\(\int \frac {48-120 x+4 x^2+(15-100 x-5 x^2) \log (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4})}{-15 x^2+100 x^3+5 x^4} \, dx\) [10183]

Optimal result

Integrand size = 63, antiderivative size = 34 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=e^4-\frac {4+x}{5 x}+\frac {\log \left (\left (-\frac {3}{x^2}+\frac {20+x}{x}\right )^2\right )}{x} \]

[Out]

exp(4)-1/5*(4+x)/x+ln(((20+x)/x-3/x^2)^2)/x

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1608, 6820, 12, 14, 1642, 646, 31, 2605, 814} \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log \left (\frac {\left (-x^2-20 x+3\right )^2}{x^4}\right )}{x}-\frac {4}{5 x} \]

[In]

Int[(48 - 120*x + 4*x^2 + (15 - 100*x - 5*x^2)*Log[(9 - 120*x + 394*x^2 + 40*x^3 + x^4)/x^4])/(-15*x^2 + 100*x
^3 + 5*x^4),x]

[Out]

-4/(5*x) + Log[(3 - 20*x - x^2)^2/x^4]/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{x^2 \left (-15+100 x+5 x^2\right )} \, dx \\ & = \int \frac {\frac {4 \left (12-30 x+x^2\right )}{-3+20 x+x^2}-5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{5 x^2} \, dx \\ & = \frac {1}{5} \int \frac {\frac {4 \left (12-30 x+x^2\right )}{-3+20 x+x^2}-5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {4 \left (12-30 x+x^2\right )}{x^2 \left (-3+20 x+x^2\right )}-\frac {5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2}\right ) \, dx \\ & = \frac {4}{5} \int \frac {12-30 x+x^2}{x^2 \left (-3+20 x+x^2\right )} \, dx-\int \frac {\log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2} \, dx \\ & = \frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}+\frac {4}{5} \int \left (-\frac {4}{x^2}-\frac {50}{3 x}+\frac {5 (203+10 x)}{3 \left (-3+20 x+x^2\right )}\right ) \, dx-\int \frac {-12+40 x}{x^2 \left (3-20 x-x^2\right )} \, dx \\ & = \frac {16}{5 x}-\frac {40 \log (x)}{3}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}+\frac {4}{3} \int \frac {203+10 x}{-3+20 x+x^2} \, dx-\int \left (-\frac {4}{x^2}-\frac {40}{3 x}+\frac {4 (203+10 x)}{3 \left (-3+20 x+x^2\right )}\right ) \, dx \\ & = -\frac {4}{5 x}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}-\frac {4}{3} \int \frac {203+10 x}{-3+20 x+x^2} \, dx+\frac {1}{3} \left (2 \left (10-\sqrt {103}\right )\right ) \int \frac {1}{10+\sqrt {103}+x} \, dx+\frac {1}{3} \left (2 \left (10+\sqrt {103}\right )\right ) \int \frac {1}{10-\sqrt {103}+x} \, dx \\ & = -\frac {4}{5 x}+\frac {2}{3} \left (10+\sqrt {103}\right ) \log \left (10-\sqrt {103}+x\right )+\frac {2}{3} \left (10-\sqrt {103}\right ) \log \left (10+\sqrt {103}+x\right )+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}-\frac {1}{3} \left (2 \left (10-\sqrt {103}\right )\right ) \int \frac {1}{10+\sqrt {103}+x} \, dx-\frac {1}{3} \left (2 \left (10+\sqrt {103}\right )\right ) \int \frac {1}{10-\sqrt {103}+x} \, dx \\ & = -\frac {4}{5 x}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {1}{5} \left (-\frac {4}{x}+\frac {5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x}\right ) \]

[In]

Integrate[(48 - 120*x + 4*x^2 + (15 - 100*x - 5*x^2)*Log[(9 - 120*x + 394*x^2 + 40*x^3 + x^4)/x^4])/(-15*x^2 +
100*x^3 + 5*x^4),x]

[Out]

(-4/x + (5*Log[(-3 + 20*x + x^2)^2/x^4])/x)/5

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

method	result	size
norman	\(\frac {-\frac {4}{5}+\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}\)	\(30\)
parallelrisch	\(-\frac {8-10 \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{10 x}\)	\(33\)
derivativedivides	\(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}\)	\(34\)
default	\(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}\)	\(34\)
risch	\(\frac {\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}-\frac {4}{5 x}\)	\(34\)
parts	\(\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}-\frac {4}{5 x}-\frac {20 \ln \left (\frac {3}{x^{2}}-\frac {20}{x}-1\right )}{3}+\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (\frac {6}{x}-20\right ) \sqrt {103}}{206}\right )}{3}-\frac {40 \ln \left (x \right )}{3}+\frac {20 \ln \left (x^{2}+20 x -3\right )}{3}-\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (2 x +20\right ) \sqrt {103}}{206}\right )}{3}\)	\(98\)

[In]

int(((-5*x^2-100*x+15)*ln((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+48)/(5*x^4+100*x^3-15*x^2),x,method=_R
ETURNVERBOSE)

[Out]

(-4/5+ln((x^4+40*x^3+394*x^2-120*x+9)/x^4))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {5 \, \log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right ) - 4}{5 \, x} \]

[In]

integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+48)/(5*x^4+100*x^3-15*x^2),x, a
lgorithm="fricas")

[Out]

1/5*(5*log((x^4 + 40*x^3 + 394*x^2 - 120*x + 9)/x^4) - 4)/x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log {\left (\frac {x^{4} + 40 x^{3} + 394 x^{2} - 120 x + 9}{x^{4}} \right )}}{x} - \frac {4}{5 x} \]

[In]

integrate(((-5*x**2-100*x+15)*ln((x**4+40*x**3+394*x**2-120*x+9)/x**4)+4*x**2-120*x+48)/(5*x**4+100*x**3-15*x*
*2),x)

[Out]

log((x**4 + 40*x**3 + 394*x**2 - 120*x + 9)/x**4)/x - 4/(5*x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=-\frac {2 \, {\left ({\left (16 \, x - 3\right )} \log \left (x^{2} + 20 \, x - 3\right ) - 2 \, {\left (16 \, x - 3\right )} \log \left (x\right ) + 6\right )}}{3 \, x} + \frac {16}{5 \, x} + \frac {32}{3} \, \log \left (x^{2} + 20 \, x - 3\right ) - \frac {64}{3} \, \log \left (x\right ) \]

[In]

integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+48)/(5*x^4+100*x^3-15*x^2),x, a
lgorithm="maxima")

[Out]

-2/3*((16*x - 3)*log(x^2 + 20*x - 3) - 2*(16*x - 3)*log(x) + 6)/x + 16/5/x + 32/3*log(x^2 + 20*x - 3) - 64/3*l
og(x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right )}{x} - \frac {4}{5 \, x} \]

[In]

integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+48)/(5*x^4+100*x^3-15*x^2),x, a
lgorithm="giac")

[Out]

log((x^4 + 40*x^3 + 394*x^2 - 120*x + 9)/x^4)/x - 4/5/x

Mupad [B] (verification not implemented)

Time = 16.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\ln \left (\frac {x^4+40\,x^3+394\,x^2-120\,x+9}{x^4}\right )-\frac {4}{5}}{x} \]

[In]

int(-(120*x - 4*x^2 + log((394*x^2 - 120*x + 40*x^3 + x^4 + 9)/x^4)*(100*x + 5*x^2 - 15) - 48)/(100*x^3 - 15*x
^2 + 5*x^4),x)

[Out]

(log((394*x^2 - 120*x + 40*x^3 + x^4 + 9)/x^4) - 4/5)/x