\(\int \frac {29 x-3 x^2+8 x^3+(3-55 x+17 x^2+18 x^3) \log (-3+x+x^2)}{-15+5 x+5 x^2} \, dx\) [3726]

Optimal result

Integrand size = 50, antiderivative size = 26 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-x^2+\frac {1}{5} \left (-x+9 x^2\right ) \log \left (-3+x+x^2\right ) \]

[Out]

1/5*(9*x^2-x)*ln(x^2+x-3)-x^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(124\) vs. \(2(26)=52\).

Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.77, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6860, 1642, 646, 31, 2605, 814} \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-x^2+\frac {1}{180} (1-18 x)^2 \log \left (x^2+x-3\right )+\frac {1}{5} \left (32-5 \sqrt {13}\right ) \log \left (2 x-\sqrt {13}+1\right )-\frac {1}{180} \left (1153-180 \sqrt {13}\right ) \log \left (2 x-\sqrt {13}+1\right )-\frac {1}{180} \left (1153+180 \sqrt {13}\right ) \log \left (2 x+\sqrt {13}+1\right )+\frac {1}{5} \left (32+5 \sqrt {13}\right ) \log \left (2 x+\sqrt {13}+1\right ) \]

[In]

Int[(29*x - 3*x^2 + 8*x^3 + (3 - 55*x + 17*x^2 + 18*x^3)*Log[-3 + x + x^2])/(-15 + 5*x + 5*x^2),x]

[Out]

-x^2 - ((1153 - 180*Sqrt[13])*Log[1 - Sqrt[13] + 2*x])/180 + ((32 - 5*Sqrt[13])*Log[1 - Sqrt[13] + 2*x])/5 + (
(32 + 5*Sqrt[13])*Log[1 + Sqrt[13] + 2*x])/5 - ((1153 + 180*Sqrt[13])*Log[1 + Sqrt[13] + 2*x])/180 + ((1 - 18*
x)^2*Log[-3 + x + x^2])/180

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x \left (29-3 x+8 x^2\right )}{5 \left (-3+x+x^2\right )}+\frac {1}{5} (-1+18 x) \log \left (-3+x+x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int \frac {x \left (29-3 x+8 x^2\right )}{-3+x+x^2} \, dx+\frac {1}{5} \int (-1+18 x) \log \left (-3+x+x^2\right ) \, dx \\ & = \frac {1}{180} (1-18 x)^2 \log \left (-3+x+x^2\right )-\frac {1}{180} \int \frac {(1+2 x) (-1+18 x)^2}{-3+x+x^2} \, dx+\frac {1}{5} \int \left (-11+8 x-\frac {33-64 x}{-3+x+x^2}\right ) \, dx \\ & = -\frac {11 x}{5}+\frac {4 x^2}{5}+\frac {1}{180} (1-18 x)^2 \log \left (-3+x+x^2\right )-\frac {1}{180} \int \left (-396+648 x-\frac {1187-2306 x}{-3+x+x^2}\right ) \, dx-\frac {1}{5} \int \frac {33-64 x}{-3+x+x^2} \, dx \\ & = -x^2+\frac {1}{180} (1-18 x)^2 \log \left (-3+x+x^2\right )+\frac {1}{180} \int \frac {1187-2306 x}{-3+x+x^2} \, dx-\frac {1}{5} \left (-32-5 \sqrt {13}\right ) \int \frac {1}{\frac {1}{2}+\frac {\sqrt {13}}{2}+x} \, dx-\frac {1}{5} \left (-32+5 \sqrt {13}\right ) \int \frac {1}{\frac {1}{2}-\frac {\sqrt {13}}{2}+x} \, dx \\ & = -x^2+\frac {1}{5} \left (32-5 \sqrt {13}\right ) \log \left (1-\sqrt {13}+2 x\right )+\frac {1}{5} \left (32+5 \sqrt {13}\right ) \log \left (1+\sqrt {13}+2 x\right )+\frac {1}{180} (1-18 x)^2 \log \left (-3+x+x^2\right )+\frac {1}{180} \left (-1153-180 \sqrt {13}\right ) \int \frac {1}{\frac {1}{2}+\frac {\sqrt {13}}{2}+x} \, dx+\frac {1}{180} \left (-1153+180 \sqrt {13}\right ) \int \frac {1}{\frac {1}{2}-\frac {\sqrt {13}}{2}+x} \, dx \\ & = -x^2-\frac {1}{180} \left (1153-180 \sqrt {13}\right ) \log \left (1-\sqrt {13}+2 x\right )+\frac {1}{5} \left (32-5 \sqrt {13}\right ) \log \left (1-\sqrt {13}+2 x\right )+\frac {1}{5} \left (32+5 \sqrt {13}\right ) \log \left (1+\sqrt {13}+2 x\right )-\frac {1}{180} \left (1153+180 \sqrt {13}\right ) \log \left (1+\sqrt {13}+2 x\right )+\frac {1}{180} (1-18 x)^2 \log \left (-3+x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=\frac {1}{5} \left (-5 x^2-x \log \left (-3+x+x^2\right )+9 x^2 \log \left (-3+x+x^2\right )\right ) \]

[In]

Integrate[(29*x - 3*x^2 + 8*x^3 + (3 - 55*x + 17*x^2 + 18*x^3)*Log[-3 + x + x^2])/(-15 + 5*x + 5*x^2),x]

[Out]

(-5*x^2 - x*Log[-3 + x + x^2] + 9*x^2*Log[-3 + x + x^2])/5

Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(\left (\frac {9}{5} x^{2}-\frac {1}{5} x \right ) \ln \left (x^{2}+x -3\right )-x^{2}\)	\(24\)
default	\(\frac {9 \ln \left (x^{2}+x -3\right ) x^{2}}{5}-x^{2}-\frac {\ln \left (x^{2}+x -3\right ) x}{5}\)	\(29\)
norman	\(\frac {9 \ln \left (x^{2}+x -3\right ) x^{2}}{5}-x^{2}-\frac {\ln \left (x^{2}+x -3\right ) x}{5}\)	\(29\)
parts	\(\frac {9 \ln \left (x^{2}+x -3\right ) x^{2}}{5}-x^{2}-\frac {\ln \left (x^{2}+x -3\right ) x}{5}\)	\(29\)
parallelrisch	\(\frac {9 \ln \left (x^{2}+x -3\right ) x^{2}}{5}-x^{2}-\frac {\ln \left (x^{2}+x -3\right ) x}{5}-5\)	\(30\)

[In]

int(((18*x^3+17*x^2-55*x+3)*ln(x^2+x-3)+8*x^3-3*x^2+29*x)/(5*x^2+5*x-15),x,method=_RETURNVERBOSE)

[Out]

(9/5*x^2-1/5*x)*ln(x^2+x-3)-x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-x^{2} + \frac {1}{5} \, {\left (9 \, x^{2} - x\right )} \log \left (x^{2} + x - 3\right ) \]

[In]

integrate(((18*x^3+17*x^2-55*x+3)*log(x^2+x-3)+8*x^3-3*x^2+29*x)/(5*x^2+5*x-15),x, algorithm="fricas")

[Out]

-x^2 + 1/5*(9*x^2 - x)*log(x^2 + x - 3)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=- x^{2} + \left (\frac {9 x^{2}}{5} - \frac {x}{5}\right ) \log {\left (x^{2} + x - 3 \right )} \]

[In]

integrate(((18*x**3+17*x**2-55*x+3)*ln(x**2+x-3)+8*x**3-3*x**2+29*x)/(5*x**2+5*x-15),x)

[Out]

-x**2 + (9*x**2/5 - x/5)*log(x**2 + x - 3)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-x^{2} + \frac {1}{5} \, {\left (9 \, x^{2} - x - 32\right )} \log \left (x^{2} + x - 3\right ) + \frac {32}{5} \, \log \left (x^{2} + x - 3\right ) \]

[In]

integrate(((18*x^3+17*x^2-55*x+3)*log(x^2+x-3)+8*x^3-3*x^2+29*x)/(5*x^2+5*x-15),x, algorithm="maxima")

[Out]

-x^2 + 1/5*(9*x^2 - x - 32)*log(x^2 + x - 3) + 32/5*log(x^2 + x - 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-x^{2} + \frac {1}{5} \, {\left (9 \, x^{2} - x\right )} \log \left (x^{2} + x - 3\right ) \]

[In]

integrate(((18*x^3+17*x^2-55*x+3)*log(x^2+x-3)+8*x^3-3*x^2+29*x)/(5*x^2+5*x-15),x, algorithm="giac")

[Out]

-x^2 + 1/5*(9*x^2 - x)*log(x^2 + x - 3)

Mupad [B] (verification not implemented)

Time = 8.84 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {29 x-3 x^2+8 x^3+\left (3-55 x+17 x^2+18 x^3\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=x^2\,\left (\frac {9\,\ln \left (x^2+x-3\right )}{5}-1\right )-\frac {x\,\ln \left (x^2+x-3\right )}{5} \]

[In]

int((29*x + log(x + x^2 - 3)*(17*x^2 - 55*x + 18*x^3 + 3) - 3*x^2 + 8*x^3)/(5*x + 5*x^2 - 15),x)

[Out]

x^2*((9*log(x + x^2 - 3))/5 - 1) - (x*log(x + x^2 - 3))/5