2 x} x^3\right ) \, dx\\ &=4 x^2+2 x^4+2 x^3 (1+\log (4))-4 \int e^{2 x} x^3 \, dx-6 \int e^{2 x} x^2 \, dx\\ &=4 x^2-3 e^{2 x} x^2-2 e^{2 x} x^3+2 x^4+2 x^3 (1+\log (4))+6 \int e^{2 x} x \, dx+6 \int e^{2 x} x^2 \, dx\\ &=3 e^{2 x} x+4 x^2-2 e^{2 x} x^3+2 x^4+2 x^3 (1+\log (4))-3 \int e^{2 x} \, dx-6 \int e^{2 x} x \, dx\\ &=-\frac {3 e^{2 x}}{2}+4 x^2-2 e^{2 x} x^3+2 x^4+2 x^3 (1+\log (4))+3 \int e^{2 x} \, dx\\ &=4 x^2-2 e^{2 x} x^3+2 x^4+2 x^3 (1+\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 1.10 \begin {gather*} 2 x^2 \left (2+x^2+x \left (1-e^{2 x}+\log (4)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8*x + 6*x^2 + 8*x^3 + E^(2*x)*(-6*x^2 - 4*x^3) + 6*x^2*Log[4],x]

[Out]

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

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fricas [A]  time = 0.74, size = 32, normalized size = 1.52 \begin {gather*} 2 \, x^{4} - 2 \, x^{3} e^{\left (2 \, x\right )} + 4 \, x^{3} \log \relax (2) + 2 \, x^{3} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-6*x^2)*exp(2*x)+12*x^2*log(2)+8*x^3+6*x^2+8*x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 32, normalized size = 1.52 \begin {gather*} 2 \, x^{4} - 2 \, x^{3} e^{\left (2 \, x\right )} + 4 \, x^{3} \log \relax (2) + 2 \, x^{3} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-6*x^2)*exp(2*x)+12*x^2*log(2)+8*x^3+6*x^2+8*x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 31, normalized size = 1.48

method	result	size
norman	\(\left (4 \ln \relax (2)+2\right ) x^{3}+4 x^{2}+2 x^{4}-2 \,{\mathrm e}^{2 x} x^{3}\)	\(31\)
derivativedivides	\(4 x^{2}+2 x^{3}+2 x^{4}-2 \,{\mathrm e}^{2 x} x^{3}+4 x^{3} \ln \relax (2)\)	\(33\)
default	\(4 x^{2}+2 x^{3}+2 x^{4}-2 \,{\mathrm e}^{2 x} x^{3}+4 x^{3} \ln \relax (2)\)	\(33\)
risch	\(4 x^{2}+2 x^{3}+2 x^{4}-2 \,{\mathrm e}^{2 x} x^{3}+4 x^{3} \ln \relax (2)\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^3-6*x^2)*exp(2*x)+12*x^2*ln(2)+8*x^3+6*x^2+8*x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 32, normalized size = 1.52 \begin {gather*} 2 \, x^{4} - 2 \, x^{3} e^{\left (2 \, x\right )} + 4 \, x^{3} \log \relax (2) + 2 \, x^{3} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-6*x^2)*exp(2*x)+12*x^2*log(2)+8*x^3+6*x^2+8*x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.56, size = 28, normalized size = 1.33 \begin {gather*} x^3\,\left (\ln \left (16\right )+2\right )-2\,x^3\,{\mathrm {e}}^{2\,x}+4\,x^2+2\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x - exp(2*x)*(6*x^2 + 4*x^3) + 12*x^2*log(2) + 6*x^2 + 8*x^3,x)

[Out]

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

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sympy [A]  time = 0.10, size = 29, normalized size = 1.38 \begin {gather*} 2 x^{4} - 2 x^{3} e^{2 x} + x^{3} \left (2 + 4 \log {\relax (2 )}\right ) + 4 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**3-6*x**2)*exp(2*x)+12*x**2*ln(2)+8*x**3+6*x**2+8*x,x)

[Out]

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

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