3.32.100 \(\int \frac {35+32 x+8 x^2+\log (3)}{4+4 x+x^2} \, dx\)

Optimal. Leaf size=24 \[ 2+4 (-1+2 x-\log (3))-\frac {3+\log (3)}{2+x} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {27, 683} \begin {gather*} 8 x-\frac {3+\log (3)}{x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(35 + 32*x + 8*x^2 + Log[3])/(4 + 4*x + x^2),x]

[Out]

8*x - (3 + Log[3])/(2 + x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {35+32 x+8 x^2+\log (3)}{(2+x)^2} \, dx\\ &=\int \left (8+\frac {3+\log (3)}{(2+x)^2}\right ) \, dx\\ &=8 x-\frac {3+\log (3)}{2+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.75 \begin {gather*} 8 (2+x)+\frac {-3-\log (3)}{2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(35 + 32*x + 8*x^2 + Log[3])/(4 + 4*x + x^2),x]

[Out]

8*(2 + x) + (-3 - Log[3])/(2 + x)

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fricas [A]  time = 0.46, size = 20, normalized size = 0.83 \begin {gather*} \frac {8 \, x^{2} + 16 \, x - \log \relax (3) - 3}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="fricas")

[Out]

(8*x^2 + 16*x - log(3) - 3)/(x + 2)

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giac [A]  time = 0.19, size = 15, normalized size = 0.62 \begin {gather*} 8 \, x - \frac {\log \relax (3) + 3}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="giac")

[Out]

8*x - (log(3) + 3)/(x + 2)

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maple [A]  time = 0.57, size = 16, normalized size = 0.67

method	result	size
default	\(8 x -\frac {3+\ln \relax (3)}{2+x}\)	\(16\)
gosper	\(-\frac {-8 x^{2}+\ln \relax (3)+35}{2+x}\)	\(17\)
norman	\(\frac {8 x^{2}-35-\ln \relax (3)}{2+x}\)	\(18\)
risch	\(8 x -\frac {3}{2+x}-\frac {\ln \relax (3)}{2+x}\)	\(21\)
meijerg	\(-\frac {29 x}{4 \left (1+\frac {x}{2}\right )}+\frac {x \ln \relax (3)}{2 x +4}+\frac {8 x \left (\frac {3 x}{2}+6\right )}{3 \left (1+\frac {x}{2}\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(3)+8*x^2+32*x+35)/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

8*x-(3+ln(3))/(2+x)

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maxima [A]  time = 0.34, size = 15, normalized size = 0.62 \begin {gather*} 8 \, x - \frac {\log \relax (3) + 3}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+8*x^2+32*x+35)/(x^2+4*x+4),x, algorithm="maxima")

[Out]

8*x - (log(3) + 3)/(x + 2)

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mupad [B]  time = 1.87, size = 15, normalized size = 0.62 \begin {gather*} 8\,x-\frac {\ln \relax (3)+3}{x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + log(3) + 8*x^2 + 35)/(4*x + x^2 + 4),x)

[Out]

8*x - (log(3) + 3)/(x + 2)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.50 \begin {gather*} 8 x + \frac {-3 - \log {\relax (3 )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(3)+8*x**2+32*x+35)/(x**2+4*x+4),x)

[Out]

8*x + (-3 - log(3))/(x + 2)

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