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]

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

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Rubi [C]  time = 1.14, 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 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 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 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 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 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 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 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 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 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]

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 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 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 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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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]

-1/4*(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])/x^4

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fricas [A]  time = 0.44, size = 115, normalized size = 1.32 \[ \frac {2 \, a b c^{2} x^{4} \arctan \left (\frac {1}{c x^{2}}\right ) - b^{2} c^{2} x^{4} \log \left (c^{2} x^{4} + 1\right ) + 4 \, b^{2} c^{2} x^{4} \log \relax (x) - 2 \, a b c x^{2} - {\left (b^{2} c^{2} x^{4} + b^{2}\right )} \arctan \left (c x^{2}\right )^{2} - a^{2} - 2 \, {\left (b^{2} c x^{2} + a b\right )} \arctan \left (c x^{2}\right )}{4 \, x^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{5}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 118, normalized size = 1.36 \[ -\frac {a^{2}}{4 x^{4}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 x^{4}}-\frac {b^{2} c^{2} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {b^{2} c \arctan \left (c \,x^{2}\right )}{2 x^{2}}+b^{2} c^{2} \ln \relax (x )-\frac {b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right )}{4}-\frac {a b \arctan \left (c \,x^{2}\right )}{2 x^{4}}-\frac {a b \,c^{2} \arctan \left (c \,x^{2}\right )}{2}-\frac {c a b}{2 x^{2}} \]

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

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maxima [A]  time = 0.52, size = 110, normalized size = 1.26 \[ -\frac {1}{2} \, {\left ({\left (c \arctan \left (c x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {\arctan \left (c x^{2}\right )}{x^{4}}\right )} a b + \frac {1}{4} \, {\left ({\left (\arctan \left (c x^{2}\right )^{2} - \log \left (c^{2} x^{4} + 1\right ) + 4 \, \log \relax (x)\right )} c^{2} - 2 \, {\left (c \arctan \left (c x^{2}\right ) + \frac {1}{x^{2}}\right )} c \arctan \left (c x^{2}\right )\right )} b^{2} - \frac {b^{2} \arctan \left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]

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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mupad [B]  time = 0.61, size = 152, normalized size = 1.75 \[ b^2\,c^2\,\ln \relax (x)-\frac {b^2\,c^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4}-\frac {b^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4\,x^4}-\frac {b^2\,c^2\,\ln \left (c^2\,x^4+1\right )}{4}-\frac {a^2}{4\,x^4}-\frac {b^2\,c\,\mathrm {atan}\left (c\,x^2\right )}{2\,x^2}-\frac {a\,b\,c}{2\,x^2}-\frac {a\,b\,c^2\,\mathrm {atan}\left (\frac {a^2\,c\,x^2}{a^2+25\,b^2}+\frac {25\,b^2\,c\,x^2}{a^2+25\,b^2}\right )}{2}-\frac {a\,b\,\mathrm {atan}\left (c\,x^2\right )}{2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))^2/x^5,x)

[Out]

b^2*c^2*log(x) - (b^2*c^2*atan(c*x^2)^2)/4 - (b^2*atan(c*x^2)^2)/(4*x^4) - (b^2*c^2*log(c^2*x^4 + 1))/4 - a^2/
(4*x^4) - (b^2*c*atan(c*x^2))/(2*x^2) - (a*b*c)/(2*x^2) - (a*b*c^2*atan((a^2*c*x^2)/(a^2 + 25*b^2) + (25*b^2*c
*x^2)/(a^2 + 25*b^2)))/2 - (a*b*atan(c*x^2))/(2*x^4)

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sympy [A]  time = 52.41, size = 167, normalized size = 1.92 \[ \begin {cases} - \frac {a^{2}}{4 x^{4}} - \frac {a b c^{2} \operatorname {atan}{\left (c x^{2} \right )}}{2} - \frac {a b c}{2 x^{2}} - \frac {a b \operatorname {atan}{\left (c x^{2} \right )}}{2 x^{4}} + b^{2} c^{2} \log {\relax (x )} - \frac {b^{2} c^{2} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{2} - \frac {b^{2} c^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4} - \frac {i b^{2} c \operatorname {atan}{\left (c x^{2} \right )}}{2 \sqrt {\frac {1}{c^{2}}}} - \frac {b^{2} c \operatorname {atan}{\left (c x^{2} \right )}}{2 x^{2}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \]

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) - a*b*c**2*atan(c*x**2)/2 - a*b*c/(2*x**2) - a*b*atan(c*x**2)/(2*x**4) + b**2*c**2*l
og(x) - b**2*c**2*log(x**2 + I*sqrt(c**(-2)))/2 - b**2*c**2*atan(c*x**2)**2/4 - I*b**2*c*atan(c*x**2)/(2*sqrt(
c**(-2))) - b**2*c*atan(c*x**2)/(2*x**2) - b**2*atan(c*x**2)**2/(4*x**4), Ne(c, 0)), (-a**2/(4*x**4), True))

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