3.43.79 \(\int \frac {-15+10 x-5 x^2+(15 x+5 x^2) \log (\frac {16}{9+6 x+x^2})}{3 x+x^2} \, dx\)

Optimal. Leaf size=19 \[ 5 \left (-2+x-\log (x)+x \log \left (\frac {16}{(3+x)^2}\right )\right ) \]

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Rubi [A]  time = 0.43, antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {1593, 6741, 12, 6688, 6742, 893, 2389, 2295} \begin {gather*} 5 x-5 \log (x)+5 (x+3) \log \left (\frac {16}{(x+3)^2}\right )+30 \log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 + 10*x - 5*x^2 + (15*x + 5*x^2)*Log[16/(9 + 6*x + x^2)])/(3*x + x^2),x]

[Out]

5*x - 5*Log[x] + 5*(3 + x)*Log[16/(3 + x)^2] + 30*Log[3 + x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 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]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15+10 x-5 x^2+\left (15 x+5 x^2\right ) \log \left (\frac {16}{9+6 x+x^2}\right )}{x (3+x)} \, dx\\ &=\int \frac {5 \left (-3+2 x-x^2+3 x \log \left (\frac {16}{(3+x)^2}\right )+x^2 \log \left (\frac {16}{(3+x)^2}\right )\right )}{x (3+x)} \, dx\\ &=5 \int \frac {-3+2 x-x^2+3 x \log \left (\frac {16}{(3+x)^2}\right )+x^2 \log \left (\frac {16}{(3+x)^2}\right )}{x (3+x)} \, dx\\ &=5 \int \frac {-3+2 x-x^2+x (3+x) \log \left (\frac {16}{(3+x)^2}\right )}{x (3+x)} \, dx\\ &=5 \int \left (\frac {-3+2 x-x^2}{x (3+x)}+\log \left (\frac {16}{(3+x)^2}\right )\right ) \, dx\\ &=5 \int \frac {-3+2 x-x^2}{x (3+x)} \, dx+5 \int \log \left (\frac {16}{(3+x)^2}\right ) \, dx\\ &=5 \int \left (-1-\frac {1}{x}+\frac {6}{3+x}\right ) \, dx+5 \operatorname {Subst}\left (\int \log \left (\frac {16}{x^2}\right ) \, dx,x,3+x\right )\\ &=5 x-5 \log (x)+5 (3+x) \log \left (\frac {16}{(3+x)^2}\right )+30 \log (3+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 1.37 \begin {gather*} 5 \left (x-\log (x)+(3+x) \log \left (\frac {16}{(3+x)^2}\right )+6 \log (3+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + 10*x - 5*x^2 + (15*x + 5*x^2)*Log[16/(9 + 6*x + x^2)])/(3*x + x^2),x]

[Out]

5*(x - Log[x] + (3 + x)*Log[16/(3 + x)^2] + 6*Log[3 + x])

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fricas [A]  time = 0.68, size = 24, normalized size = 1.26 \begin {gather*} 5 \, x \log \left (\frac {16}{x^{2} + 6 \, x + 9}\right ) + 5 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+15*x)*log(16/(x^2+6*x+9))-5*x^2+10*x-15)/(x^2+3*x),x, algorithm="fricas")

[Out]

5*x*log(16/(x^2 + 6*x + 9)) + 5*x - 5*log(x)

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giac [A]  time = 0.17, size = 24, normalized size = 1.26 \begin {gather*} 5 \, x \log \left (\frac {16}{x^{2} + 6 \, x + 9}\right ) + 5 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+15*x)*log(16/(x^2+6*x+9))-5*x^2+10*x-15)/(x^2+3*x),x, algorithm="giac")

[Out]

5*x*log(16/(x^2 + 6*x + 9)) + 5*x - 5*log(x)

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maple [A]  time = 0.10, size = 25, normalized size = 1.32

method	result	size
norman	\(5 x +5 \ln \left (\frac {16}{x^{2}+6 x +9}\right ) x -5 \ln \relax (x )\)	\(25\)
risch	\(5 x +5 \ln \left (\frac {16}{x^{2}+6 x +9}\right ) x -5 \ln \relax (x )\)	\(25\)
derivativedivides	\(20 \left (3+x \right ) \ln \relax (2)+5 \left (3+x \right ) \ln \left (\frac {1}{\left (3+x \right )^{2}}\right )+15+5 x -25 \ln \left (\frac {1}{3+x}\right )-5 \ln \left (-1+\frac {3}{3+x}\right )\)	\(44\)
default	\(20 \left (3+x \right ) \ln \relax (2)+5 \left (3+x \right ) \ln \left (\frac {1}{\left (3+x \right )^{2}}\right )+15+5 x -25 \ln \left (\frac {1}{3+x}\right )-5 \ln \left (-1+\frac {3}{3+x}\right )\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+15*x)*ln(16/(x^2+6*x+9))-5*x^2+10*x-15)/(x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

5*x+5*ln(16/(x^2+6*x+9))*x-5*ln(x)

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maxima [B]  time = 0.36, size = 50, normalized size = 2.63 \begin {gather*} 5 \, {\left (x - 3 \, \log \left (x + 3\right )\right )} \log \left (\frac {16}{x^{2} + 6 \, x + 9}\right ) + 15 \, \log \left (x + 3\right ) \log \left (\frac {16}{x^{2} + 6 \, x + 9}\right ) + 5 \, x - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+15*x)*log(16/(x^2+6*x+9))-5*x^2+10*x-15)/(x^2+3*x),x, algorithm="maxima")

[Out]

5*(x - 3*log(x + 3))*log(16/(x^2 + 6*x + 9)) + 15*log(x + 3)*log(16/(x^2 + 6*x + 9)) + 5*x - 5*log(x)

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mupad [B]  time = 3.31, size = 24, normalized size = 1.26 \begin {gather*} 5\,x-5\,\ln \relax (x)+5\,x\,\ln \left (\frac {16}{x^2+6\,x+9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log(16/(6*x + x^2 + 9))*(15*x + 5*x^2) - 5*x^2 - 15)/(3*x + x^2),x)

[Out]

5*x - 5*log(x) + 5*x*log(16/(6*x + x^2 + 9))

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sympy [A]  time = 0.15, size = 22, normalized size = 1.16 \begin {gather*} 5 x \log {\left (\frac {16}{x^{2} + 6 x + 9} \right )} + 5 x - 5 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+15*x)*ln(16/(x**2+6*x+9))-5*x**2+10*x-15)/(x**2+3*x),x)

[Out]

5*x*log(16/(x**2 + 6*x + 9)) + 5*x - 5*log(x)

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