3.76 \(\int x^3 (a+b \tan ^{-1}(c x^2))^2 \, dx\)

Optimal. Leaf size=90 \[ \frac {\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{4 c^2}-\frac {a b x^2}{2 c}+\frac {1}{4} x^4 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (c^2 x^4+1\right )}{4 c^2}-\frac {b^2 x^2 \tan ^{-1}\left (c x^2\right )}{2 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [C]  time = 1.04, antiderivative size = 612, normalized size of antiderivative = 6.80, number of steps used = 44, number of rules used = 16, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5035, 2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304, 2395, 43, 2439, 2416, 2394, 2393, 2391} \[ \frac {b^2 \text {PolyLog}\left (2,\frac {1}{2} \left (1-i c x^2\right )\right )}{8 c^2}+\frac {b^2 \text {PolyLog}\left (2,\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c^2}-\frac {\left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c^2}+\frac {\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c^2}+\frac {i b \left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )}{16 c^2}-\frac {b \log \left (\frac {1}{2} \left (1+i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{8 c^2}-\frac {3 a b x^2}{4 c}+\frac {1}{16} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right )-\frac {1}{8} b x^4 \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac {b^2 \left (1-i c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+i c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+i c x^2\right )^2 \log ^2\left (1+i c x^2\right )}{16 c^2}-\frac {b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c^2}-\frac {b^2 \log \left (-c x^2+i\right )}{16 c^2}+\frac {3 b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{8 c^2}-\frac {b^2 \left (1+i c x^2\right )^2 \log \left (1+i c x^2\right )}{16 c^2}+\frac {3 b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{8 c^2}+\frac {b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {b^2 \log \left (c x^2+i\right )}{16 c^2}-\frac {1}{16} b^2 x^4 \log \left (1+i c x^2\right )+\frac {b^2 x^4}{16} \]

Warning: Unable to verify antiderivative.

[In]

Int[x^3*(a + b*ArcTan[c*x^2])^2,x]

[Out]

(-3*a*b*x^2)/(4*c) + (b^2*x^4)/16 + (b^2*(1 - I*c*x^2)^2)/(32*c^2) + (b^2*(1 + I*c*x^2)^2)/(32*c^2) - (b^2*Log
[I - c*x^2])/(16*c^2) + (3*b^2*(1 - I*c*x^2)*Log[1 - I*c*x^2])/(8*c^2) + (b*x^4*((2*I)*a - b*Log[1 - I*c*x^2])
)/16 + ((I/16)*b*(1 - I*c*x^2)^2*(2*a + I*b*Log[1 - I*c*x^2]))/c^2 + ((1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2
])^2)/(8*c^2) - ((1 - I*c*x^2)^2*(2*a + I*b*Log[1 - I*c*x^2])^2)/(16*c^2) - (b*((2*I)*a - b*Log[1 - I*c*x^2])*
Log[(1 + I*c*x^2)/2])/(8*c^2) - (b^2*x^4*Log[1 + I*c*x^2])/16 + (3*b^2*(1 + I*c*x^2)*Log[1 + I*c*x^2])/(8*c^2)
- (b^2*(1 + I*c*x^2)^2*Log[1 + I*c*x^2])/(16*c^2) + (b^2*Log[(1 - I*c*x^2)/2]*Log[1 + I*c*x^2])/(8*c^2) - (b*
x^4*((2*I)*a - b*Log[1 - I*c*x^2])*Log[1 + I*c*x^2])/8 - (b^2*(1 + I*c*x^2)*Log[1 + I*c*x^2]^2)/(8*c^2) + (b^2
*(1 + I*c*x^2)^2*Log[1 + I*c*x^2]^2)/(16*c^2) - (b^2*Log[I + c*x^2])/(16*c^2) + (b^2*PolyLog[2, (1 - I*c*x^2)/
2])/(8*c^2) + (b^2*PolyLog[2, (1 + I*c*x^2)/2])/(8*c^2)

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

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 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 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 2401

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

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 x^3 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^3 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2+\frac {1}{2} b x^3 \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {1}{4} b^2 x^3 \log ^2\left (1+i c x^2\right )\right ) \, dx\\ &=\frac {1}{4} \int x^3 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2 \, dx+\frac {1}{2} b \int x^3 \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right ) \, dx-\frac {1}{4} b^2 \int x^3 \log ^2\left (1+i c x^2\right ) \, dx\\ &=\frac {1}{8} \operatorname {Subst}\left (\int x (2 a+i b \log (1-i c x))^2 \, dx,x,x^2\right )+\frac {1}{4} b \operatorname {Subst}\left (\int x (-2 i a+b \log (1-i c x)) \log (1+i c x) \, dx,x,x^2\right )-\frac {1}{8} b^2 \operatorname {Subst}\left (\int x \log ^2(1+i c x) \, dx,x,x^2\right )\\ &=-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {1}{8} \operatorname {Subst}\left (\int \left (-\frac {i (2 a+i b \log (1-i c x))^2}{c}+\frac {i (1-i c x) (2 a+i b \log (1-i c x))^2}{c}\right ) \, dx,x,x^2\right )-\frac {1}{8} b^2 \operatorname {Subst}\left (\int \left (\frac {i \log ^2(1+i c x)}{c}-\frac {i (1+i c x) \log ^2(1+i c x)}{c}\right ) \, dx,x,x^2\right )-\frac {1}{8} (i b c) \operatorname {Subst}\left (\int \frac {x^2 (-2 i a+b \log (1-i c x))}{1+i c x} \, dx,x,x^2\right )+\frac {1}{8} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {x^2 \log (1+i c x)}{1-i c x} \, dx,x,x^2\right )\\ &=-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {i \operatorname {Subst}\left (\int (2 a+i b \log (1-i c x))^2 \, dx,x,x^2\right )}{8 c}+\frac {i \operatorname {Subst}\left (\int (1-i c x) (2 a+i b \log (1-i c x))^2 \, dx,x,x^2\right )}{8 c}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \log ^2(1+i c x) \, dx,x,x^2\right )}{8 c}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int (1+i c x) \log ^2(1+i c x) \, dx,x,x^2\right )}{8 c}-\frac {1}{8} (i b c) \operatorname {Subst}\left (\int \left (\frac {-2 i a+b \log (1-i c x)}{c^2}-\frac {i x (-2 i a+b \log (1-i c x))}{c}+\frac {i (-2 i a+b \log (1-i c x))}{c^2 (-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)}{c^2}+\frac {i x \log (1+i c x)}{c}-\frac {i \log (1+i c x)}{c^2 (i+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {1}{8} b \operatorname {Subst}\left (\int x (-2 i a+b \log (1-i c x)) \, dx,x,x^2\right )-\frac {1}{8} b^2 \operatorname {Subst}\left (\int x \log (1+i c x) \, dx,x,x^2\right )+\frac {\operatorname {Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-i c x^2\right )}{8 c^2}-\frac {\operatorname {Subst}\left (\int x (2 a+i b \log (x))^2 \, dx,x,1-i c x^2\right )}{8 c^2}-\frac {b^2 \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^2\right )}{8 c^2}+\frac {b^2 \operatorname {Subst}\left (\int x \log ^2(x) \, dx,x,1+i c x^2\right )}{8 c^2}-\frac {(i b) \operatorname {Subst}\left (\int (-2 i a+b \log (1-i c x)) \, dx,x,x^2\right )}{8 c}+\frac {b \operatorname {Subst}\left (\int \frac {-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^2\right )}{8 c}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \log (1+i c x) \, dx,x,x^2\right )}{8 c}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (1+i c x)}{i+c x} \, dx,x,x^2\right )}{8 c}\\ &=-\frac {a b x^2}{4 c}+\frac {1}{16} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac {\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c^2}-\frac {\left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c^2}-\frac {b \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 )}{8 c^2}-\frac {1}{16} b^2 x^4 \log \left (1+i c x^2\right )+\frac {b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c^2}+\frac {b^2 \left (1+i c x^2\right )^2 \log ^2\left (1+i c x^2\right )}{16 c^2}+\frac {(i b) \operatorname {Subst}\left (\int x (2 a+i b \log (x)) \, dx,x,1-i c x^2\right )}{8 c^2}-\frac {(i b) \operatorname {Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-i c x^2\right )}{4 c^2}+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{8 c^2}-\frac {b^2 \operatorname {Subst}\left (\int x \log (x) \, dx,x,1+i c x^2\right )}{8 c^2}+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{4 c^2}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \log (1-i c x) \, dx,x,x^2\right )}{8 c}+\frac {\left (i b^2\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 )}{8 c}-\frac {\left (i b^2\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 )}{8 c}-\frac {1}{16} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-i c x} \, dx,x,x^2\right )+\frac {1}{16} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+i c x} \, dx,x,x^2\right )\\ &=-\frac {3 a b x^2}{4 c}-\frac {3 i b^2 x^2}{8 c}+\frac {b^2 \left (1-i c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+i c x^2\right )^2}{32 c^2}+\frac {1}{16} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac {i b \left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )}{16 c^2}+\frac {\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c^2}-\frac {\left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c^2}-\frac {b \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 )}{8 c^2}-\frac {1}{16} b^2 x^4 \log \left (1+i c x^2\right )+\frac {3 b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {b^2 \left (1+i c x^2\right )^2 \log \left (1+i c x^2\right )}{16 c^2}+\frac {b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c^2}+\frac {b^2 \left (1+i c x^2\right )^2 \log ^2\left (1+i c x^2\right )}{16 c^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-i c x^2\right )}{8 c^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+i c x^2\right )}{8 c^2}+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{8 c^2}+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{4 c^2}+\frac {1}{16} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2}-\frac {i x}{c}+\frac {i}{c^2 (-i+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2}+\frac {i x}{c}-\frac {i}{c^2 (i+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 a b x^2}{4 c}+\frac {b^2 x^4}{16}+\frac {b^2 \left (1-i c x^2\right )^2}{32 c^2}+\frac {b^2 \left (1+i c x^2\right )^2}{32 c^2}-\frac {b^2 \log \left (i-c x^2\right )}{16 c^2}+\frac {3 b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{8 c^2}+\frac {1}{16} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac {i b \left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )}{16 c^2}+\frac {\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c^2}-\frac {\left (1-i c x^2\right )^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c^2}-\frac {b \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 )}{8 c^2}-\frac {1}{16} b^2 x^4 \log \left (1+i c x^2\right )+\frac {3 b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {b^2 \left (1+i c x^2\right )^2 \log \left (1+i c x^2\right )}{16 c^2}+\frac {b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{8 c^2}-\frac {1}{8} b x^4 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c^2}+\frac {b^2 \left (1+i c x^2\right )^2 \log ^2\left (1+i c x^2\right )}{16 c^2}-\frac {b^2 \log \left (i+c x^2\right )}{16 c^2}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{8 c^2}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 85, normalized size = 0.94 \[ \frac {2 b \tan ^{-1}\left (c x^2\right ) \left (a c^2 x^4+a-b c x^2\right )+a c x^2 \left (a c x^2-2 b\right )+b^2 \log \left (c^2 x^4+1\right )+b^2 \left (c^2 x^4+1\right ) \tan ^{-1}\left (c x^2\right )^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTan[c*x^2])^2,x]

[Out]

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

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fricas [A]  time = 0.45, size = 100, normalized size = 1.11 \[ \frac {a^{2} c^{2} x^{4} - 2 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} + b^{2}\right )} \arctan \left (c x^{2}\right )^{2} - 2 \, a b \arctan \left (\frac {1}{c x^{2}}\right ) + b^{2} \log \left (c^{2} x^{4} + 1\right ) + 2 \, {\left (a b c^{2} x^{4} - b^{2} c x^{2}\right )} \arctan \left (c x^{2}\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.33, size = 100, normalized size = 1.11 \[ \frac {a^{2} c x^{4} + \frac {2 \, {\left (c^{2} x^{4} \arctan \left (c x^{2}\right ) - c x^{2} + \arctan \left (c x^{2}\right )\right )} a b}{c} + \frac {{\left (c^{2} x^{4} \arctan \left (c x^{2}\right )^{2} - 2 \, c x^{2} \arctan \left (c x^{2}\right ) + \arctan \left (c x^{2}\right )^{2} + \log \left (c^{2} x^{4} + 1\right )\right )} b^{2}}{c}}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 113, normalized size = 1.26 \[ \frac {a^{2} x^{4}}{4}+\frac {b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {b^{2} x^{2} \arctan \left (c \,x^{2}\right )}{2 c}+\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 c^{2}}+\frac {b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {a b \,x^{4} \arctan \left (c \,x^{2}\right )}{2}-\frac {a b \,x^{2}}{2 c}+\frac {a b \arctan \left (c \,x^{2}\right )}{2 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x^2))^2,x)

[Out]

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

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maxima [A]  time = 0.49, size = 126, normalized size = 1.40 \[ \frac {1}{4} \, b^{2} x^{4} \arctan \left (c x^{2}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{2} \, {\left (x^{4} \arctan \left (c x^{2}\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\arctan \left (c x^{2}\right )}{c^{3}}\right )}\right )} a b - \frac {1}{4} \, {\left (2 \, c {\left (\frac {x^{2}}{c^{2}} - \frac {\arctan \left (c x^{2}\right )}{c^{3}}\right )} \arctan \left (c x^{2}\right ) + \frac {\arctan \left (c x^{2}\right )^{2} - \log \left (4 \, c^{5} x^{4} + 4 \, c^{3}\right )}{c^{2}}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x^2))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arctan(c*x^2)^2 + 1/4*a^2*x^4 + 1/2*(x^4*arctan(c*x^2) - c*(x^2/c^2 - arctan(c*x^2)/c^3))*a*b - 1/
4*(2*c*(x^2/c^2 - arctan(c*x^2)/c^3)*arctan(c*x^2) + (arctan(c*x^2)^2 - log(4*c^5*x^4 + 4*c^3))/c^2)*b^2

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mupad [B]  time = 0.68, size = 112, normalized size = 1.24 \[ \frac {a^2\,x^4}{4}+\frac {b^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4\,c^2}+\frac {b^2\,x^4\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4}+\frac {b^2\,\ln \left (c^2\,x^4+1\right )}{4\,c^2}-\frac {b^2\,x^2\,\mathrm {atan}\left (c\,x^2\right )}{2\,c}-\frac {a\,b\,x^2}{2\,c}+\frac {a\,b\,\mathrm {atan}\left (c\,x^2\right )}{2\,c^2}+\frac {a\,b\,x^4\,\mathrm {atan}\left (c\,x^2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atan(c*x^2))^2,x)

[Out]

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

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sympy [A]  time = 34.91, size = 155, normalized size = 1.72 \[ \begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atan}{\left (c x^{2} \right )}}{2} - \frac {a b x^{2}}{2 c} + \frac {a b \operatorname {atan}{\left (c x^{2} \right )}}{2 c^{2}} + \frac {b^{2} x^{4} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4} - \frac {b^{2} x^{2} \operatorname {atan}{\left (c x^{2} \right )}}{2 c} + \frac {i b^{2} \sqrt {\frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{2 c} + \frac {b^{2} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{2 c^{2}} + \frac {b^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x**2))**2,x)

[Out]

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

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