\(\int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx\) [68]

Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {i b^2 d x^3}{30 c}+\frac {3 i b^2 d \arctan (c x)}{10 c^4}+\frac {b^2 d x \arctan (c x)}{2 c^3}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}-\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {2 i b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{3 c^4}-\frac {b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^4} \]

[Out]

1/2*a*b*d*x/c^3-3/10*I*b^2*d*x/c^3+1/12*b^2*d*x^2/c^2+1/30*I*b^2*d*x^3/c+3/10*I*b^2*d*arctan(c*x)/c^4+1/2*b^2*
d*x*arctan(c*x)/c^3+1/5*I*b*d*x^2*(a+b*arctan(c*x))/c^2-1/6*b*d*x^3*(a+b*arctan(c*x))/c-1/10*I*b*d*x^4*(a+b*ar
ctan(c*x))-9/20*d*(a+b*arctan(c*x))^2/c^4+1/4*d*x^4*(a+b*arctan(c*x))^2+1/5*I*c*d*x^5*(a+b*arctan(c*x))^2+2/5*
I*b*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4-1/3*b^2*d*ln(c^2*x^2+1)/c^4-1/5*b^2*d*polylog(2,1-2/(1+I*c*x))/c^4

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {4996, 4946, 5036, 272, 45, 4930, 266, 5004, 308, 209, 327, 5040, 4964, 2449, 2352} \[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {2 i b d \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{5 c^4}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {1}{4} d x^4 (a+b \arctan (c x))^2-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {b d x^3 (a+b \arctan (c x))}{6 c}+\frac {a b d x}{2 c^3}+\frac {3 i b^2 d \arctan (c x)}{10 c^4}+\frac {b^2 d x \arctan (c x)}{2 c^3}-\frac {b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{5 c^4}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}-\frac {b^2 d \log \left (c^2 x^2+1\right )}{3 c^4}+\frac {i b^2 d x^3}{30 c} \]

[In]

Int[x^3*(d + I*c*d*x)*(a + b*ArcTan[c*x])^2,x]

[Out]

(a*b*d*x)/(2*c^3) - (((3*I)/10)*b^2*d*x)/c^3 + (b^2*d*x^2)/(12*c^2) + ((I/30)*b^2*d*x^3)/c + (((3*I)/10)*b^2*d
*ArcTan[c*x])/c^4 + (b^2*d*x*ArcTan[c*x])/(2*c^3) + ((I/5)*b*d*x^2*(a + b*ArcTan[c*x]))/c^2 - (b*d*x^3*(a + b*
ArcTan[c*x]))/(6*c) - (I/10)*b*d*x^4*(a + b*ArcTan[c*x]) - (9*d*(a + b*ArcTan[c*x])^2)/(20*c^4) + (d*x^4*(a +
b*ArcTan[c*x])^2)/4 + (I/5)*c*d*x^5*(a + b*ArcTan[c*x])^2 + (((2*I)/5)*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*
x)])/c^4 - (b^2*d*Log[1 + c^2*x^2])/(3*c^4) - (b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)])/(5*c^4)

Rule 45

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 327

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

Rule 2352

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

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

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

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 4964

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

Rule 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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 5036

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

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (d x^3 (a+b \arctan (c x))^2+i c d x^4 (a+b \arctan (c x))^2\right ) \, dx \\ & = d \int x^3 (a+b \arctan (c x))^2 \, dx+(i c d) \int x^4 (a+b \arctan (c x))^2 \, dx \\ & = \frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2-\frac {1}{2} (b c d) \int \frac {x^4 (a+b \arctan (c x))}{1+c^2 x^2} \, dx-\frac {1}{5} \left (2 i b c^2 d\right ) \int \frac {x^5 (a+b \arctan (c x))}{1+c^2 x^2} \, dx \\ & = \frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2-\frac {1}{5} (2 i b d) \int x^3 (a+b \arctan (c x)) \, dx+\frac {1}{5} (2 i b d) \int \frac {x^3 (a+b \arctan (c x))}{1+c^2 x^2} \, dx-\frac {(b d) \int x^2 (a+b \arctan (c x)) \, dx}{2 c}+\frac {(b d) \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{2 c} \\ & = -\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {1}{6} \left (b^2 d\right ) \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {(b d) \int (a+b \arctan (c x)) \, dx}{2 c^3}-\frac {(b d) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{2 c^3}+\frac {(2 i b d) \int x (a+b \arctan (c x)) \, dx}{5 c^2}-\frac {(2 i b d) \int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{5 c^2}+\frac {1}{10} \left (i b^2 c d\right ) \int \frac {x^4}{1+c^2 x^2} \, dx \\ & = \frac {a b d x}{2 c^3}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}-\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {1}{12} \left (b^2 d\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {(2 i b d) \int \frac {a+b \arctan (c x)}{i-c x} \, dx}{5 c^3}+\frac {\left (b^2 d\right ) \int \arctan (c x) \, dx}{2 c^3}-\frac {\left (i b^2 d\right ) \int \frac {x^2}{1+c^2 x^2} \, dx}{5 c}+\frac {1}{10} \left (i b^2 c d\right ) \int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {i b^2 d x^3}{30 c}+\frac {b^2 d x \arctan (c x)}{2 c^3}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}-\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {2 i b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}+\frac {1}{12} \left (b^2 d\right ) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {\left (i b^2 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{10 c^3}+\frac {\left (i b^2 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (2 i b^2 d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (b^2 d\right ) \int \frac {x}{1+c^2 x^2} \, dx}{2 c^2} \\ & = \frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {i b^2 d x^3}{30 c}+\frac {3 i b^2 d \arctan (c x)}{10 c^4}+\frac {b^2 d x \arctan (c x)}{2 c^3}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}-\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {2 i b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{3 c^4}-\frac {\left (2 b^2 d\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{5 c^4} \\ & = \frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {i b^2 d x^3}{30 c}+\frac {3 i b^2 d \arctan (c x)}{10 c^4}+\frac {b^2 d x \arctan (c x)}{2 c^3}+\frac {i b d x^2 (a+b \arctan (c x))}{5 c^2}-\frac {b d x^3 (a+b \arctan (c x))}{6 c}-\frac {1}{10} i b d x^4 (a+b \arctan (c x))-\frac {9 d (a+b \arctan (c x))^2}{20 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))^2+\frac {1}{5} i c d x^5 (a+b \arctan (c x))^2+\frac {2 i b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{3 c^4}-\frac {b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.99 \[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\frac {d \left (18 i a b+5 b^2+30 a b c x-18 i b^2 c x+12 i a b c^2 x^2+5 b^2 c^2 x^2-10 a b c^3 x^3+2 i b^2 c^3 x^3+15 a^2 c^4 x^4-6 i a b c^4 x^4+12 i a^2 c^5 x^5+3 b^2 \left (-1+5 c^4 x^4+4 i c^5 x^5\right ) \arctan (c x)^2+2 b \arctan (c x) \left (b \left (9 i+15 c x+6 i c^2 x^2-5 c^3 x^3-3 i c^4 x^4\right )+3 a \left (-5+5 c^4 x^4+4 i c^5 x^5\right )+12 i b \log \left (1+e^{2 i \arctan (c x)}\right )\right )-12 i a b \log \left (1+c^2 x^2\right )-20 b^2 \log \left (1+c^2 x^2\right )+12 b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )}{60 c^4} \]

[In]

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

[Out]

(d*((18*I)*a*b + 5*b^2 + 30*a*b*c*x - (18*I)*b^2*c*x + (12*I)*a*b*c^2*x^2 + 5*b^2*c^2*x^2 - 10*a*b*c^3*x^3 + (
2*I)*b^2*c^3*x^3 + 15*a^2*c^4*x^4 - (6*I)*a*b*c^4*x^4 + (12*I)*a^2*c^5*x^5 + 3*b^2*(-1 + 5*c^4*x^4 + (4*I)*c^5
*x^5)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(b*(9*I + 15*c*x + (6*I)*c^2*x^2 - 5*c^3*x^3 - (3*I)*c^4*x^4) + 3*a*(-5
+ 5*c^4*x^4 + (4*I)*c^5*x^5) + (12*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (12*I)*a*b*Log[1 + c^2*x^2] - 20*b^2
*Log[1 + c^2*x^2] + 12*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(60*c^4)

Maple [A] (verified)

Time = 1.48 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.31

method	result	size
parts	\(a^{2} d \left (\frac {1}{5} i c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {d \,b^{2} \left (-\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{10}+\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}+\frac {c x \arctan \left (c x \right )}{2}-\frac {i \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {i c^{3} x^{3}}{30}+\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{5}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right )^{2}}{20}-\frac {\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}+\frac {\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right )^{2}}{20}+\frac {3 i \arctan \left (c x \right )}{10}-\frac {3 i c x}{10}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}+\frac {i \arctan \left (c x \right )^{2} c^{5} x^{5}}{5}\right )}{c^{4}}+\frac {2 a b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(375\)
derivativedivides	\(\frac {a^{2} d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d \,b^{2} \left (-\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{10}+\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}+\frac {c x \arctan \left (c x \right )}{2}-\frac {i \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {i c^{3} x^{3}}{30}+\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{5}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right )^{2}}{20}-\frac {\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}+\frac {\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right )^{2}}{20}+\frac {3 i \arctan \left (c x \right )}{10}-\frac {3 i c x}{10}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}+\frac {i \arctan \left (c x \right )^{2} c^{5} x^{5}}{5}\right )+2 a b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(378\)
default	\(\frac {a^{2} d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d \,b^{2} \left (-\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{10}+\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}+\frac {c x \arctan \left (c x \right )}{2}-\frac {i \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {i c^{3} x^{3}}{30}+\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{5}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {\ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{10}-\frac {\ln \left (c x -i\right )^{2}}{20}-\frac {\ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{10}+\frac {\operatorname {dilog}\left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{10}+\frac {\ln \left (c x +i\right )^{2}}{20}+\frac {3 i \arctan \left (c x \right )}{10}-\frac {3 i c x}{10}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}+\frac {i \arctan \left (c x \right )^{2} c^{5} x^{5}}{5}\right )+2 a b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(378\)
risch	\(\frac {b^{2} d \,x^{2}}{12 c^{2}}-\frac {b^{2} d \ln \left (c^{2} x^{2}+1\right )}{3 c^{4}}+\frac {a b d x}{2 c^{3}}+\frac {d \,x^{4} a^{2}}{4}-\frac {9 d \,a^{2}}{20 c^{4}}-\frac {b d a \arctan \left (c x \right )}{2 c^{4}}-\frac {d a b \,x^{3}}{6 c}+\frac {5 b^{2} d}{12 c^{4}}+\left (\frac {i d \,b^{2} \left (4 x^{5} c -5 i x^{4}\right ) \ln \left (-i c x +1\right )}{40}+\frac {b d \left (24 a \,c^{5} x^{5}-30 i a \,c^{4} x^{4}-6 b \,c^{4} x^{4}+10 i b \,c^{3} x^{3}+12 b \,c^{2} x^{2}-30 i b c x -27 b \ln \left (-i c x +1\right )\right )}{120 c^{4}}\right ) \ln \left (i c x +1\right )+\frac {i b^{2} d \,x^{3}}{30 c}+\frac {3 i b^{2} d \arctan \left (c x \right )}{10 c^{4}}+\frac {29 i b d a}{30 c^{4}}-\frac {i b d \,x^{4} a}{10}-\frac {d \,b^{2} \ln \left (-i c x +1\right ) x^{2}}{10 c^{2}}+\frac {d \,b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{5 c^{4}}-\frac {d \,b^{2} \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{5 c^{4}}+\frac {i d c \,a^{2} x^{5}}{5}-\frac {3 i b^{2} d x}{10 c^{3}}-\frac {d \,b^{2} \ln \left (-i c x +1\right )^{2} x^{4}}{16}+\frac {d \,b^{2} \ln \left (-i c x +1\right ) x^{4}}{20}+\frac {9 d \,b^{2} \ln \left (-i c x +1\right )^{2}}{80 c^{4}}-\frac {d \,b^{2} \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{5 c^{4}}-\frac {i d c \,b^{2} \ln \left (-i c x +1\right )^{2} x^{5}}{20}+\frac {i d a b \ln \left (-i c x +1\right ) x^{4}}{4}-\frac {i d \,b^{2} \left (4 c^{5} x^{5}-5 i c^{4} x^{4}+i\right ) \ln \left (i c x +1\right )^{2}}{80 c^{4}}-\frac {i b d a \ln \left (c^{2} x^{2}+1\right )}{5 c^{4}}-\frac {d c a b \ln \left (-i c x +1\right ) x^{5}}{5}+\frac {i b d \,x^{2} a}{5 c^{2}}+\frac {i d \,b^{2} \ln \left (-i c x +1\right ) x}{4 c^{3}}-\frac {i d \,b^{2} \ln \left (-i c x +1\right ) x^{3}}{12 c}\)	\(577\)

[In]

int(x^3*(d+I*c*d*x)*(a+b*arctan(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

a^2*d*(1/5*I*c*x^5+1/4*x^4)+d*b^2/c^4*(-1/10*I*arctan(c*x)*c^4*x^4+1/4*c^4*x^4*arctan(c*x)^2+1/2*c*x*arctan(c*
x)-1/5*I*arctan(c*x)*ln(c^2*x^2+1)-1/6*c^3*x^3*arctan(c*x)+1/30*I*c^3*x^3+1/5*I*arctan(c*x)*c^2*x^2-1/4*arctan
(c*x)^2+1/10*ln(c*x-I)*ln(c^2*x^2+1)-1/10*dilog(-1/2*I*(c*x+I))-1/10*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/20*ln(c*x-
I)^2-1/10*ln(c*x+I)*ln(c^2*x^2+1)+1/10*dilog(1/2*I*(c*x-I))+1/10*ln(c*x+I)*ln(1/2*I*(c*x-I))+1/20*ln(c*x+I)^2+
3/10*I*arctan(c*x)-3/10*I*c*x+1/12*c^2*x^2-1/3*ln(c^2*x^2+1)+1/5*I*arctan(c*x)^2*c^5*x^5)+2*a*b*d/c^4*(1/5*I*a
rctan(c*x)*c^5*x^5+1/4*c^4*x^4*arctan(c*x)+1/4*c*x-1/20*I*c^4*x^4-1/12*c^3*x^3+1/10*I*c^2*x^2-1/10*I*ln(c^2*x^
2+1)-1/4*arctan(c*x))

Fricas [F]

\[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3} \,d x } \]

[In]

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

[Out]

1/80*(-4*I*b^2*c*d*x^5 - 5*b^2*d*x^4)*log(-(c*x + I)/(c*x - I))^2 + integral(1/20*(20*I*a^2*c^3*d*x^6 + 20*a^2
*c^2*d*x^5 + 20*I*a^2*c*d*x^4 + 20*a^2*d*x^3 - (20*a*b*c^3*d*x^6 + 4*(-5*I*a*b - b^2)*c^2*d*x^5 + 5*(4*a*b + I
*b^2)*c*d*x^4 - 20*I*a*b*d*x^3)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

Sympy [F(-1)]

Timed out. \[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3} \,d x } \]

[In]

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

[Out]

1/5*I*a^2*c*d*x^5 + 1/4*b^2*d*x^4*arctan(c*x)^2 + 1/4*a^2*d*x^4 + 1/10*I*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*
x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*a*b*c*d + 1/80*I*(4*x^5*arctan(c*x)^2 - x^5*log(c^2*x^2 + 1)^2 + 80*integr
ate(1/80*(4*c^2*x^6*log(c^2*x^2 + 1) - 8*c*x^5*arctan(c*x) + 60*(c^2*x^6 + x^4)*arctan(c*x)^2 + 5*(c^2*x^6 + x
^4)*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x))*b^2*c*d + 1/6*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arcta
n(c*x)/c^5))*a*b*d - 1/12*(2*c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5)*arctan(c*x) - (c^2*x^2 + 3*arctan(c*x
)^2 - 4*log(c^2*x^2 + 1))/c^4)*b^2*d

Giac [F]

\[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int x^3 (d+i c d x) (a+b \arctan (c x))^2 \, dx=\int x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right ) \,d x \]

[In]

int(x^3*(a + b*atan(c*x))^2*(d + c*d*x*1i),x)

[Out]

int(x^3*(a + b*atan(c*x))^2*(d + c*d*x*1i), x)