3.96.73 \(\int (12-16 x+6 x^2+e^{3/4} (4-8 x+3 x^2)+(4-8 x+3 x^2) \log (2)) \, dx\)

Optimal. Leaf size=29 \[ x \left (x-(2-x)^2 \left (-2-e^{3/4}+\frac {1}{x}-\log (2)\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.72, number of steps used = 3, number of rules used = 1, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6} \begin {gather*} 2 x^3+x^3 \left (e^{3/4}+\log (2)\right )-8 x^2-4 x^2 \left (e^{3/4}+\log (2)\right )+12 x+4 x \left (e^{3/4}+\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[12 - 16*x + 6*x^2 + E^(3/4)*(4 - 8*x + 3*x^2) + (4 - 8*x + 3*x^2)*Log[2],x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (12-16 x+6 x^2+\left (4-8 x+3 x^2\right ) \left (e^{3/4}+\log (2)\right )\right ) \, dx\\ &=12 x-8 x^2+2 x^3+\left (e^{3/4}+\log (2)\right ) \int \left (4-8 x+3 x^2\right ) \, dx\\ &=12 x-8 x^2+2 x^3+4 x \left (e^{3/4}+\log (2)\right )-4 x^2 \left (e^{3/4}+\log (2)\right )+x^3 \left (e^{3/4}+\log (2)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 1.24 \begin {gather*} \frac {1}{3} x \left (36+3 e^{3/4} (-2+x)^2-12 x (2+\log (2))+x^2 (6+\log (8))+\log (4096)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[12 - 16*x + 6*x^2 + E^(3/4)*(4 - 8*x + 3*x^2) + (4 - 8*x + 3*x^2)*Log[2],x]

[Out]

(x*(36 + 3*E^(3/4)*(-2 + x)^2 - 12*x*(2 + Log[2]) + x^2*(6 + Log[8]) + Log[4096]))/3

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fricas [A]  time = 0.49, size = 44, normalized size = 1.52 \begin {gather*} 2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \relax (2) + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-8*x+4)*log(2)+(3*x^2-8*x+4)*exp(3/4)+6*x^2-16*x+12,x, algorithm="fricas")

[Out]

2*x^3 - 8*x^2 + (x^3 - 4*x^2 + 4*x)*e^(3/4) + (x^3 - 4*x^2 + 4*x)*log(2) + 12*x

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giac [A]  time = 0.12, size = 44, normalized size = 1.52 \begin {gather*} 2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \relax (2) + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-8*x+4)*log(2)+(3*x^2-8*x+4)*exp(3/4)+6*x^2-16*x+12,x, algorithm="giac")

[Out]

2*x^3 - 8*x^2 + (x^3 - 4*x^2 + 4*x)*e^(3/4) + (x^3 - 4*x^2 + 4*x)*log(2) + 12*x

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maple [A]  time = 0.03, size = 38, normalized size = 1.31

method	result	size
norman	\(\left ({\mathrm e}^{\frac {3}{4}}+\ln \relax (2)+2\right ) x^{3}+\left (-4 \ln \relax (2)-4 \,{\mathrm e}^{\frac {3}{4}}-8\right ) x^{2}+\left (4 \ln \relax (2)+4 \,{\mathrm e}^{\frac {3}{4}}+12\right ) x\)	\(38\)
gosper	\(x \left (x^{2} \ln \relax (2)+x^{2} {\mathrm e}^{\frac {3}{4}}-4 x \ln \relax (2)-4 x \,{\mathrm e}^{\frac {3}{4}}+2 x^{2}+4 \ln \relax (2)+4 \,{\mathrm e}^{\frac {3}{4}}-8 x +12\right )\)	\(43\)
default	\(\ln \relax (2) \left (x^{3}-4 x^{2}+4 x \right )+{\mathrm e}^{\frac {3}{4}} \left (x^{3}-4 x^{2}+4 x \right )+2 x^{3}-8 x^{2}+12 x\)	\(45\)
risch	\(x^{3} \ln \relax (2)-4 x^{2} \ln \relax (2)+4 x \ln \relax (2)+{\mathrm e}^{\frac {3}{4}} x^{3}-4 x^{2} {\mathrm e}^{\frac {3}{4}}+4 x \,{\mathrm e}^{\frac {3}{4}}+2 x^{3}-8 x^{2}+12 x\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2-8*x+4)*ln(2)+(3*x^2-8*x+4)*exp(3/4)+6*x^2-16*x+12,x,method=_RETURNVERBOSE)

[Out]

(exp(3/4)+ln(2)+2)*x^3+(-4*ln(2)-4*exp(3/4)-8)*x^2+(4*ln(2)+4*exp(3/4)+12)*x

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maxima [A]  time = 0.37, size = 44, normalized size = 1.52 \begin {gather*} 2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \relax (2) + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-8*x+4)*log(2)+(3*x^2-8*x+4)*exp(3/4)+6*x^2-16*x+12,x, algorithm="maxima")

[Out]

2*x^3 - 8*x^2 + (x^3 - 4*x^2 + 4*x)*e^(3/4) + (x^3 - 4*x^2 + 4*x)*log(2) + 12*x

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mupad [B]  time = 0.08, size = 34, normalized size = 1.17 \begin {gather*} \left ({\mathrm {e}}^{3/4}+\ln \relax (2)+2\right )\,x^3+\left (-4\,{\mathrm {e}}^{3/4}-\ln \left (16\right )-8\right )\,x^2+\left (4\,{\mathrm {e}}^{3/4}+\ln \left (16\right )+12\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3/4)*(3*x^2 - 8*x + 4) - 16*x + log(2)*(3*x^2 - 8*x + 4) + 6*x^2 + 12,x)

[Out]

x*(4*exp(3/4) + log(16) + 12) + x^3*(exp(3/4) + log(2) + 2) - x^2*(4*exp(3/4) + log(16) + 8)

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sympy [B]  time = 0.06, size = 46, normalized size = 1.59 \begin {gather*} x^{3} \left (\log {\relax (2 )} + 2 + e^{\frac {3}{4}}\right ) + x^{2} \left (- 4 e^{\frac {3}{4}} - 8 - 4 \log {\relax (2 )}\right ) + x \left (4 \log {\relax (2 )} + 4 e^{\frac {3}{4}} + 12\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2-8*x+4)*ln(2)+(3*x**2-8*x+4)*exp(3/4)+6*x**2-16*x+12,x)

[Out]

x**3*(log(2) + 2 + exp(3/4)) + x**2*(-4*exp(3/4) - 8 - 4*log(2)) + x*(4*log(2) + 4*exp(3/4) + 12)

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