3.35.5 \(\int \frac {17 x+8 x^2+(2+x) \log (2+x)}{2+x} \, dx\)

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

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6688, 77, 2389, 2295} \begin {gather*} 4 x^2+(x+2) \log (x+2)-2 \log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(17*x + 8*x^2 + (2 + x)*Log[2 + x])/(2 + x),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x (17+8 x)}{2+x}+\log (2+x)\right ) \, dx\\ &=\int \frac {x (17+8 x)}{2+x} \, dx+\int \log (2+x) \, dx\\ &=\int \left (1+8 x-\frac {2}{2+x}\right ) \, dx+\operatorname {Subst}(\int \log (x) \, dx,x,2+x)\\ &=4 x^2-2 \log (2+x)+(2+x) \log (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.10 \begin {gather*} -x-15 (2+x)+4 (2+x)^2+x \log (2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(17*x + 8*x^2 + (2 + x)*Log[2 + x])/(2 + x),x]

[Out]

-x - 15*(2 + x) + 4*(2 + x)^2 + x*Log[2 + x]

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fricas [A]  time = 0.71, size = 12, normalized size = 0.60 \begin {gather*} 4 \, x^{2} + x \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)+8*x^2+17*x)/(2+x),x, algorithm="fricas")

[Out]

4*x^2 + x*log(x + 2)

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giac [A]  time = 0.14, size = 12, normalized size = 0.60 \begin {gather*} 4 \, x^{2} + x \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)+8*x^2+17*x)/(2+x),x, algorithm="giac")

[Out]

4*x^2 + x*log(x + 2)

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maple [A]  time = 0.46, size = 13, normalized size = 0.65

method	result	size
norman	\(x \ln \left (2+x \right )+4 x^{2}\)	\(13\)
risch	\(x \ln \left (2+x \right )+4 x^{2}\)	\(13\)
derivativedivides	\(\left (2+x \right ) \ln \left (2+x \right )-32-16 x +4 \left (2+x \right )^{2}-2 \ln \left (2+x \right )\)	\(27\)
default	\(\left (2+x \right ) \ln \left (2+x \right )-32-16 x +4 \left (2+x \right )^{2}-2 \ln \left (2+x \right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2+x)*ln(2+x)+8*x^2+17*x)/(2+x),x,method=_RETURNVERBOSE)

[Out]

x*ln(2+x)+4*x^2

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maxima [A]  time = 0.48, size = 27, normalized size = 1.35 \begin {gather*} 4 \, x^{2} + {\left (x - 2 \, \log \left (x + 2\right )\right )} \log \left (x + 2\right ) + 2 \, \log \left (x + 2\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)+8*x^2+17*x)/(2+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.03, size = 10, normalized size = 0.50 \begin {gather*} x\,\left (4\,x+\ln \left (x+2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((17*x + log(x + 2)*(x + 2) + 8*x^2)/(x + 2),x)

[Out]

x*(4*x + log(x + 2))

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sympy [A]  time = 0.10, size = 10, normalized size = 0.50 \begin {gather*} 4 x^{2} + x \log {\left (x + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*ln(2+x)+8*x**2+17*x)/(2+x),x)

[Out]

4*x**2 + x*log(x + 2)

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