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

Optimal. Leaf size=226 \[ \frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {4 b d^3 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {7}{2} i a b d^3 x+\frac {11 i b^2 d^3 \log \left (c^2 x^2+1\right )}{6 c}-\frac {2 i b^2 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}-\frac {1}{12} i b^2 c d^3 x^2+\frac {b^2 d^3 \tan ^{-1}(c x)}{c}-\frac {7}{2} i b^2 d^3 x \tan ^{-1}(c x)-b^2 d^3 x \]

[Out]

-7/2*I*a*b*d^3*x-b^2*d^3*x-1/12*I*b^2*c*d^3*x^2+b^2*d^3*arctan(c*x)/c-7/2*I*b^2*d^3*x*arctan(c*x)+b*c*d^3*x^2*
(a+b*arctan(c*x))+1/6*I*b*c^2*d^3*x^3*(a+b*arctan(c*x))-1/4*I*d^3*(1+I*c*x)^4*(a+b*arctan(c*x))^2/c+4*b*d^3*(a
+b*arctan(c*x))*ln(2/(1-I*c*x))/c+11/6*I*b^2*d^3*ln(c^2*x^2+1)/c-2*I*b^2*d^3*polylog(2,1-2/(1-I*c*x))/c

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Rubi [A]  time = 0.21, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4864, 4846, 260, 4852, 321, 203, 266, 43, 1586, 4854, 2402, 2315} \[ -\frac {2 i b^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c}+\frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {4 b d^3 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {7}{2} i a b d^3 x+\frac {11 i b^2 d^3 \log \left (c^2 x^2+1\right )}{6 c}-\frac {1}{12} i b^2 c d^3 x^2+\frac {b^2 d^3 \tan ^{-1}(c x)}{c}-\frac {7}{2} i b^2 d^3 x \tan ^{-1}(c x)-b^2 d^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*I)/2)*a*b*d^3*x - b^2*d^3*x - (I/12)*b^2*c*d^3*x^2 + (b^2*d^3*ArcTan[c*x])/c - ((7*I)/2)*b^2*d^3*x*ArcTan
[c*x] + b*c*d^3*x^2*(a + b*ArcTan[c*x]) + (I/6)*b*c^2*d^3*x^3*(a + b*ArcTan[c*x]) - ((I/4)*d^3*(1 + I*c*x)^4*(
a + b*ArcTan[c*x])^2)/c + (4*b*d^3*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/c + (((11*I)/6)*b^2*d^3*Log[1 + c^2
*x^2])/c - ((2*I)*b^2*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/c

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 203

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

Rule 260

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 266

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 321

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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 2402

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 4846

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

Rule 4852

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4864

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

Rubi steps

\begin {align*} \int (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {(i b) \int \left (-7 d^4 \left (a+b \tan ^{-1}(c x)\right )-4 i c d^4 x \left (a+b \tan ^{-1}(c x)\right )+c^2 d^4 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {8 i \left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{2 d}\\ &=-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {(4 b) \int \frac {\left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{d}-\frac {1}{2} \left (7 i b d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx+\left (2 b c d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {1}{2} \left (i b c^2 d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=-\frac {7}{2} i a b d^3 x+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {(4 b) \int \frac {a+b \tan ^{-1}(c x)}{-\frac {i}{d^4}-\frac {c x}{d^4}} \, dx}{d}-\frac {1}{2} \left (7 i b^2 d^3\right ) \int \tan ^{-1}(c x) \, dx-\left (b^2 c^2 d^3\right ) \int \frac {x^2}{1+c^2 x^2} \, dx-\frac {1}{6} \left (i b^2 c^3 d^3\right ) \int \frac {x^3}{1+c^2 x^2} \, dx\\ &=-\frac {7}{2} i a b d^3 x-b^2 d^3 x-\frac {7}{2} i b^2 d^3 x \tan ^{-1}(c x)+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {4 b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{c}+\left (b^2 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx-\left (4 b^2 d^3\right ) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx+\frac {1}{2} \left (7 i b^2 c d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx-\frac {1}{12} \left (i b^2 c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {7}{2} i a b d^3 x-b^2 d^3 x+\frac {b^2 d^3 \tan ^{-1}(c x)}{c}-\frac {7}{2} i b^2 d^3 x \tan ^{-1}(c x)+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {4 b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{c}+\frac {7 i b^2 d^3 \log \left (1+c^2 x^2\right )}{4 c}-\frac {\left (4 i b^2 d^3\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )}{c}-\frac {1}{12} \left (i b^2 c^3 d^3\right ) \operatorname {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 {7}{2} i a b d^3 x-b^2 d^3 x-\frac {1}{12} i b^2 c d^3 x^2+\frac {b^2 d^3 \tan ^{-1}(c x)}{c}-\frac {7}{2} i b^2 d^3 x \tan ^{-1}(c x)+b c d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{6} i b c^2 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c}+\frac {4 b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{c}+\frac {11 i b^2 d^3 \log \left (1+c^2 x^2\right )}{6 c}-\frac {2 i b^2 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 267, normalized size = 1.18 \[ -\frac {i d^3 \left (3 a^2 c^4 x^4-12 i a^2 c^3 x^3-18 a^2 c^2 x^2+12 i a^2 c x-2 a b c^3 x^3+12 i a b c^2 x^2-24 i a b \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (3 a \left (c^4 x^4-4 i c^3 x^3-6 c^2 x^2+4 i c x-7\right )+b \left (-c^3 x^3+6 i c^2 x^2+21 c x+6 i\right )+24 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+42 a b c x+b^2 c^2 x^2-22 b^2 \log \left (c^2 x^2+1\right )+24 b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-12 i b^2 c x+3 b^2 (c x-i)^4 \tan ^{-1}(c x)^2+b^2\right )}{12 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/12*I)*d^3*(b^2 + (12*I)*a^2*c*x + 42*a*b*c*x - (12*I)*b^2*c*x - 18*a^2*c^2*x^2 + (12*I)*a*b*c^2*x^2 + b^2
*c^2*x^2 - (12*I)*a^2*c^3*x^3 - 2*a*b*c^3*x^3 + 3*a^2*c^4*x^4 + 3*b^2*(-I + c*x)^4*ArcTan[c*x]^2 + 2*b*ArcTan[
c*x]*(b*(6*I + 21*c*x + (6*I)*c^2*x^2 - c^3*x^3) + 3*a*(-7 + (4*I)*c*x - 6*c^2*x^2 - (4*I)*c^3*x^3 + c^4*x^4)
+ (24*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (24*I)*a*b*Log[1 + c^2*x^2] - 22*b^2*Log[1 + c^2*x^2] + 24*b^2*Po
lyLog[2, -E^((2*I)*ArcTan[c*x])]))/c

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, {\left (i \, b^{2} c^{3} d^{3} x^{4} + 4 \, b^{2} c^{2} d^{3} x^{3} - 6 i \, b^{2} c d^{3} x^{2} - 4 \, b^{2} d^{3} x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + {\rm integral}\left (\frac {-4 i \, a^{2} c^{5} d^{3} x^{5} - 12 \, a^{2} c^{4} d^{3} x^{4} + 8 i \, a^{2} c^{3} d^{3} x^{3} - 8 \, a^{2} c^{2} d^{3} x^{2} + 12 i \, a^{2} c d^{3} x + 4 \, a^{2} d^{3} + {\left (4 \, a b c^{5} d^{3} x^{5} + {\left (-12 i \, a b - b^{2}\right )} c^{4} d^{3} x^{4} - {\left (8 \, a b - 4 i \, b^{2}\right )} c^{3} d^{3} x^{3} - 2 \, {\left (4 i \, a b - 3 \, b^{2}\right )} c^{2} d^{3} x^{2} - {\left (12 \, a b + 4 i \, b^{2}\right )} c d^{3} x + 4 i \, a b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, {\left (c^{2} x^{2} + 1\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(I*b^2*c^3*d^3*x^4 + 4*b^2*c^2*d^3*x^3 - 6*I*b^2*c*d^3*x^2 - 4*b^2*d^3*x)*log(-(c*x + I)/(c*x - I))^2 + i
ntegral(1/4*(-4*I*a^2*c^5*d^3*x^5 - 12*a^2*c^4*d^3*x^4 + 8*I*a^2*c^3*d^3*x^3 - 8*a^2*c^2*d^3*x^2 + 12*I*a^2*c*
d^3*x + 4*a^2*d^3 + (4*a*b*c^5*d^3*x^5 + (-12*I*a*b - b^2)*c^4*d^3*x^4 - (8*a*b - 4*I*b^2)*c^3*d^3*x^3 - 2*(4*
I*a*b - 3*b^2)*c^2*d^3*x^2 - (12*a*b + 4*I*b^2)*c*d^3*x + 4*I*a*b*d^3)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1
), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.11, size = 620, normalized size = 2.74 \[ -\frac {i d^{3} a^{2}}{4 c}+d^{3} b^{2} \arctan \left (c x \right )^{2} x -c^{2} d^{3} a^{2} x^{3}+\frac {i c^{2} d^{3} a b \,x^{3}}{6}+\frac {i c^{2} d^{3} b^{2} \arctan \left (c x \right ) x^{3}}{6}+\frac {i d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{c}-\frac {i c^{3} d^{3} b^{2} \arctan \left (c x \right )^{2} x^{4}}{4}+\frac {3 i c \,d^{3} b^{2} \arctan \left (c x \right )^{2} x^{2}}{2}-\frac {i d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{c}+\frac {7 i d^{3} a b \arctan \left (c x \right )}{2 c}+\frac {i d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{c}-\frac {i d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{c}-2 c^{2} d^{3} a b \arctan \left (c x \right ) x^{3}+\frac {b^{2} d^{3} \arctan \left (c x \right )}{c}-\frac {7 i b^{2} d^{3} x \arctan \left (c x \right )}{2}-\frac {i d^{3} b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{c}+\frac {i d^{3} b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{c}+\frac {7 i d^{3} b^{2} \arctan \left (c x \right )^{2}}{4 c}+\frac {i d^{3} b^{2} \ln \left (c x -i\right )^{2}}{2 c}-\frac {i d^{3} b^{2} \ln \left (c x +i\right )^{2}}{2 c}+\frac {3 i c \,x^{2} a^{2} d^{3}}{2}-\frac {i c^{3} x^{4} a^{2} d^{3}}{4}-\frac {7 i a b \,d^{3} x}{2}-\frac {i b^{2} c \,d^{3} x^{2}}{12}+\frac {11 i b^{2} d^{3} \ln \left (c^{2} x^{2}+1\right )}{6 c}-\frac {2 d^{3} a b \ln \left (c^{2} x^{2}+1\right )}{c}-\frac {2 d^{3} b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{c}+c \,d^{3} b^{2} \arctan \left (c x \right ) x^{2}-c^{2} d^{3} b^{2} \arctan \left (c x \right )^{2} x^{3}+2 d^{3} a b \arctan \left (c x \right ) x +c \,d^{3} a b \,x^{2}-b^{2} d^{3} x +x \,a^{2} d^{3}+3 i c \,d^{3} a b \arctan \left (c x \right ) x^{2}-\frac {i c^{3} d^{3} a b \arctan \left (c x \right ) x^{4}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*b^2*arctan(c*x)^2*x-c^2*d^3*a^2*x^3-1/4*I/c*d^3*a^2-I/c*d^3*b^2*ln(c*x-I)*ln(c^2*x^2+1)-I/c*d^3*b^2*ln(I+c
*x)*ln(1/2*I*(c*x-I))+7/2*I/c*d^3*a*b*arctan(c*x)+I/c*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))+1/6*I*c^2*d^3*a*b*x
^3+1/6*I*c^2*d^3*b^2*arctan(c*x)*x^3-2*c^2*d^3*a*b*arctan(c*x)*x^3+I/c*d^3*b^2*ln(I+c*x)*ln(c^2*x^2+1)-1/4*I*c
^3*d^3*b^2*arctan(c*x)^2*x^4+3/2*I*c*d^3*b^2*arctan(c*x)^2*x^2+11/6*I*b^2*d^3*ln(c^2*x^2+1)/c+b^2*d^3*arctan(c
*x)/c-7/2*I*a*b*d^3*x-1/12*I*b^2*c*d^3*x^2-7/2*I*b^2*d^3*x*arctan(c*x)-2/c*d^3*a*b*ln(c^2*x^2+1)-2/c*d^3*b^2*a
rctan(c*x)*ln(c^2*x^2+1)+c*d^3*b^2*arctan(c*x)*x^2-c^2*d^3*b^2*arctan(c*x)^2*x^3+2*d^3*a*b*arctan(c*x)*x+c*d^3
*a*b*x^2+7/4*I/c*d^3*b^2*arctan(c*x)^2+1/2*I/c*d^3*b^2*ln(c*x-I)^2-1/2*I/c*d^3*b^2*ln(I+c*x)^2-I/c*d^3*b^2*dil
og(1/2*I*(c*x-I))+3/2*I*c*x^2*a^2*d^3+I/c*d^3*b^2*dilog(-1/2*I*(I+c*x))-1/4*I*c^3*x^4*a^2*d^3-b^2*d^3*x+x*a^2*
d^3+3*I*c*d^3*a*b*arctan(c*x)*x^2-1/2*I*c^3*d^3*a*b*arctan(c*x)*x^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*a^2*c^3*d^3*x^4 - 4*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - 2*b
^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)/(c^2*x^2 + 1), x) - a^2*c^2*d^3*x^3 - 36*b^2*c^4*d^3*integrate(1/16*
x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) - 3*b^2*c^4*d^3*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) -
5*b^2*c^4*d^3*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - 1/6*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 -
3*x)/c^4 + 3*arctan(c*x)/c^5))*a*b*c^3*d^3 + 8*b^2*c^3*d^3*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c
^2*x^2 + 1), x) + 20*b^2*c^3*d^3*integrate(1/16*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - (2*x^3*arctan(c*x) - c*(x^
2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*c^2*d^3 + 3/2*I*a^2*c*d^3*x^2 - 24*b^2*c^2*d^3*integrate(1/16*x^2*arctan(c*
x)^2/(c^2*x^2 + 1), x) - 2*b^2*c^2*d^3*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 10*b^2*c^2*d^
3*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3*I*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*
a*b*c*d^3 + 1/4*b^2*d^3*arctan(c*x)^3/c + 12*b^2*c*d^3*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2
+ 1), x) - 8*b^2*c*d^3*integrate(1/16*x*arctan(c*x)/(c^2*x^2 + 1), x) + a^2*d^3*x + b^2*d^3*integrate(1/16*log
(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d^3/c - 1/64*(4*I*b^2*c^3*d^3*x
^4 + 16*b^2*c^2*d^3*x^3 - 24*I*b^2*c*d^3*x^2 - 16*b^2*d^3*x)*arctan(c*x)^2 + 1/64*(4*b^2*c^3*d^3*x^4 - 16*I*b^
2*c^2*d^3*x^3 - 24*b^2*c*d^3*x^2 + 16*I*b^2*d^3*x)*arctan(c*x)*log(c^2*x^2 + 1) - 1/64*(-I*b^2*c^3*d^3*x^4 - 4
*b^2*c^2*d^3*x^3 + 6*I*b^2*c*d^3*x^2 + 4*b^2*d^3*x)*log(c^2*x^2 + 1)^2 - I*integrate(1/16*(12*(b^2*c^5*d^3*x^5
- 2*b^2*c^3*d^3*x^3 - 3*b^2*c*d^3*x)*arctan(c*x)^2 + (b^2*c^5*d^3*x^5 - 2*b^2*c^3*d^3*x^3 - 3*b^2*c*d^3*x)*lo
g(c^2*x^2 + 1)^2 - 10*(b^2*c^4*d^3*x^4 - 2*b^2*c^2*d^3*x^2)*arctan(c*x) + (b^2*c^5*d^3*x^5 - 10*b^2*c^3*d^3*x^
3 + 4*b^2*c*d^3*x - 4*(3*b^2*c^4*d^3*x^4 + 2*b^2*c^2*d^3*x^2 - b^2*d^3)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^
2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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