3.24.35 \(\int \frac {-x+(10+x) \log (10+x)+(20 x+2 x^2+100 x^4+10 x^5+80 x^7+8 x^8) \log ^2(10+x)}{(10+x) \log ^2(10+x)} \, dx\)

Optimal. Leaf size=17 \[ 1+\left (x+x^4\right )^2+\frac {x}{\log (10+x)} \]

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Rubi [A]  time = 0.20, antiderivative size = 30, normalized size of antiderivative = 1.76, number of steps used = 13, number of rules used = 9, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6688, 14, 2411, 2353, 2297, 2298, 2302, 30, 2389} \begin {gather*} x^8+2 x^5+x^2+\frac {x+10}{\log (x+10)}-\frac {10}{\log (x+10)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + (10 + x)*Log[10 + x] + (20*x + 2*x^2 + 100*x^4 + 10*x^5 + 80*x^7 + 8*x^8)*Log[10 + x]^2)/((10 + x)*L
og[10 + x]^2),x]

[Out]

x^2 + 2*x^5 + x^8 - 10/Log[10 + x] + (10 + x)/Log[10 + 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 30

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

Rule 2297

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 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

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 2353

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 2389

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 2411

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 x \left (1+5 x^3+4 x^6\right )-\frac {x}{(10+x) \log ^2(10+x)}+\frac {1}{\log (10+x)}\right ) \, dx\\ &=2 \int x \left (1+5 x^3+4 x^6\right ) \, dx-\int \frac {x}{(10+x) \log ^2(10+x)} \, dx+\int \frac {1}{\log (10+x)} \, dx\\ &=2 \int \left (x+5 x^4+4 x^7\right ) \, dx-\operatorname {Subst}\left (\int \frac {-10+x}{x \log ^2(x)} \, dx,x,10+x\right )+\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,10+x\right )\\ &=x^2+2 x^5+x^8+\text {li}(10+x)-\operatorname {Subst}\left (\int \left (\frac {1}{\log ^2(x)}-\frac {10}{x \log ^2(x)}\right ) \, dx,x,10+x\right )\\ &=x^2+2 x^5+x^8+\text {li}(10+x)+10 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,10+x\right )-\operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,10+x\right )\\ &=x^2+2 x^5+x^8+\frac {10+x}{\log (10+x)}+\text {li}(10+x)+10 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (10+x)\right )-\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,10+x\right )\\ &=x^2+2 x^5+x^8-\frac {10}{\log (10+x)}+\frac {10+x}{\log (10+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 20, normalized size = 1.18 \begin {gather*} x^2+2 x^5+x^8+\frac {x}{\log (10+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + (10 + x)*Log[10 + x] + (20*x + 2*x^2 + 100*x^4 + 10*x^5 + 80*x^7 + 8*x^8)*Log[10 + x]^2)/((10
+ x)*Log[10 + x]^2),x]

[Out]

x^2 + 2*x^5 + x^8 + x/Log[10 + x]

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fricas [A]  time = 0.49, size = 26, normalized size = 1.53 \begin {gather*} \frac {{\left (x^{8} + 2 \, x^{5} + x^{2}\right )} \log \left (x + 10\right ) + x}{\log \left (x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^8+80*x^7+10*x^5+100*x^4+2*x^2+20*x)*log(x+10)^2+(x+10)*log(x+10)-x)/(x+10)/log(x+10)^2,x, algo
rithm="fricas")

[Out]

((x^8 + 2*x^5 + x^2)*log(x + 10) + x)/log(x + 10)

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giac [A]  time = 0.23, size = 20, normalized size = 1.18 \begin {gather*} x^{8} + 2 \, x^{5} + x^{2} + \frac {x}{\log \left (x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^8+80*x^7+10*x^5+100*x^4+2*x^2+20*x)*log(x+10)^2+(x+10)*log(x+10)-x)/(x+10)/log(x+10)^2,x, algo
rithm="giac")

[Out]

x^8 + 2*x^5 + x^2 + x/log(x + 10)

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maple [A]  time = 0.40, size = 21, normalized size = 1.24

method	result	size
risch	\(x^{8}+2 x^{5}+x^{2}+\frac {x}{\ln \left (x +10\right )}\)	\(21\)
derivativedivides	\(\left (x +10\right )^{8}-80 \left (x +10\right )^{7}+2800 \left (x +10\right )^{6}-55998 \left (x +10\right )^{5}+699900 \left (x +10\right )^{4}-5598000 \left (x +10\right )^{3}[phone] \left (x +10\right )^{2}-79900020 x -799000200+\frac {x +10}{\ln \left (x +10\right )}-\frac {10}{\ln \left (x +10\right )}\)	\(71\)
default	\(\left (x +10\right )^{8}-80 \left (x +10\right )^{7}+2800 \left (x +10\right )^{6}-55998 \left (x +10\right )^{5}+699900 \left (x +10\right )^{4}-5598000 \left (x +10\right )^{3}[phone] \left (x +10\right )^{2}-79900020 x -799000200+\frac {x +10}{\ln \left (x +10\right )}-\frac {10}{\ln \left (x +10\right )}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^8+80*x^7+10*x^5+100*x^4+2*x^2+20*x)*ln(x+10)^2+(x+10)*ln(x+10)-x)/(x+10)/ln(x+10)^2,x,method=_RETURN
VERBOSE)

[Out]

x^8+2*x^5+x^2+x/ln(x+10)

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maxima [A]  time = 0.77, size = 26, normalized size = 1.53 \begin {gather*} \frac {{\left (x^{8} + 2 \, x^{5} + x^{2}\right )} \log \left (x + 10\right ) + x}{\log \left (x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^8+80*x^7+10*x^5+100*x^4+2*x^2+20*x)*log(x+10)^2+(x+10)*log(x+10)-x)/(x+10)/log(x+10)^2,x, algo
rithm="maxima")

[Out]

((x^8 + 2*x^5 + x^2)*log(x + 10) + x)/log(x + 10)

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mupad [B]  time = 1.46, size = 20, normalized size = 1.18 \begin {gather*} \frac {x}{\ln \left (x+10\right )}+x^2+2\,x^5+x^8 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 10)^2*(20*x + 2*x^2 + 100*x^4 + 10*x^5 + 80*x^7 + 8*x^8) - x + log(x + 10)*(x + 10))/(log(x + 10)
^2*(x + 10)),x)

[Out]

x/log(x + 10) + x^2 + 2*x^5 + x^8

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sympy [A]  time = 0.11, size = 17, normalized size = 1.00 \begin {gather*} x^{8} + 2 x^{5} + x^{2} + \frac {x}{\log {\left (x + 10 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**8+80*x**7+10*x**5+100*x**4+2*x**2+20*x)*ln(x+10)**2+(x+10)*ln(x+10)-x)/(x+10)/ln(x+10)**2,x)

[Out]

x**8 + 2*x**5 + x**2 + x/log(x + 10)

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