\(\int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx\) [3992]

Optimal result

Integrand size = 43, antiderivative size = 19 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=1+\frac {-4-x+\log ^2(5+x)}{-3+x} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.84, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6874, 2458, 2379, 2438, 2444, 2441, 2440} \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {8}{x+5}\right )-\frac {1}{4} \operatorname {PolyLog}\left (2,\frac {x+5}{8}\right )+\frac {7}{3-x}-\frac {(x+5) \log ^2(x+5)}{8 (3-x)}-\frac {1}{4} \log \left (\frac {3-x}{8}\right ) \log (x+5)+\frac {1}{4} \log \left (1-\frac {8}{x+5}\right ) \log (x+5) \]

[In]

Int[(35 + 7*x + (-6 + 2*x)*Log[5 + x] + (-5 - x)*Log[5 + x]^2)/(45 - 21*x - x^2 + x^3),x]

[Out]

7/(3 - x) - (Log[(3 - x)/8]*Log[5 + x])/4 - ((5 + x)*Log[5 + x]^2)/(8*(3 - x)) + (Log[5 + x]*Log[1 - 8/(5 + x)
])/4 - PolyLog[2, 8/(5 + x)]/4 - PolyLog[2, (5 + x)/8]/4

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 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 2440

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

Rule 2441

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

Rule 2444

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[(d + e
*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f - d*g)*(f + g*x))), x] - Dist[b*e*n*(p/(e*f - d*g)), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {7}{(-3+x)^2}+\frac {2 \log (5+x)}{(-3+x) (5+x)}-\frac {\log ^2(5+x)}{(-3+x)^2}\right ) \, dx \\ & = \frac {7}{3-x}+2 \int \frac {\log (5+x)}{(-3+x) (5+x)} \, dx-\int \frac {\log ^2(5+x)}{(-3+x)^2} \, dx \\ & = \frac {7}{3-x}-\frac {(5+x) \log ^2(5+x)}{8 (3-x)}-\frac {1}{4} \int \frac {\log (5+x)}{-3+x} \, dx+2 \text {Subst}\left (\int \frac {\log (x)}{(-8+x) x} \, dx,x,5+x\right ) \\ & = \frac {7}{3-x}-\frac {1}{4} \log \left (\frac {3-x}{8}\right ) \log (5+x)-\frac {(5+x) \log ^2(5+x)}{8 (3-x)}+\frac {1}{4} \log (5+x) \log \left (1-\frac {8}{5+x}\right )+\frac {1}{4} \int \frac {\log \left (\frac {3-x}{8}\right )}{5+x} \, dx-\frac {1}{4} \text {Subst}\left (\int \frac {\log \left (1-\frac {8}{x}\right )}{x} \, dx,x,5+x\right ) \\ & = \frac {7}{3-x}-\frac {1}{4} \log \left (\frac {3-x}{8}\right ) \log (5+x)-\frac {(5+x) \log ^2(5+x)}{8 (3-x)}+\frac {1}{4} \log (5+x) \log \left (1-\frac {8}{5+x}\right )-\frac {1}{4} \text {Li}_2\left (\frac {8}{5+x}\right )+\frac {1}{4} \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,5+x\right ) \\ & = \frac {7}{3-x}-\frac {1}{4} \log \left (\frac {3-x}{8}\right ) \log (5+x)-\frac {(5+x) \log ^2(5+x)}{8 (3-x)}+\frac {1}{4} \log (5+x) \log \left (1-\frac {8}{5+x}\right )-\frac {1}{4} \text {Li}_2\left (\frac {8}{5+x}\right )-\frac {1}{4} \text {Li}_2\left (\frac {5+x}{8}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {-7+\log ^2(5+x)}{-3+x} \]

[In]

Integrate[(35 + 7*x + (-6 + 2*x)*Log[5 + x] + (-5 - x)*Log[5 + x]^2)/(45 - 21*x - x^2 + x^3),x]

[Out]

(-7 + Log[5 + x]^2)/(-3 + x)

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
norman	\(\frac {\ln \left (5+x \right )^{2}-7}{-3+x}\)	\(15\)
parallelrisch	\(\frac {\ln \left (5+x \right )^{2}-7}{-3+x}\)	\(15\)
risch	\(\frac {\ln \left (5+x \right )^{2}}{-3+x}-\frac {7}{-3+x}\)	\(21\)

[In]

int(((-x-5)*ln(5+x)^2+(2*x-6)*ln(5+x)+7*x+35)/(x^3-x^2-21*x+45),x,method=_RETURNVERBOSE)

[Out]

(ln(5+x)^2-7)/(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {\log \left (x + 5\right )^{2} - 7}{x - 3} \]

[In]

integrate(((-x-5)*log(5+x)^2+(2*x-6)*log(5+x)+7*x+35)/(x^3-x^2-21*x+45),x, algorithm="fricas")

[Out]

(log(x + 5)^2 - 7)/(x - 3)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {\log {\left (x + 5 \right )}^{2}}{x - 3} - \frac {7}{x - 3} \]

[In]

integrate(((-x-5)*ln(5+x)**2+(2*x-6)*ln(5+x)+7*x+35)/(x**3-x**2-21*x+45),x)

[Out]

log(x + 5)**2/(x - 3) - 7/(x - 3)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {\log \left (x + 5\right )^{2}}{x - 3} - \frac {7}{x - 3} \]

[In]

integrate(((-x-5)*log(5+x)^2+(2*x-6)*log(5+x)+7*x+35)/(x^3-x^2-21*x+45),x, algorithm="maxima")

[Out]

log(x + 5)^2/(x - 3) - 7/(x - 3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {\log \left (x + 5\right )^{2}}{x - 3} - \frac {7}{x - 3} \]

[In]

integrate(((-x-5)*log(5+x)^2+(2*x-6)*log(5+x)+7*x+35)/(x^3-x^2-21*x+45),x, algorithm="giac")

[Out]

log(x + 5)^2/(x - 3) - 7/(x - 3)

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {35+7 x+(-6+2 x) \log (5+x)+(-5-x) \log ^2(5+x)}{45-21 x-x^2+x^3} \, dx=\frac {{\ln \left (x+5\right )}^2-7}{x-3} \]

[In]

int(-(7*x + log(x + 5)*(2*x - 6) - log(x + 5)^2*(x + 5) + 35)/(21*x + x^2 - x^3 - 45),x)

[Out]

(log(x + 5)^2 - 7)/(x - 3)