3.241 \(\int x \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx\)
Optimal. Leaf size=133 \[ -\frac{\text{PolyLog}\left (3,-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^2}+\frac{i x \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{2} x^2 \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{6} i b x^3 \]
[Out]
(I/6)*b*x^3 + (x^2*ArcCoth[1 - I*d + d*Tan[a + b*x]])/2 - (x^2*Log[1 + (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/4 +
((I/4)*x*PolyLog[2, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b - PolyLog[3, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x)
)]/(8*b^2)
________________________________________________________________________________________
Rubi [A] time = 0.253474, antiderivative size = 133, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{6264, 2184, 2190, 2531, 2282, 6589} \[ -\frac{\text{PolyLog}\left (3,-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^2}+\frac{i x \text{PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{2} x^2 \coth ^{-1}(d \tan (a+b x)-i d+1)+\frac{1}{6} i b x^3 \]
Antiderivative was successfully verified.
[In]
Int[x*ArcCoth[1 - I*d + d*Tan[a + b*x]],x]
[Out]
(I/6)*b*x^3 + (x^2*ArcCoth[1 - I*d + d*Tan[a + b*x]])/2 - (x^2*Log[1 + (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/4 +
((I/4)*x*PolyLog[2, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b - PolyLog[3, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x)
)]/(8*b^2)
Rule 6264
Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCoth[c + d*Tan[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 2184
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 2190
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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2531
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 2282
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 6589
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]
Rubi steps
\begin{align*} \int x \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx &=\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))+\frac{1}{2} (i b) \int \frac{x^2}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{2} (b (i+d)) \int \frac{e^{2 i a+2 i b x} x^2}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{1}{2} \int x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right ) \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{i \int \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right ) \, dx}{4 b}\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(i (i+d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \coth ^{-1}(1-i d+d \tan (a+b x))-\frac{1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac{i x \text{Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac{\text{Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.10564, size = 119, normalized size = 0.89 \[ \frac{1}{2} x^2 \coth ^{-1}(d \tan (a+b x)-i d+1)-\frac{2 i b x \text{PolyLog}\left (2,-\frac{i e^{-2 i (a+b x)}}{d+i}\right )+\text{PolyLog}\left (3,-\frac{i e^{-2 i (a+b x)}}{d+i}\right )+2 b^2 x^2 \log \left (1+\frac{i e^{-2 i (a+b x)}}{d+i}\right )}{8 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*ArcCoth[1 - I*d + d*Tan[a + b*x]],x]
[Out]
(x^2*ArcCoth[1 - I*d + d*Tan[a + b*x]])/2 - (2*b^2*x^2*Log[1 + I/((I + d)*E^((2*I)*(a + b*x)))] + (2*I)*b*x*Po
lyLog[2, (-I)/((I + d)*E^((2*I)*(a + b*x)))] + PolyLog[3, (-I)/((I + d)*E^((2*I)*(a + b*x)))])/(8*b^2)
________________________________________________________________________________________
Maple [C] time = 17.08, size = 2249, normalized size = 16.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*arccoth(1-I*d+d*tan(b*x+a)),x)
[Out]
1/8*I*x^2*Pi*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))^3-1/8*I*x^2*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+1/8*I*x^2*Pi*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^3+1/4*x^2*ln(I
*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)-1/8*I*x^2*Pi*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+1/8*I*x^2
*Pi*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^3-1/8*I/b^2/(I+d)*polylog(3,I*(I+d)*exp(2*I*(b*x+a)))-1/4*d/
(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^2+1/2/b^2*a/(I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/2/b^2*a/
(I+d)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/6*I*b*x^3-1/4*I*x^2*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*
I*(b*x+a)))^2-1/8*I*x^2*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))-1/4/b/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*x-1/4
/b^2/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*a-1/8/b^2*d/(I+d)*polylog(3,I*(I+d)*exp(2*I*(b*x+a)))-1/8*I*x^2
*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+1/8*I*x^2*Pi*csgn((I*exp(2*I*(b*x
+a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))^2-1/4*x^2*ln(d)+1/8*I*x^2*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*
exp(2*I*(b*x+a)))+1/8*I*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))^3-1/4*I/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x^2-1/2*x
^2*ln(exp(I*(b*x+a)))+1/2*I/b*a/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x+1/2*I/b*a/(I+d)*ln(1+I*exp(I*(
b*x+a))*(-I*(I+d))^(1/2))*x-1/2/b*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x*a+1/2/b*a*d/(I+d)*ln(1-I*exp(I*(b*x
+a))*(-I*(I+d))^(1/2))*x+1/2/b*a*d/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x-1/8*I*x^2*Pi*csgn(I*d/(exp(
2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+1/8*I*x^2*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+1/8*I*x^2*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))^2-1/2*I/b^2*a*d/(I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2*
I/b^2*a*d/(I+d)*dilog(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/8*I*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*
I*(b*x+a))/(exp(2*I*(b*x+a))+1))^2-1/8*I*x^2*Pi*csgn(I*d)*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+1/
8*I*x^2*Pi*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))*csgn(d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))-1/8*I
*x^2*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))^2-1/4/b^2*a^2*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)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^2+1/2/b^2*a^2*d/(I+d)*
ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/2/b^2*a^2*d/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/4*I/b^2*
a^2/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)-1/2*I/b/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*x*a+1/4*I/b
*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*x+1/4*I/b^2*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*a+1/8*I*x
^2*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))+1/8*
I*x^2*Pi*csgn(I*d)*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)
))-1/4*I/b^2/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^2+1/2*I/b^2*a^2/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/
2))-1/8*I*x^2*Pi*csgn(I*exp(2*I*(b*x+a))/(exp(2*I*(b*x+a))+1))*csgn(I*d/(exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))
^2+1/8*I*x^2*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+1/2*I/b^2*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/8*I*x^2*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))
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Maxima [B] time = 1.1443, size = 333, normalized size = 2.5 \begin{align*} \frac{\frac{12 \,{\left ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\right )} \operatorname{arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{b} - \frac{-4 i \,{\left (b x + a\right )}^{3} + 12 i \,{\left (b x + a\right )}^{2} a - 6 i \, b x{\rm Li}_2\left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (6 i \,{\left (b x + a\right )}^{2} - 12 i \,{\left (b x + a\right )} a\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 ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\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 \,{\rm Li}_{3}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b}}{24 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="maxima")
[Out]
1/24*(12*((b*x + a)^2 - 2*(b*x + a)*a)*arccoth(d*tan(b*x + a) - I*d + 1)/b - (-4*I*(b*x + a)^3 + 12*I*(b*x + a
)^2*a - 6*I*b*x*dilog((I*d - 1)*e^(2*I*b*x + 2*I*a)) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a)*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*((b*x + a)^2 - 2*(b*x + a)*a)*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*cos(2*b*x + 2*a) + 1)
+ 3*polylog(3, (I*d - 1)*e^(2*I*b*x + 2*I*a)))/b)/b
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Fricas [C] time = 1.73063, size = 811, normalized size = 6.1 \begin{align*} \frac{2 i \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (\frac{{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 2 i \, a^{3} + 6 i \, b x{\rm Li}_2\left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b x{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 3 \, a^{2} \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) - 3 \, a^{2} \log \left (\frac{{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{4 i \, d - 4}}{2 \, d + 2 i}\right ) - 3 \,{\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \,{\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 6 \,{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 6 \,{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="fricas")
[Out]
1/12*(2*I*b^3*x^3 + 3*b^2*x^2*log(((d + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/d) + 2*I*a^3 + 6*I*b*
x*dilog(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) + 6*I*b*x*dilog(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 3*a^2*log
(((2*d + 2*I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(2*d + 2*I)) - 3*a^2*log(((2*d + 2*I)*e^(I*b*x + I*a) - I*s
qrt(4*I*d - 4))/(2*d + 2*I)) - 3*(b^2*x^2 - a^2)*log(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 3*(b^2*x^2 - a
^2)*log(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 6*polylog(3, 1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 6*poly
log(3, -1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)))/b^2
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*acoth(1-I*d+d*tan(b*x+a)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1-I*d+d*tan(b*x+a)),x, algorithm="giac")
[Out]
integrate(x*arccoth(d*tan(b*x + a) - I*d + 1), x)