\(\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx\) [261]
Optimal result
Integrand size = 21, antiderivative size = 169 \[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}
\]
[Out]
1/12*I*b*x^4+1/3*x^3*arccoth(1-I*d-d*cot(b*x+a))-1/6*x^3*ln(1-(1-I*d)*exp(2*I*a+2*I*b*x))+1/4*I*x^2*polylog(2,
(1-I*d)*exp(2*I*a+2*I*b*x))/b-1/4*x*polylog(3,(1-I*d)*exp(2*I*a+2*I*b*x))/b^2-1/8*I*polylog(4,(1-I*d)*exp(2*I*
a+2*I*b*x))/b^3
Rubi [A] (verified)
Time = 0.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6401, 2215, 2221, 2611, 6744,
2320, 6724} \[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=-\frac {i \operatorname {PolyLog}\left (4,(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {x \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \coth ^{-1}(d (-\cot (a+b x))-i d+1)+\frac {1}{12} i b x^4
\]
[In]
Int[x^2*ArcCoth[1 - I*d - d*Cot[a + b*x]],x]
[Out]
(I/12)*b*x^4 + (x^3*ArcCoth[1 - I*d - d*Cot[a + b*x]])/3 - (x^3*Log[1 - (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/6
+ ((I/4)*x^2*PolyLog[2, (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/b - (x*PolyLog[3, (1 - I*d)*E^((2*I)*a + (2*I)*b*x
)])/(4*b^2) - ((I/8)*PolyLog[4, (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/b^3
Rule 2215
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Rule 2221
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Rule 2320
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 2611
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]
Rule 6401
Int[ArcCoth[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[c + d*Cot[a + b*x]]/(f*(m + 1))), x] + Dist[I*(b/(f*(m + 1))), Int[(e + f*x)^(m + 1)/(c - I*d -
c*E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - I*d)^2, 1]
Rule 6724
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]
Rule 6744
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]
Rubi steps \begin{align*}
\text {integral}& = \frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))+\frac {1}{3} (i b) \int \frac {x^3}{1+(-1+i d) e^{2 i a+2 i b x}} \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))+\frac {1}{3} (b (i+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+(-1+i d) e^{2 i a+2 i b x}} \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {1}{2} \int x^2 \log \left (1+(-1+i d) e^{2 i a+2 i b x}\right ) \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {i \int x \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right ) \, dx}{2 b} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {\int \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right ) \, dx}{4 b^2} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,(1-i d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.14 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92
\[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=\frac {1}{3} x^3 \coth ^{-1}(1-i d-d \cot (a+b x))-\frac {4 b^3 x^3 \log \left (1+\frac {e^{-2 i (a+b x)}}{-1+i d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{i+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{i+d}\right )-3 i \operatorname {PolyLog}\left (4,\frac {i e^{-2 i (a+b x)}}{i+d}\right )}{24 b^3}
\]
[In]
Integrate[x^2*ArcCoth[1 - I*d - d*Cot[a + b*x]],x]
[Out]
(x^3*ArcCoth[1 - I*d - d*Cot[a + b*x]])/3 - (4*b^3*x^3*Log[1 + 1/((-1 + I*d)*E^((2*I)*(a + b*x)))] + (6*I)*b^2
*x^2*PolyLog[2, I/((I + d)*E^((2*I)*(a + b*x)))] + 6*b*x*PolyLog[3, I/((I + d)*E^((2*I)*(a + b*x)))] - (3*I)*P
olyLog[4, I/((I + d)*E^((2*I)*(a + b*x)))])/(24*b^3)
Maple [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.49 (sec) , antiderivative size = 2273, normalized size of antiderivative =
13.45
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method | result | size |
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risch |
\(\text {Expression too large to display}\) |
\(2273\) |
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[In]
int(x^2*arccoth(1-I*d-d*cot(b*x+a)),x,method=_RETURNVERBOSE)
[Out]
1/6*x^3*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)-1/2*I/b^2*a^2/(I+d)*ln(1-I*exp(I*(b*x+a))*(I*(I+d))^(1/2))
*x+1/12*I*b*x^4-1/4/b/(I+d)*polylog(2,-I*(I+d)*exp(2*I*(b*x+a)))*x^2+1/4/b^3/(I+d)*polylog(2,-I*(I+d)*exp(2*I*
(b*x+a)))*a^2-1/2/b^3*a^2/(I+d)*dilog(1+I*exp(I*(b*x+a))*(I*(I+d))^(1/2))-1/2/b^3*a^2/(I+d)*dilog(1-I*exp(I*(b
*x+a))*(I*(I+d))^(1/2))-1/6*d/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a)))*x^3-1/6*I/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a
)))*x^3-1/3*x^3*ln(exp(I*(b*x+a)))+1/6/b^3*a^3*d/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)-1/4/b^2*d/(
I+d)*polylog(3,-I*(I+d)*exp(2*I*(b*x+a)))*x+1/3/b^3*d/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a)))*a^3-1/2/b^3*d*a^3/(
I+d)*ln(1+I*exp(I*(b*x+a))*(I*(I+d))^(1/2))-1/2/b^3*d*a^3/(I+d)*ln(1-I*exp(I*(b*x+a))*(I*(I+d))^(1/2))-1/8*I/b
^3*d/(I+d)*polylog(4,-I*(I+d)*exp(2*I*(b*x+a)))-1/2*I/b^3*a^3/(I+d)*ln(1+I*exp(I*(b*x+a))*(I*(I+d))^(1/2))-1/2
*I/b^3*a^3/(I+d)*ln(1-I*exp(I*(b*x+a))*(I*(I+d))^(1/2))+1/6*I/b^3*a^3/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x
+a))*d-I)+1/3*I/b^3/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a)))*a^3-1/4*I/b^2/(I+d)*polylog(3,-I*(I+d)*exp(2*I*(b*x+a
)))*x-1/2/b^2*d*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(I*(I+d))^(1/2))*x-1/2/b^2*d*a^2/(I+d)*ln(1-I*exp(I*(b*x+a))*(
I*(I+d))^(1/2))*x+1/2/b^2*d/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a)))*a^2*x+1/2*I/b^3*d*a^2/(I+d)*dilog(1+I*exp(I*(
b*x+a))*(I*(I+d))^(1/2))+1/2*I/b^3*d*a^2/(I+d)*dilog(1-I*exp(I*(b*x+a))*(I*(I+d))^(1/2))-1/4*I/b^3*a^2*d/(I+d)
*polylog(2,-I*(I+d)*exp(2*I*(b*x+a)))+1/2*I/b^2/(I+d)*ln(1+I*(I+d)*exp(2*I*(b*x+a)))*a^2*x+1/4*I/b*d/(I+d)*pol
ylog(2,-I*(I+d)*exp(2*I*(b*x+a)))*x^2-1/2*I/b^2*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(I*(I+d))^(1/2))*x-1/12*(-I*Pi
*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x+
a))*d-I)/(exp(2*I*(b*x+a))-1))^2+I*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))^2
-I*Pi*csgn(I*exp(2*I*(b*x+a)))^3+2*I*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^2-I*Pi*csgn(I/(exp(2*I
*(b*x+a))-1)*exp(2*I*(b*x+a)))^3+I*Pi*csgn(I/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))*csgn(I*d/(exp(2*I*(b*x+a))
-1)*exp(2*I*(b*x+a)))^2-I*Pi*csgn(I*d/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))^3+I*Pi*csgn(I/(exp(2*I*(b*x+a))-1
))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*I
*(b*x+a))-1))+I*Pi*csgn(I*d)*csgn(I*d/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))^2+I*Pi*csgn(I*d/(exp(2*I*(b*x+a))
-1)*exp(2*I*(b*x+a)))*csgn(1/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)^2+I*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*
I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))^3+I*Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I/(exp(2*I*(b*x+a))-1)*exp(2*I*
(b*x+a)))^2+I*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))*csgn((I*exp(2*I*(b*x+a
))+exp(2*I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))-I*Pi*csgn(1/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)^3-I*Pi*csg
n(I*exp(I*(b*x+a)))^2*csgn(I*exp(2*I*(b*x+a)))-I*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn
(I/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))+I*Pi*csgn(1/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)^2-I*Pi*csgn((I*
exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))^2-I*Pi*csgn(I*d/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x
+a)))*csgn(1/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)-I*Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(I*exp(2*I*(b*x
+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*I*(b*x+a))-1))^2+I*Pi*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I)/(exp(2*
I*(b*x+a))-1))^3-I*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d-I))*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*
x+a))*d-I)/(exp(2*I*(b*x+a))-1))^2-I*Pi*csgn(I*d)*csgn(I/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))*csgn(I*d/(exp(
2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))+2*ln(d))*x^3+1/8/b^3/(I+d)*polylog(4,-I*(I+d)*exp(2*I*(b*x+a)))
Fricas [A] (verification not implemented)
none
Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06
\[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=\frac {2 i \, b^{4} x^{4} - 4 \, b^{3} x^{3} \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-{\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 2 i \, a^{4} + 4 \, a^{3} \log \left (\frac {{\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}{d + i}\right ) - 6 \, b x {\rm polylog}\left (3, {\left (-i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 3 i \, {\rm polylog}\left (4, {\left (-i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{3}}
\]
[In]
integrate(x^2*arccoth(1-I*d-d*cot(b*x+a)),x, algorithm="fricas")
[Out]
1/24*(2*I*b^4*x^4 - 4*b^3*x^3*log(d*e^(2*I*b*x + 2*I*a)/((d + I)*e^(2*I*b*x + 2*I*a) - I)) + 6*I*b^2*x^2*dilog
(-(I*d - 1)*e^(2*I*b*x + 2*I*a)) - 2*I*a^4 + 4*a^3*log(((d + I)*e^(2*I*b*x + 2*I*a) - I)/(d + I)) - 6*b*x*poly
log(3, (-I*d + 1)*e^(2*I*b*x + 2*I*a)) - 4*(b^3*x^3 + a^3)*log((I*d - 1)*e^(2*I*b*x + 2*I*a) + 1) - 3*I*polylo
g(4, (-I*d + 1)*e^(2*I*b*x + 2*I*a)))/b^3
Sympy [F]
\[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=- \int x^{2} \operatorname {acoth}{\left (d \cot {\left (a + b x \right )} + i d - 1 \right )}\, dx
\]
[In]
integrate(x**2*acoth(1-I*d-d*cot(b*x+a)),x)
[Out]
-Integral(x**2*acoth(d*cot(a + b*x) + I*d - 1), x)
Maxima [B] (verification not implemented)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf
count of optimal. 345 vs. \(2 (120) = 240\).
Time = 0.24 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.04
\[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=-\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {arcoth}\left (d \cot \left (b x + a\right ) + i \, d - 1\right )}{b^{2}} + \frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), -d \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left ({\left (-i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, d \sin \left (2 \, b x + 2 \, a\right ) - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (-i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (-i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b}
\]
[In]
integrate(x^2*arccoth(1-I*d-d*cot(b*x+a)),x, algorithm="maxima")
[Out]
-1/36*(12*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arccoth(d*cot(b*x + a) + I*d - 1)/b^2 + (-3*I*(b*x
+ a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b*x + a)^2*a^2 - 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*I*(b*x + a)*a
^2)*arctan2(-d*cos(2*b*x + 2*a) + sin(2*b*x + 2*a), -d*sin(2*b*x + 2*a) - cos(2*b*x + 2*a) + 1) - 3*(4*I*(b*x
+ a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilog((-I*d + 1)*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a)^3 - 9*(b*x + a)^2*a +
9*(b*x + a)*a^2)*log((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 + 1)*sin(2*b*x + 2*a)^2 - 2*d*sin(2*b*x + 2*a) - 2*c
os(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3, (-I*d + 1)*e^(2*I*b*x + 2*I*a)) + 6*I*polylog(4, (-I*d + 1)*e^
(2*I*b*x + 2*I*a)))/b^2)/b
Giac [F]
\[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=\int { x^{2} \operatorname {arcoth}\left (-d \cot \left (b x + a\right ) - i \, d + 1\right ) \,d x }
\]
[In]
integrate(x^2*arccoth(1-I*d-d*cot(b*x+a)),x, algorithm="giac")
[Out]
integrate(x^2*arccoth(-d*cot(b*x + a) - I*d + 1), x)
Mupad [F(-1)]
Timed out. \[
\int x^2 \coth ^{-1}(1-i d-d \cot (a+b x)) \, dx=\int -x^2\,\mathrm {acoth}\left (d\,\mathrm {cot}\left (a+b\,x\right )-1+d\,1{}\mathrm {i}\right ) \,d x
\]
[In]
int(-x^2*acoth(d*1i + d*cot(a + b*x) - 1),x)
[Out]
int(-x^2*acoth(d*1i + d*cot(a + b*x) - 1), x)