\(\int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log (16+8 x+x^2)}{4+x} \, dx\) [3199]

Optimal result

Integrand size = 41, antiderivative size = 29 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=2 x+x^2-(x-4 (i \pi +\log (3))) \log \left ((-4-x)^2\right ) \]

[Out]

2*x+x^2-(x-4*ln(3)-4*I*Pi)*ln((-4-x)^2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {6873, 6874, 712, 2436, 2332} \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=x^2+2 x+2 (4+4 i \pi +\log (81)) \log (x+4)-(x+4) \log \left ((x+4)^2\right ) \]

[In]

Int[(8 + 8*x + 2*x^2 + 8*(I*Pi + Log[3]) + (-4 - x)*Log[16 + 8*x + x^2])/(4 + x),x]

[Out]

2*x + x^2 + 2*(4 + (4*I)*Pi + Log[81])*Log[4 + x] - (4 + x)*Log[(4 + x)^2]

Rule 712

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] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2332

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

Rule 2436

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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {8 x+2 x^2+8 (1+i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx \\ & = \int \left (\frac {2 i \left (4 \pi -4 i x-i x^2-i (4+\log (81))\right )}{4+x}-\log \left ((4+x)^2\right )\right ) \, dx \\ & = 2 i \int \frac {4 \pi -4 i x-i x^2-i (4+\log (81))}{4+x} \, dx-\int \log \left ((4+x)^2\right ) \, dx \\ & = 2 i \int \left (-i x+\frac {4 \pi -i (4+\log (81))}{4+x}\right ) \, dx-\text {Subst}\left (\int \log \left (x^2\right ) \, dx,x,4+x\right ) \\ & = 2 x+x^2+2 (4+4 i \pi +\log (81)) \log (4+x)-(4+x) \log \left ((4+x)^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=-16+2 x+x^2+2 (4+4 i \pi +\log (81)) \log (4+x)-(4+x) \log \left ((4+x)^2\right ) \]

[In]

Integrate[(8 + 8*x + 2*x^2 + 8*(I*Pi + Log[3]) + (-4 - x)*Log[16 + 8*x + x^2])/(4 + x),x]

[Out]

-16 + 2*x + x^2 + 2*(4 + (4*I)*Pi + Log[81])*Log[4 + x] - (4 + x)*Log[(4 + x)^2]

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24

method	result	size
risch	\(-\ln \left (x^{2}+8 x +16\right ) x +x^{2}+2 x +8 \ln \left (3\right ) \ln \left (4+x \right )+8 i \pi \ln \left (4+x \right )\)	\(36\)
norman	\(x^{2}+\left (4 i \pi +4 \ln \left (3\right )\right ) \ln \left (x^{2}+8 x +16\right )+2 x -\ln \left (x^{2}+8 x +16\right ) x\)	\(39\)
parts	\(x^{2}+2 \left (4 i \pi +4 \ln \left (3\right )+4\right ) \ln \left (4+x \right )-\ln \left (x^{2}+8 x +16\right ) x +2 x -8 \ln \left (4+x \right )\)	\(42\)
default	\(8 \ln \left (3\right ) \ln \left (4+x \right )+x^{2}+2 \left (4 i \pi +4\right ) \ln \left (4+x \right )-\ln \left (x^{2}+8 x +16\right ) x +2 x -8 \ln \left (4+x \right )\)	\(46\)
parallelrisch	\(4 i \pi \ln \left (x^{2}+8 x +16\right )+4 \ln \left (3\right ) \ln \left (x^{2}+8 x +16\right )-32+x^{2}-\ln \left (x^{2}+8 x +16\right ) x +2 x\)	\(47\)

[In]

int(((-4-x)*ln(x^2+8*x+16)+8*ln(3)+8*I*Pi+2*x^2+8*x+8)/(4+x),x,method=_RETURNVERBOSE)

[Out]

-ln(x^2+8*x+16)*x+x^2+2*x+8*ln(3)*ln(4+x)+8*I*Pi*ln(4+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=x^{2} + {\left (4 i \, \pi - x + 4 \, \log \left (3\right )\right )} \log \left (x^{2} + 8 \, x + 16\right ) + 2 \, x \]

[In]

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

[Out]

x^2 + (4*I*pi - x + 4*log(3))*log(x^2 + 8*x + 16) + 2*x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=x^{2} - x \log {\left (x^{2} + 8 x + 16 \right )} + 2 x + 8 \left (\log {\left (3 \right )} + i \pi \right ) \log {\left (x + 4 \right )} \]

[In]

integrate(((-4-x)*ln(x**2+8*x+16)+8*ln(3)+8*I*pi+2*x**2+8*x+8)/(4+x),x)

[Out]

x**2 - x*log(x**2 + 8*x + 16) + 2*x + 8*(log(3) + I*pi)*log(x + 4)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=x^{2} - {\left (x - 4 \, \log \left (x + 4\right )\right )} \log \left (x^{2} + 8 \, x + 16\right ) + 8 i \, \pi \log \left (x + 4\right ) + 8 \, \log \left (3\right ) \log \left (x + 4\right ) - 8 \, \log \left (x + 4\right )^{2} + 2 \, x \]

[In]

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

[Out]

x^2 - (x - 4*log(x + 4))*log(x^2 + 8*x + 16) + 8*I*pi*log(x + 4) + 8*log(3)*log(x + 4) - 8*log(x + 4)^2 + 2*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=x^{2} - x \log \left (x^{2} + 8 \, x + 16\right ) - 8 \, {\left (-i \, \pi - \log \left (3\right )\right )} \log \left (x + 4\right ) + 2 \, x \]

[In]

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

[Out]

x^2 - x*log(x^2 + 8*x + 16) - 8*(-I*pi - log(3))*log(x + 4) + 2*x

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {8+8 x+2 x^2+8 (i \pi +\log (3))+(-4-x) \log \left (16+8 x+x^2\right )}{4+x} \, dx=2\,x+4\,\ln \left (3\right )\,\ln \left ({\left (x+4\right )}^2\right )-x\,\ln \left ({\left (x+4\right )}^2\right )+x^2+\Pi \,\ln \left ({\left (x+4\right )}^2\right )\,4{}\mathrm {i} \]

[In]

int((Pi*8i + 8*x + 8*log(3) - log(8*x + x^2 + 16)*(x + 4) + 2*x^2 + 8)/(x + 4),x)

[Out]

2*x + Pi*log((x + 4)^2)*4i + 4*log(3)*log((x + 4)^2) - x*log((x + 4)^2) + x^2