\(\int \frac {(a+b \arctan (c x)) (d+e \log (1+c^2 x^2))}{x^6} \, dx\) [1296]

Optimal result

Integrand size = 26, antiderivative size = 248 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=-\frac {7 b c^3 e}{60 x^2}-\frac {2 c^2 e (a+b \arctan (c x))}{15 x^3}+\frac {2 c^4 e (a+b \arctan (c x))}{5 x}+\frac {c^5 e (a+b \arctan (c x))^2}{5 b}-\frac {5}{6} b c^5 e \log (x)+\frac {19}{60} b c^5 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac {b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac {1}{10} b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right ) \log \left (1-\frac {1}{1+c^2 x^2}\right )-\frac {1}{10} b c^5 e \operatorname {PolyLog}\left (2,\frac {1}{1+c^2 x^2}\right ) \]

[Out]

-7/60*b*c^3*e/x^2-2/15*c^2*e*(a+b*arctan(c*x))/x^3+2/5*c^4*e*(a+b*arctan(c*x))/x+1/5*c^5*e*(a+b*arctan(c*x))^2
/b-5/6*b*c^5*e*ln(x)+19/60*b*c^5*e*ln(c^2*x^2+1)-1/20*b*c*(d+e*ln(c^2*x^2+1))/x^4+1/10*b*c^3*(c^2*x^2+1)*(d+e*
ln(c^2*x^2+1))/x^2-1/5*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5+1/10*b*c^5*(d+e*ln(c^2*x^2+1))*ln(1-1/(c^2*x^
2+1))-1/10*b*c^5*e*polylog(2,1/(c^2*x^2+1))

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {5137, 2525, 2458, 2389, 2379, 2438, 2351, 31, 2356, 46, 5038, 4946, 272, 36, 29, 5004} \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\frac {c^5 e (a+b \arctan (c x))^2}{5 b}+\frac {2 c^4 e (a+b \arctan (c x))}{5 x}-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{5 x^5}-\frac {2 c^2 e (a+b \arctan (c x))}{15 x^3}-\frac {5}{6} b c^5 e \log (x)-\frac {7 b c^3 e}{60 x^2}-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 x^4}+\frac {1}{10} b c^5 \log \left (1-\frac {1}{c^2 x^2+1}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {1}{10} b c^5 e \operatorname {PolyLog}\left (2,\frac {1}{c^2 x^2+1}\right )+\frac {19}{60} b c^5 e \log \left (c^2 x^2+1\right )+\frac {b c^3 \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 x^2} \]

[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) - (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

Rule 29

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

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 272

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 2351

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 2356

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 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 2389

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 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 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 2525

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 4946

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

Rule 5004

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]

Rule 5038

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 5137

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]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\frac {1}{60} \left (-\frac {8 c^2 e (a+b \arctan (c x))}{x^3}-24 c^4 e \left (-\frac {a+b \arctan (c x)}{x}-\frac {c (a+b \arctan (c x))^2}{2 b}+b c \left (\log (x)-\frac {1}{2} \log \left (1+c^2 x^2\right )\right )\right )-6 b c^5 e \left (2 \log (x)-\log \left (1+c^2 x^2\right )\right )+7 b c^3 e \left (-\frac {1}{x^2}-2 c^2 \log (x)+c^2 \log \left (1+c^2 x^2\right )\right )-\frac {3 b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^4}+\frac {6 b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2}-\frac {12 (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^5}-\frac {3 b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{e}+6 b c^5 \left (\log \left (-c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )+e \operatorname {PolyLog}\left (2,1+c^2 x^2\right )\right )\right ) \]

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

Maple [F]

\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{6}}d x\]

[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)

Fricas [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{6}} \,d x } \]

[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 [F(-2)]

Exception generated. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

Maxima [F]

\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{6}} \,d x } \]

[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

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^6} \,d x \]

[In]

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

[Out]

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