\(\int \frac {(20+6 x) \log (5 x^2+x^3)+(-15-3 x) \log ^2(5 x^2+x^3)}{5 x^4+x^5} \, dx\) [7937]

Optimal result

Integrand size = 47, antiderivative size = 14 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log ^2\left (x^2 (5+x)\right )}{x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 54, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.404, Rules used = {1607, 6873, 6820, 6874, 78, 2581, 30, 46, 2594, 36, 29, 31, 2580, 2338, 2354, 2438, 2439, 2437, 2584} \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log ^2\left (x^2 (x+5)\right )}{x^3} \]

[In]

Int[((20 + 6*x)*Log[5*x^2 + x^3] + (-15 - 3*x)*Log[5*x^2 + x^3]^2)/(5*x^4 + x^5),x]

[Out]

Log[x^2*(5 + x)]^2/x^3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 78

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 1607

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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + e*(x/d)]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2580

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[Log[g + h*x]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/h), x] + (-Dist[b*p*(r/h), Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[d*q*(r/h), Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 2581

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(h*(m + 1))), x] + (-Dist[b*p*(r/(h
*(m + 1))), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[d*q*(r/(h*(m + 1))), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 2584

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_)*((g_.) + (h_.)*(x_))^(
m_.), x_Symbol] :> Simp[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s/(h*(m + 1))), x] + (-Dist[b*
p*r*(s/(h*(m + 1))), Int[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/(a + b*x)), x], x] -
Dist[d*q*r*(s/(h*(m + 1))), Int[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/(c + d*x)), x]
, x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && NeQ[m, -1]

Rule 2594

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(RFx_), x_Symbol] :>
With[{u = ExpandIntegrand[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, p, q, r, s}, x] && RationalFunctionQ[RFx, x] && IGtQ[s, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{x^4 (5+x)} \, dx \\ & = \int \frac {\log \left (x^2 (5+x)\right ) \left (20+6 x-15 \log \left (x^2 (5+x)\right )-3 x \log \left (x^2 (5+x)\right )\right )}{x^4 (5+x)} \, dx \\ & = \int \frac {\log \left (x^2 (5+x)\right ) \left (20+6 x-3 (5+x) \log \left (x^2 (5+x)\right )\right )}{x^4 (5+x)} \, dx \\ & = \int \left (\frac {2 (10+3 x) \log \left (x^2 (5+x)\right )}{x^4 (5+x)}-\frac {3 \log ^2\left (x^2 (5+x)\right )}{x^4}\right ) \, dx \\ & = 2 \int \frac {(10+3 x) \log \left (x^2 (5+x)\right )}{x^4 (5+x)} \, dx-3 \int \frac {\log ^2\left (x^2 (5+x)\right )}{x^4} \, dx \\ & = \frac {\log ^2\left (x^2 (5+x)\right )}{x^3}-2 \int \frac {\log \left (x^2 (5+x)\right )}{x^3 (5+x)} \, dx+2 \int \left (\frac {2 \log \left (x^2 (5+x)\right )}{x^4}+\frac {\log \left (x^2 (5+x)\right )}{5 x^3}-\frac {\log \left (x^2 (5+x)\right )}{25 x^2}+\frac {\log \left (x^2 (5+x)\right )}{125 x}-\frac {\log \left (x^2 (5+x)\right )}{125 (5+x)}\right ) \, dx-4 \int \frac {\log \left (x^2 (5+x)\right )}{x^4} \, dx \\ & = \frac {4 \log \left (x^2 (5+x)\right )}{3 x^3}+\frac {\log ^2\left (x^2 (5+x)\right )}{x^3}+\frac {2}{125} \int \frac {\log \left (x^2 (5+x)\right )}{x} \, dx-\frac {2}{125} \int \frac {\log \left (x^2 (5+x)\right )}{5+x} \, dx-\frac {2}{25} \int \frac {\log \left (x^2 (5+x)\right )}{x^2} \, dx+\frac {2}{5} \int \frac {\log \left (x^2 (5+x)\right )}{x^3} \, dx-\frac {4}{3} \int \frac {1}{x^3 (5+x)} \, dx-2 \int \left (\frac {\log \left (x^2 (5+x)\right )}{5 x^3}-\frac {\log \left (x^2 (5+x)\right )}{25 x^2}+\frac {\log \left (x^2 (5+x)\right )}{125 x}-\frac {\log \left (x^2 (5+x)\right )}{125 (5+x)}\right ) \, dx-\frac {8}{3} \int \frac {1}{x^4} \, dx+4 \int \frac {\log \left (x^2 (5+x)\right )}{x^4} \, dx \\ & = \frac {8}{9 x^3}-\frac {\log \left (x^2 (5+x)\right )}{5 x^2}+\frac {2 \log \left (x^2 (5+x)\right )}{25 x}+\frac {2}{125} \log (x) \log \left (x^2 (5+x)\right )-\frac {2}{125} \log (5+x) \log \left (x^2 (5+x)\right )+\frac {\log ^2\left (x^2 (5+x)\right )}{x^3}-\frac {2}{125} \int \frac {\log (x)}{5+x} \, dx+\frac {2}{125} \int \frac {\log (5+x)}{5+x} \, dx-\frac {2}{125} \int \frac {\log \left (x^2 (5+x)\right )}{x} \, dx+\frac {2}{125} \int \frac {\log \left (x^2 (5+x)\right )}{5+x} \, dx-\frac {4}{125} \int \frac {\log (x)}{x} \, dx+\frac {4}{125} \int \frac {\log (5+x)}{x} \, dx-\frac {2}{25} \int \frac {1}{x (5+x)} \, dx+\frac {2}{25} \int \frac {\log \left (x^2 (5+x)\right )}{x^2} \, dx-\frac {4}{25} \int \frac {1}{x^2} \, dx+\frac {1}{5} \int \frac {1}{x^2 (5+x)} \, dx+\frac {2}{5} \int \frac {1}{x^3} \, dx-\frac {2}{5} \int \frac {\log \left (x^2 (5+x)\right )}{x^3} \, dx+\frac {4}{3} \int \frac {1}{x^3 (5+x)} \, dx-\frac {4}{3} \int \left (\frac {1}{5 x^3}-\frac {1}{25 x^2}+\frac {1}{125 x}-\frac {1}{125 (5+x)}\right ) \, dx+\frac {8}{3} \int \frac {1}{x^4} \, dx \\ & = -\frac {1}{15 x^2}+\frac {8}{75 x}-\frac {4 \log (x)}{375}+\frac {4}{125} \log (5) \log (x)-\frac {2}{125} \log \left (1+\frac {x}{5}\right ) \log (x)-\frac {2 \log ^2(x)}{125}+\frac {4}{375} \log (5+x)+\frac {\log ^2\left (x^2 (5+x)\right )}{x^3}-\frac {2}{125} \int \frac {1}{x} \, dx+\frac {2}{125} \int \frac {1}{5+x} \, dx+\frac {2}{125} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx+\frac {2}{125} \int \frac {\log (x)}{5+x} \, dx-\frac {2}{125} \int \frac {\log (5+x)}{5+x} \, dx+\frac {2}{125} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )+\frac {4}{125} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx+\frac {4}{125} \int \frac {\log (x)}{x} \, dx-\frac {4}{125} \int \frac {\log (5+x)}{x} \, dx+\frac {2}{25} \int \frac {1}{x (5+x)} \, dx+\frac {4}{25} \int \frac {1}{x^2} \, dx-\frac {1}{5} \int \frac {1}{x^2 (5+x)} \, dx+\frac {1}{5} \int \left (\frac {1}{5 x^2}-\frac {1}{25 x}+\frac {1}{25 (5+x)}\right ) \, dx-\frac {2}{5} \int \frac {1}{x^3} \, dx+\frac {4}{3} \int \left (\frac {1}{5 x^3}-\frac {1}{25 x^2}+\frac {1}{125 x}-\frac {1}{125 (5+x)}\right ) \, dx \\ & = -\frac {1}{25 x}-\frac {3 \log (x)}{125}+\frac {3}{125} \log (5+x)+\frac {1}{125} \log ^2(5+x)+\frac {\log ^2\left (x^2 (5+x)\right )}{x^3}-\frac {6 \operatorname {PolyLog}\left (2,-\frac {x}{5}\right )}{125}+\frac {2}{125} \int \frac {1}{x} \, dx-\frac {2}{125} \int \frac {1}{5+x} \, dx-\frac {2}{125} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx-\frac {2}{125} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )-\frac {4}{125} \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx-\frac {1}{5} \int \left (\frac {1}{5 x^2}-\frac {1}{25 x}+\frac {1}{25 (5+x)}\right ) \, dx \\ & = \frac {\log ^2\left (x^2 (5+x)\right )}{x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log ^2\left (x^2 (5+x)\right )}{x^3} \]

[In]

Integrate[((20 + 6*x)*Log[5*x^2 + x^3] + (-15 - 3*x)*Log[5*x^2 + x^3]^2)/(5*x^4 + x^5),x]

[Out]

Log[x^2*(5 + x)]^2/x^3

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\ln \left (x^{2} \left (5+x \right )\right )^{2}}{x^{3}}\)	\(15\)
norman	\(\frac {\ln \left (x^{3}+5 x^{2}\right )^{2}}{x^{3}}\)	\(17\)
risch	\(\frac {\ln \left (x^{3}+5 x^{2}\right )^{2}}{x^{3}}\)	\(17\)

[In]

int(((-3*x-15)*ln(x^3+5*x^2)^2+(6*x+20)*ln(x^3+5*x^2))/(x^5+5*x^4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log \left (x^{3} + 5 \, x^{2}\right )^{2}}{x^{3}} \]

[In]

integrate(((-3*x-15)*log(x^3+5*x^2)^2+(6*x+20)*log(x^3+5*x^2))/(x^5+5*x^4),x, algorithm="fricas")

[Out]

log(x^3 + 5*x^2)^2/x^3

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log {\left (x^{3} + 5 x^{2} \right )}^{2}}{x^{3}} \]

[In]

integrate(((-3*x-15)*ln(x**3+5*x**2)**2+(6*x+20)*ln(x**3+5*x**2))/(x**5+5*x**4),x)

[Out]

log(x**3 + 5*x**2)**2/x**3

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log \left (x + 5\right )^{2} + 4 \, \log \left (x + 5\right ) \log \left (x\right ) + 4 \, \log \left (x\right )^{2}}{x^{3}} \]

[In]

integrate(((-3*x-15)*log(x^3+5*x^2)^2+(6*x+20)*log(x^3+5*x^2))/(x^5+5*x^4),x, algorithm="maxima")

[Out]

(log(x + 5)^2 + 4*log(x + 5)*log(x) + 4*log(x)^2)/x^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {\log \left (x^{3} + 5 \, x^{2}\right )^{2}}{x^{3}} \]

[In]

integrate(((-3*x-15)*log(x^3+5*x^2)^2+(6*x+20)*log(x^3+5*x^2))/(x^5+5*x^4),x, algorithm="giac")

[Out]

log(x^3 + 5*x^2)^2/x^3

Mupad [B] (verification not implemented)

Time = 13.90 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(20+6 x) \log \left (5 x^2+x^3\right )+(-15-3 x) \log ^2\left (5 x^2+x^3\right )}{5 x^4+x^5} \, dx=\frac {{\ln \left (x^2\,\left (x+5\right )\right )}^2}{x^3} \]

[In]

int((log(5*x^2 + x^3)*(6*x + 20) - log(5*x^2 + x^3)^2*(3*x + 15))/(5*x^4 + x^5),x)

[Out]

log(x^2*(x + 5))^2/x^3