\(\int \frac {4+37 x+32 x^2+(-1-17 x-16 x^2) \log (1+x)+(2 x+2 x^2) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx\) [6825]

Optimal result

Integrand size = 69, antiderivative size = 16 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=18+x^2-\frac {x}{-4+\log (1+x)} \]

[Out]

18+x^2-x/(ln(1+x)-4)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {6873, 6874, 2436, 2336, 2209, 2458, 2395, 2334, 2339, 30} \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=x^2+\frac {x+1}{4-\log (x+1)}+\frac {1}{\log (x+1)-4} \]

[In]

Int[(4 + 37*x + 32*x^2 + (-1 - 17*x - 16*x^2)*Log[1 + x] + (2*x + 2*x^2)*Log[1 + x]^2)/(16 + 16*x + (-8 - 8*x)
*Log[1 + x] + (1 + x)*Log[1 + x]^2),x]

[Out]

x^2 + (1 + x)/(4 - Log[1 + x]) + (-4 + Log[1 + x])^(-1)

Rule 30

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2436

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 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 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 {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{(1+x) (4-\log (1+x))^2} \, dx \\ & = \int \left (2 x+\frac {1}{4-\log (1+x)}+\frac {x}{(1+x) (-4+\log (1+x))^2}\right ) \, dx \\ & = x^2+\int \frac {1}{4-\log (1+x)} \, dx+\int \frac {x}{(1+x) (-4+\log (1+x))^2} \, dx \\ & = x^2+\text {Subst}\left (\int \frac {1}{4-\log (x)} \, dx,x,1+x\right )+\text {Subst}\left (\int \frac {-1+x}{x (-4+\log (x))^2} \, dx,x,1+x\right ) \\ & = x^2+\text {Subst}\left (\int \frac {e^x}{4-x} \, dx,x,\log (1+x)\right )+\text {Subst}\left (\int \left (\frac {1}{(-4+\log (x))^2}-\frac {1}{x (-4+\log (x))^2}\right ) \, dx,x,1+x\right ) \\ & = x^2-e^4 \text {Ei}(-4+\log (1+x))+\text {Subst}\left (\int \frac {1}{(-4+\log (x))^2} \, dx,x,1+x\right )-\text {Subst}\left (\int \frac {1}{x (-4+\log (x))^2} \, dx,x,1+x\right ) \\ & = x^2-e^4 \text {Ei}(-4+\log (1+x))+\frac {1+x}{4-\log (1+x)}-\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,-4+\log (1+x)\right )+\text {Subst}\left (\int \frac {1}{-4+\log (x)} \, dx,x,1+x\right ) \\ & = x^2-e^4 \text {Ei}(-4+\log (1+x))+\frac {1+x}{4-\log (1+x)}+\frac {1}{-4+\log (1+x)}+\text {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (1+x)\right ) \\ & = x^2+\frac {1+x}{4-\log (1+x)}+\frac {1}{-4+\log (1+x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=-1+x^2-\frac {x}{-4+\log (1+x)} \]

[In]

Integrate[(4 + 37*x + 32*x^2 + (-1 - 17*x - 16*x^2)*Log[1 + x] + (2*x + 2*x^2)*Log[1 + x]^2)/(16 + 16*x + (-8
- 8*x)*Log[1 + x] + (1 + x)*Log[1 + x]^2),x]

[Out]

-1 + x^2 - x/(-4 + Log[1 + x])

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
risch	\(x^{2}-\frac {x}{\ln \left (1+x \right )-4}\)	\(16\)
norman	\(\frac {x^{2} \ln \left (1+x \right )-x -4 x^{2}}{\ln \left (1+x \right )-4}\)	\(27\)
derivativedivides	\(\frac {1}{\ln \left (1+x \right )-4}-\frac {1+x}{\ln \left (1+x \right )-4}+\left (1+x \right )^{2}-2-2 x\)	\(32\)
default	\(\frac {1}{\ln \left (1+x \right )-4}-\frac {1+x}{\ln \left (1+x \right )-4}+\left (1+x \right )^{2}-2-2 x\)	\(32\)
parallelrisch	\(\frac {x^{2} \ln \left (1+x \right )+4-4 x^{2}-x -\ln \left (1+x \right )}{\ln \left (1+x \right )-4}\)	\(34\)

[In]

int(((2*x^2+2*x)*ln(1+x)^2+(-16*x^2-17*x-1)*ln(1+x)+32*x^2+37*x+4)/((1+x)*ln(1+x)^2+(-8*x-8)*ln(1+x)+16*x+16),
x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=\frac {x^{2} \log \left (x + 1\right ) - 4 \, x^{2} - x}{\log \left (x + 1\right ) - 4} \]

[In]

integrate(((2*x^2+2*x)*log(1+x)^2+(-16*x^2-17*x-1)*log(1+x)+32*x^2+37*x+4)/((1+x)*log(1+x)^2+(-8*x-8)*log(1+x)
+16*x+16),x, algorithm="fricas")

[Out]

(x^2*log(x + 1) - 4*x^2 - x)/(log(x + 1) - 4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=x^{2} - \frac {x}{\log {\left (x + 1 \right )} - 4} \]

[In]

integrate(((2*x**2+2*x)*ln(1+x)**2+(-16*x**2-17*x-1)*ln(1+x)+32*x**2+37*x+4)/((1+x)*ln(1+x)**2+(-8*x-8)*ln(1+x
)+16*x+16),x)

[Out]

x**2 - x/(log(x + 1) - 4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.38 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=\frac {x^{2} \log \left (x + 1\right ) - 4 \, x^{2} - x + 4}{\log \left (x + 1\right ) - 4} - \frac {4}{\log \left (x + 1\right ) - 4} \]

[In]

integrate(((2*x^2+2*x)*log(1+x)^2+(-16*x^2-17*x-1)*log(1+x)+32*x^2+37*x+4)/((1+x)*log(1+x)^2+(-8*x-8)*log(1+x)
+16*x+16),x, algorithm="maxima")

[Out]

(x^2*log(x + 1) - 4*x^2 - x + 4)/(log(x + 1) - 4) - 4/(log(x + 1) - 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=x^{2} - \frac {x}{\log \left (x + 1\right ) - 4} \]

[In]

integrate(((2*x^2+2*x)*log(1+x)^2+(-16*x^2-17*x-1)*log(1+x)+32*x^2+37*x+4)/((1+x)*log(1+x)^2+(-8*x-8)*log(1+x)
+16*x+16),x, algorithm="giac")

[Out]

x^2 - x/(log(x + 1) - 4)

Mupad [B] (verification not implemented)

Time = 11.49 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {4+37 x+32 x^2+\left (-1-17 x-16 x^2\right ) \log (1+x)+\left (2 x+2 x^2\right ) \log ^2(1+x)}{16+16 x+(-8-8 x) \log (1+x)+(1+x) \log ^2(1+x)} \, dx=x^2-\frac {x}{\ln \left (x+1\right )-4} \]

[In]

int((37*x - log(x + 1)*(17*x + 16*x^2 + 1) + log(x + 1)^2*(2*x + 2*x^2) + 32*x^2 + 4)/(16*x - log(x + 1)*(8*x
+ 8) + log(x + 1)^2*(x + 1) + 16),x)

[Out]

x^2 - x/(log(x + 1) - 4)