3.1296 \(\int \frac{(a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2))}{x^6} \, dx\)

Optimal. Leaf size=248 \[ -\frac{1}{10} b c^5 e \text{PolyLog}\left (2,\frac{1}{c^2 x^2+1}\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{5 x^5}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{1}{10} b c^5 \log \left (1-\frac{1}{c^2 x^2+1}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac{b c^3 \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 x^2}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 x^4}-\frac{7 b c^3 e}{60 x^2}+\frac{19}{60} b c^5 e \log \left (c^2 x^2+1\right )-\frac{5}{6} b c^5 e \log (x) \]

[Out]

(-7*b*c^3*e)/(60*x^2) - (2*c^2*e*(a + b*ArcTan[c*x]))/(15*x^3) + (2*c^4*e*(a + b*ArcTan[c*x]))/(5*x) + (c^5*e*
(a + b*ArcTan[c*x])^2)/(5*b) - (5*b*c^5*e*Log[x])/6 + (19*b*c^5*e*Log[1 + c^2*x^2])/60 - (b*c*(d + e*Log[1 + c
^2*x^2]))/(20*x^4) + (b*c^3*(1 + c^2*x^2)*(d + e*Log[1 + c^2*x^2]))/(10*x^2) - ((a + b*ArcTan[c*x])*(d + e*Log
[1 + c^2*x^2]))/(5*x^5) + (b*c^5*(d + e*Log[1 + c^2*x^2])*Log[1 - (1 + c^2*x^2)^(-1)])/10 - (b*c^5*e*PolyLog[2
, (1 + c^2*x^2)^(-1)])/10

________________________________________________________________________________________

Rubi [A]  time = 0.625143, antiderivative size = 245, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 18, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {5017, 2475, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 2319, 44, 4918, 4852, 266, 36, 29, 4884} \[ -\frac{1}{10} b c^5 e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{5 x^5}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}-\frac{b c^5 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{20 e}+\frac{b c^3 \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 x^2}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 x^4}+\frac{1}{5} b c^5 d \log (x)-\frac{7 b c^3 e}{60 x^2}+\frac{19}{60} b c^5 e \log \left (c^2 x^2+1\right )-\frac{5}{6} b c^5 e \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^6,x]

[Out]

(-7*b*c^3*e)/(60*x^2) - (2*c^2*e*(a + b*ArcTan[c*x]))/(15*x^3) + (2*c^4*e*(a + b*ArcTan[c*x]))/(5*x) + (c^5*e*
(a + b*ArcTan[c*x])^2)/(5*b) + (b*c^5*d*Log[x])/5 - (5*b*c^5*e*Log[x])/6 + (19*b*c^5*e*Log[1 + c^2*x^2])/60 -
(b*c*(d + e*Log[1 + c^2*x^2]))/(20*x^4) + (b*c^3*(1 + c^2*x^2)*(d + e*Log[1 + c^2*x^2]))/(10*x^2) - ((a + b*Ar
cTan[c*x])*(d + e*Log[1 + c^2*x^2]))/(5*x^5) - (b*c^5*(d + e*Log[1 + c^2*x^2])^2)/(20*e) - (b*c^5*e*PolyLog[2,
-(c^2*x^2)])/10

Rule 5017

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

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

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 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2344

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

Rule 2301

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 2316

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2314

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

Rule 31

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

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

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

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 29

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

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{5} (b c) \int \frac{d+e \log \left (1+c^2 x^2\right )}{x^5 \left (1+c^2 x^2\right )} \, dx+\frac{1}{5} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log \left (1+c^2 x\right )}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{5} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx-\frac{1}{5} \left (2 c^4 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^3} \, dx,x,1+c^2 x^2\right )}{10 c}+\frac{1}{15} \left (2 b c^3 e\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac{1}{5} \left (2 c^4 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx+\frac{1}{5} \left (2 c^6 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^3} \, dx,x,1+c^2 x^2\right )}{10 c}-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{15} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{5} \left (2 b c^5 e\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{10} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )} \, dx,x,1+c^2 x^2\right )+\frac{1}{20} (b c e) \operatorname{Subst}\left (\int \frac{1}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{15} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}-\frac{c^2}{x}+\frac{c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{5} \left (b c^5 e\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{b c^3 e}{15 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{2}{15} b c^5 e \log (x)+\frac{1}{15} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{10} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )-\frac{1}{10} \left (b c^5\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x} \, dx,x,1+c^2 x^2\right )+\frac{1}{20} (b c e) \operatorname{Subst}\left (\int \left (\frac{c^4}{(-1+x)^2}-\frac{c^4}{-1+x}+\frac{c^4}{x}\right ) \, dx,x,1+c^2 x^2\right )-\frac{1}{10} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )-\frac{1}{5} \left (b c^5 e\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{5} \left (b c^7 e\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{7 b c^3 e}{60 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{1}{5} b c^5 d \log (x)-\frac{5}{6} b c^5 e \log (x)+\frac{19}{60} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{20 e}+\frac{1}{10} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )\\ &=-\frac{7 b c^3 e}{60 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{1}{5} b c^5 d \log (x)-\frac{5}{6} b c^5 e \log (x)+\frac{19}{60} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{20 e}-\frac{1}{10} b c^5 e \text{Li}_2\left (-c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.246785, size = 259, normalized size = 1.04 \[ \frac{1}{60} \left (6 b c^5 e \text{PolyLog}\left (2,c^2 x^2+1\right )-\frac{12 \left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^5}-\frac{8 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac{12 c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{b}+\frac{24 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{3 b c^5 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{e}+6 b c^5 \log \left (-c^2 x^2\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac{6 b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^2}-\frac{3 b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^4}-18 b c^5 e \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )+7 b c^3 e \left (c^2 \log \left (c^2 x^2+1\right )-2 c^2 \log (x)-\frac{1}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^6,x]

[Out]

((-8*c^2*e*(a + b*ArcTan[c*x]))/x^3 + (24*c^4*e*(a + b*ArcTan[c*x]))/x + (12*c^5*e*(a + b*ArcTan[c*x])^2)/b -
18*b*c^5*e*(2*Log[x] - Log[1 + c^2*x^2]) + 7*b*c^3*e*(-x^(-2) - 2*c^2*Log[x] + c^2*Log[1 + c^2*x^2]) - (3*b*c*
(d + e*Log[1 + c^2*x^2]))/x^4 + (6*b*c^3*(d + e*Log[1 + c^2*x^2]))/x^2 - (12*(a + b*ArcTan[c*x])*(d + e*Log[1
+ c^2*x^2]))/x^5 + 6*b*c^5*Log[-(c^2*x^2)]*(d + e*Log[1 + c^2*x^2]) - (3*b*c^5*(d + e*Log[1 + c^2*x^2])^2)/e +
6*b*c^5*e*PolyLog[2, 1 + c^2*x^2])/60

________________________________________________________________________________________

Maple [F]  time = 14.653, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( cx \right ) \right ) \left ( d+e\ln \left ({c}^{2}{x}^{2}+1 \right ) \right ) }{{x}^{6}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^6,x)

[Out]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^6,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac{2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac{4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d + \frac{1}{15} \,{\left (2 \,{\left (3 \, c^{3} \arctan \left (c x\right ) + \frac{3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c^{2} - \frac{3 \, \log \left (c^{2} x^{2} + 1\right )}{x^{5}}\right )} a e + b e \int \frac{\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{x^{6}}\,{d x} - \frac{a d}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d + 1/15*(2*(3
*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c^2 - 3*log(c^2*x^2 + 1)/x^5)*a*e + b*e*integrate(arctan(c*x)*log(c^2*
x^2 + 1)/x^6, x) - 1/5*a*d/x^5

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \arctan \left (c x\right ) + a d +{\left (b e \arctan \left (c x\right ) + a e\right )} \log \left (c^{2} x^{2} + 1\right )}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^6,x, algorithm="fricas")

[Out]

integral((b*d*arctan(c*x) + a*d + (b*e*arctan(c*x) + a*e)*log(c^2*x^2 + 1))/x^6, x)

________________________________________________________________________________________

Sympy [A]  time = 87.3639, size = 474, normalized size = 1.91 \begin{align*} \frac{2 a c^{4} e \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5 \sqrt{\frac{1}{c^{2}}}} + \frac{2 a c^{4} e}{5 x} - \frac{2 a c^{2} e}{15 x^{3}} - \frac{a d}{5 x^{5}} - \frac{a e \log{\left (c^{2} x^{2} + 1 \right )}}{5 x^{5}} + 4 b c^{9} e \left (\begin{cases} \frac{x^{2}}{40 c^{2}} - \frac{\log{\left (c^{2} x^{2} + 1 \right )}}{40 c^{4}} & \text{for}\: c = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}^{2}}{80 c^{4}} & \text{otherwise} \end{cases}\right ) - \frac{b c^{7} d \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right )}{10} - \frac{b c^{7} e \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right ) \log{\left (c^{2} x^{2} + 1 \right )}}{10} + \frac{b c^{5} d \log{\left (x^{2} \right )}}{10} - \frac{5 b c^{5} e \log{\left (x \right )}}{6} + \frac{5 b c^{5} e \log{\left (c^{2} x^{2} + 1 \right )}}{12} - \frac{b c^{5} e \operatorname{atan}^{2}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5} - \frac{b c^{5} e \operatorname{Li}_{2}\left (c^{2} x^{2} e^{i \pi }\right )}{10} + \frac{2 b c^{4} e \operatorname{atan}{\left (c x \right )} \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5 \sqrt{\frac{1}{c^{2}}}} + \frac{2 b c^{4} e \operatorname{atan}{\left (c x \right )}}{5 x} + \frac{b c^{3} d}{10 x^{2}} + \frac{b c^{3} e \log{\left (c^{2} x^{2} + 1 \right )}}{10 x^{2}} - \frac{7 b c^{3} e}{60 x^{2}} - \frac{2 b c^{2} e \operatorname{atan}{\left (c x \right )}}{15 x^{3}} - \frac{b c d}{20 x^{4}} - \frac{b c e \log{\left (c^{2} x^{2} + 1 \right )}}{20 x^{4}} - \frac{b d \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{b e \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1))/x**6,x)

[Out]

2*a*c**4*e*atan(x/sqrt(c**(-2)))/(5*sqrt(c**(-2))) + 2*a*c**4*e/(5*x) - 2*a*c**2*e/(15*x**3) - a*d/(5*x**5) -
a*e*log(c**2*x**2 + 1)/(5*x**5) + 4*b*c**9*e*Piecewise((x**2/(40*c**2) - log(c**2*x**2 + 1)/(40*c**4), Eq(c, 0
)), (log(c**2*x**2 + 1)**2/(80*c**4), True)) - b*c**7*d*Piecewise((x**2, Eq(c**2, 0)), (log(c**2*x**2 + 1)/c**
2, True))/10 - b*c**7*e*Piecewise((x**2, Eq(c**2, 0)), (log(c**2*x**2 + 1)/c**2, True))*log(c**2*x**2 + 1)/10
+ b*c**5*d*log(x**2)/10 - 5*b*c**5*e*log(x)/6 + 5*b*c**5*e*log(c**2*x**2 + 1)/12 - b*c**5*e*atan(x/sqrt(c**(-2
)))**2/5 - b*c**5*e*polylog(2, c**2*x**2*exp_polar(I*pi))/10 + 2*b*c**4*e*atan(c*x)*atan(x/sqrt(c**(-2)))/(5*s
qrt(c**(-2))) + 2*b*c**4*e*atan(c*x)/(5*x) + b*c**3*d/(10*x**2) + b*c**3*e*log(c**2*x**2 + 1)/(10*x**2) - 7*b*
c**3*e/(60*x**2) - 2*b*c**2*e*atan(c*x)/(15*x**3) - b*c*d/(20*x**4) - b*c*e*log(c**2*x**2 + 1)/(20*x**4) - b*d
*atan(c*x)/(5*x**5) - b*e*log(c**2*x**2 + 1)*atan(c*x)/(5*x**5)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}{\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^6,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*(e*log(c^2*x^2 + 1) + d)/x^6, x)