3.80 \(\int \frac{(a+b \tan ^{-1}(c x^2))^2}{x^5} \, dx\)

Optimal. Leaf size=87 \[ -\frac{1}{4} c^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-\frac{b c \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{2 x^2}-\frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{4 x^4}-\frac{1}{4} b^2 c^2 \log \left (c^2 x^4+1\right )+b^2 c^2 \log (x) \]

[Out]

-(b*c*(a + b*ArcTan[c*x^2]))/(2*x^2) - (c^2*(a + b*ArcTan[c*x^2])^2)/4 - (a + b*ArcTan[c*x^2])^2/(4*x^4) + b^2
*c^2*Log[x] - (b^2*c^2*Log[1 + c^2*x^4])/4

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Rubi [C]  time = 1.13623, antiderivative size = 419, normalized size of antiderivative = 4.82, number of steps used = 46, number of rules used = 23, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.438, Rules used = {5035, 2454, 2398, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 2395, 44, 2439, 2416, 36, 29, 2392, 2391, 2394, 2393, 2410, 2390} \[ -\frac{1}{8} b^2 c^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-i c x^2\right )\right )-\frac{1}{8} b^2 c^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1+i c x^2\right )\right )+\frac{1}{8} b c^2 \log \left (\frac{1}{2} \left (1+i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )-\frac{1}{16} c^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2+\frac{i b c \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )}{8 x^2}+\frac{b \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 x^4}-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (1+i c x^2\right )-\frac{1}{4} b^2 c^2 \log \left (-c x^2+i\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac{1}{8} b^2 c^2 \log \left (c x^2+i\right )+b^2 c^2 \log (x)+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}+\frac{i b^2 c \log \left (1+i c x^2\right )}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c*x^2])^2/x^5,x]

[Out]

b^2*c^2*Log[x] - (b^2*c^2*Log[I - c*x^2])/4 + ((I/8)*b*c*((2*I)*a - b*Log[1 - I*c*x^2]))/x^2 - (b*c*(1 - I*c*x
^2)*(2*a + I*b*Log[1 - I*c*x^2]))/(8*x^2) - (c^2*(2*a + I*b*Log[1 - I*c*x^2])^2)/16 - (2*a + I*b*Log[1 - I*c*x
^2])^2/(16*x^4) + (b*c^2*((2*I)*a - b*Log[1 - I*c*x^2])*Log[(1 + I*c*x^2)/2])/8 + ((I/4)*b^2*c*Log[1 + I*c*x^2
])/x^2 - (b^2*c^2*Log[(1 - I*c*x^2)/2]*Log[1 + I*c*x^2])/8 + (b*((2*I)*a - b*Log[1 - I*c*x^2])*Log[1 + I*c*x^2
])/(8*x^4) + (b^2*c^2*Log[1 + I*c*x^2]^2)/16 + (b^2*Log[1 + I*c*x^2]^2)/(16*x^4) - (b^2*c^2*Log[I + c*x^2])/8
- (b^2*c^2*PolyLog[2, (1 - I*c*x^2)/2])/8 - (b^2*c^2*PolyLog[2, (1 + I*c*x^2)/2])/8

Rule 5035

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2398

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

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 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && 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 2439

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2391

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 2394

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 2393

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 2410

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[Log[c*(d
+ e*x)], x^m/(f + g*x), x], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m
]

Rule 2390

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{x^5} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{4 x^5}+\frac{b \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{2 x^5}-\frac{b^2 \log ^2\left (1+i c x^2\right )}{4 x^5}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{x^5} \, dx+\frac{1}{2} b \int \frac{\left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{x^5} \, dx-\frac{1}{4} b^2 \int \frac{\log ^2\left (1+i c x^2\right )}{x^5} \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^2}{x^3} \, dx,x,x^2\right )+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{x^3} \, dx,x,x^2\right )-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{\log ^2(1+i c x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}+\frac{1}{8} (i b c) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{x^2 (1+i c x)} \, dx,x,x^2\right )+\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a+i b \log (1-i c x)}{x^2 (1-i c x)} \, dx,x,x^2\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x^2 (1-i c x)} \, dx,x,x^2\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x^2 (1+i c x)} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}+\frac{1}{8} (i b) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{x \left (-\frac{i}{c}+\frac{i x}{c}\right )^2} \, dx,x,1-i c x^2\right )+\frac{1}{8} (i b c) \operatorname{Subst}\left (\int \left (\frac{-2 i a+b \log (1-i c x)}{x^2}-\frac{i c (-2 i a+b \log (1-i c x))}{x}+\frac{i c^2 (-2 i a+b \log (1-i c x))}{-i+c x}\right ) \, dx,x,x^2\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+i c x)}{x^2}-\frac{i c \log (1+i c x)}{x}+\frac{i c^2 \log (1+i c x)}{-i+c x}\right ) \, dx,x,x^2\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+i c x)}{x^2}+\frac{i c \log (1+i c x)}{x}-\frac{i c^2 \log (1+i c x)}{i+c x}\right ) \, dx,x,x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}+\frac{1}{8} (i b) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{\left (-\frac{i}{c}+\frac{i x}{c}\right )^2} \, dx,x,1-i c x^2\right )+\frac{1}{8} (i b c) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{x \left (-\frac{i}{c}+\frac{i x}{c}\right )} \, dx,x,1-i c x^2\right )-2 \left (\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x^2} \, dx,x,x^2\right )\right )+\frac{1}{8} \left (b c^2\right ) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{x} \, dx,x,x^2\right )-\frac{1}{8} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{-i+c x} \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{i+c x} \, dx,x,x^2\right )\\ &=-\frac{1}{2} i a b c^2 \log (x)+\frac{i b c \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )-\frac{1}{8} \left (i b c^2\right ) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{x} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (1-i c x)} \, dx,x,x^2\right )-2 \left (-\frac{i b^2 c \log \left (1+i c x^2\right )}{8 x^2}-\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (1+i c x)} \, dx,x,x^2\right )\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1+i c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i c x)}{x} \, dx,x,x^2\right )-\frac{1}{8} \left (i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^2\right )+\frac{1}{8} \left (i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^2\right )\\ &=\frac{1}{4} b^2 c^2 \log (x)+\frac{i b c \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{1}{16} c^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (1+i c x^2\right )+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}-\frac{1}{8} b^2 c^2 \text{Li}_2\left (i c x^2\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+i c x^2\right )+\frac{1}{8} \left (i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-i c x} \, dx,x,x^2\right )-2 \left (-\frac{i b^2 c \log \left (1+i c x^2\right )}{8 x^2}-\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{8} \left (i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+i c x} \, dx,x,x^2\right )\right )\\ &=\frac{1}{2} b^2 c^2 \log (x)+\frac{i b c \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )}{8 x^2}-\frac{1}{16} c^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (1+i c x^2\right )+\frac{b^2 \log ^2\left (1+i c x^2\right )}{16 x^4}-2 \left (-\frac{1}{4} b^2 c^2 \log (x)+\frac{1}{8} b^2 c^2 \log \left (i-c x^2\right )-\frac{i b^2 c \log \left (1+i c x^2\right )}{8 x^2}\right )-\frac{1}{8} b^2 c^2 \log \left (i+c x^2\right )-\frac{1}{8} b^2 c^2 \text{Li}_2\left (\frac{1}{2} \left (1-i c x^2\right )\right )-\frac{1}{8} b^2 c^2 \text{Li}_2\left (\frac{1}{2} \left (1+i c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0769828, size = 98, normalized size = 1.13 \[ -\frac{a^2+2 b \tan ^{-1}\left (c x^2\right ) \left (a c^2 x^4+a+b c x^2\right )+2 a b c x^2-4 b^2 c^2 x^4 \log (x)+b^2 c^2 x^4 \log \left (c^2 x^4+1\right )+b^2 \left (c^2 x^4+1\right ) \tan ^{-1}\left (c x^2\right )^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^2])^2/x^5,x]

[Out]

-(a^2 + 2*a*b*c*x^2 + 2*b*(a + b*c*x^2 + a*c^2*x^4)*ArcTan[c*x^2] + b^2*(1 + c^2*x^4)*ArcTan[c*x^2]^2 - 4*b^2*
c^2*x^4*Log[x] + b^2*c^2*x^4*Log[1 + c^2*x^4])/(4*x^4)

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Maple [A]  time = 0.037, size = 118, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}}{4\,{x}^{4}}}-{\frac{{b}^{2} \left ( \arctan \left ( c{x}^{2} \right ) \right ) ^{2}}{4\,{x}^{4}}}-{\frac{{b}^{2}c\arctan \left ( c{x}^{2} \right ) }{2\,{x}^{2}}}-{\frac{{b}^{2}{c}^{2} \left ( \arctan \left ( c{x}^{2} \right ) \right ) ^{2}}{4}}-{\frac{{b}^{2}{c}^{2}\ln \left ({c}^{2}{x}^{4}+1 \right ) }{4}}+{b}^{2}{c}^{2}\ln \left ( x \right ) -{\frac{ab\arctan \left ( c{x}^{2} \right ) }{2\,{x}^{4}}}-{\frac{abc}{2\,{x}^{2}}}-{\frac{ab{c}^{2}\arctan \left ( c{x}^{2} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))^2/x^5,x)

[Out]

-1/4*a^2/x^4-1/4*b^2/x^4*arctan(c*x^2)^2-1/2*b^2*c*arctan(c*x^2)/x^2-1/4*b^2*c^2*arctan(c*x^2)^2-1/4*b^2*c^2*l
n(c^2*x^4+1)+b^2*c^2*ln(x)-1/2*a*b/x^4*arctan(c*x^2)-1/2*c*a*b/x^2-1/2*a*b*c^2*arctan(c*x^2)

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Maxima [A]  time = 1.8367, size = 149, normalized size = 1.71 \begin{align*} -\frac{1}{2} \,{\left ({\left (c \arctan \left (c x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{\arctan \left (c x^{2}\right )}{x^{4}}\right )} a b + \frac{1}{4} \,{\left ({\left (\arctan \left (c x^{2}\right )^{2} - \log \left (c^{2} x^{4} + 1\right ) + 4 \, \log \left (x\right )\right )} c^{2} - 2 \,{\left (c \arctan \left (c x^{2}\right ) + \frac{1}{x^{2}}\right )} c \arctan \left (c x^{2}\right )\right )} b^{2} - \frac{b^{2} \arctan \left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac{a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="maxima")

[Out]

-1/2*((c*arctan(c*x^2) + 1/x^2)*c + arctan(c*x^2)/x^4)*a*b + 1/4*((arctan(c*x^2)^2 - log(c^2*x^4 + 1) + 4*log(
x))*c^2 - 2*(c*arctan(c*x^2) + 1/x^2)*c*arctan(c*x^2))*b^2 - 1/4*b^2*arctan(c*x^2)^2/x^4 - 1/4*a^2/x^4

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Fricas [A]  time = 2.73242, size = 258, normalized size = 2.97 \begin{align*} \frac{2 \, a b c^{2} x^{4} \arctan \left (\frac{1}{c x^{2}}\right ) - b^{2} c^{2} x^{4} \log \left (c^{2} x^{4} + 1\right ) + 4 \, b^{2} c^{2} x^{4} \log \left (x\right ) - 2 \, a b c x^{2} -{\left (b^{2} c^{2} x^{4} + b^{2}\right )} \arctan \left (c x^{2}\right )^{2} - a^{2} - 2 \,{\left (b^{2} c x^{2} + a b\right )} \arctan \left (c x^{2}\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="fricas")

[Out]

1/4*(2*a*b*c^2*x^4*arctan(1/(c*x^2)) - b^2*c^2*x^4*log(c^2*x^4 + 1) + 4*b^2*c^2*x^4*log(x) - 2*a*b*c*x^2 - (b^
2*c^2*x^4 + b^2)*arctan(c*x^2)^2 - a^2 - 2*(b^2*c*x^2 + a*b)*arctan(c*x^2))/x^4

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Sympy [A]  time = 86.8163, size = 1187, normalized size = 13.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))**2/x**5,x)

[Out]

Piecewise((-a**2/(4*x**4), Eq(c, 0)), (-(a - oo*I*b)**2/(4*x**4), Eq(c, -I/x**2)), (-(a + oo*I*b)**2/(4*x**4),
Eq(c, I/x**2)), (3*I*a**2*c**3*x**4*(c**(-2))**(3/2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2)
)**(3/2)) + 3*I*a**2*c*(c**(-2))**(3/2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*
I*a*b*c**5*x**8*(c**(-2))**(3/2)*atan(c*x**2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)
) + 6*I*a*b*c**4*x**6*(c**(-2))**(3/2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 12*
I*a*b*c**3*x**4*(c**(-2))**(3/2)*atan(c*x**2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)
) + 6*I*a*b*c**2*x**2*(c**(-2))**(3/2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*I
*a*b*c*(c**(-2))**(3/2)*atan(c*x**2)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) - 12*I*
b**2*c**5*x**8*(c**(-2))**(3/2)*log(x)/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*I
*b**2*c**5*x**8*(c**(-2))**(3/2)*log(x**2 + I*sqrt(c**(-2)))/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(
c**(-2))**(3/2)) + 3*I*b**2*c**5*x**8*(c**(-2))**(3/2)*atan(c*x**2)**2/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*
I*c*x**4*(c**(-2))**(3/2)) + 6*I*b**2*c**4*x**6*(c**(-2))**(3/2)*atan(c*x**2)/(-12*I*c**3*x**8*(c**(-2))**(3/2
) - 12*I*c*x**4*(c**(-2))**(3/2)) - 12*I*b**2*c**3*x**4*(c**(-2))**(3/2)*log(x)/(-12*I*c**3*x**8*(c**(-2))**(3
/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*I*b**2*c**3*x**4*(c**(-2))**(3/2)*log(x**2 + I*sqrt(c**(-2)))/(-12*I*c
**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*I*b**2*c**3*x**4*(c**(-2))**(3/2)*atan(c*x**2)**
2/(-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) - 6*b**2*c**2*x**8*atan(c*x**2)/(-12*I*c**
3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 6*I*b**2*c**2*x**2*(c**(-2))**(3/2)*atan(c*x**2)/(-1
2*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) + 3*I*b**2*c*(c**(-2))**(3/2)*atan(c*x**2)**2/(
-12*I*c**3*x**8*(c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)) - 6*b**2*x**4*atan(c*x**2)/(-12*I*c**3*x**8*(
c**(-2))**(3/2) - 12*I*c*x**4*(c**(-2))**(3/2)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^2) + a)^2/x^5, x)