∫ x cot−1(c − (i − c) coth(a + bx))dx

3.214 \(\int x \cot ^{-1}(c-(i-c) \coth (a+b x)) \, dx\)

Optimal. Leaf size=116 \[ -\frac{i \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2}+\frac{i x \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \cot ^{-1}(c-(-c+i) \coth (a+b x))-\frac{1}{6} i b x^3 \]

[Out]

(-I/6)*b*x^3 + (x^2*ArcCot[c - (I - c)*Coth[a + b*x]])/2 + (I/4)*x^2*Log[1 + I*c*E^(2*a + 2*b*x)] + ((I/4)*x*P

olyLog[2, (-I)*c*E^(2*a + 2*b*x)])/b - ((I/8)*PolyLog[3, (-I)*c*E^(2*a + 2*b*x)])/b^2

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Rubi [A] time = 0.205318, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5198, 2184, 2190, 2531, 2282, 6589} \[ -\frac{i \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2}+\frac{i x \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \cot ^{-1}(c-(-c+i) \coth (a+b x))-\frac{1}{6} i b x^3 \]

Antiderivative was successfully veriﬁed.

[In]

Int[x*ArcCot[c - (I - c)*Coth[a + b*x]],x]

[Out]

(-I/6)*b*x^3 + (x^2*ArcCot[c - (I - c)*Coth[a + b*x]])/2 + (I/4)*x^2*Log[1 + I*c*E^(2*a + 2*b*x)] + ((I/4)*x*P

olyLog[2, (-I)*c*E^(2*a + 2*b*x)])/b - ((I/8)*PolyLog[3, (-I)*c*E^(2*a + 2*b*x)])/b^2

Rule 5198

Int[ArcCot[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m

+ 1)*ArcCot[c + d*Coth[a + b*x]])/(f*(m + 1)), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d - c*E^(2

*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - 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 \cot ^{-1}(c-(i-c) \coth (a+b x)) \, dx &=\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))+\frac{1}{2} b \int \frac{x^2}{i-c e^{2 a+2 b x}} \, dx\\ &=-\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))-\frac{1}{2} (i b c) \int \frac{e^{2 a+2 b x} x^2}{i-c e^{2 a+2 b x}} \, dx\\ &=-\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac{1}{2} i \int x \log \left (1+i c e^{2 a+2 b x}\right ) \, dx\\ &=-\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i \int \text{Li}_2\left (-i c e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=-\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=-\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c-(i-c) \coth (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i \text{Li}_3\left (-i c e^{2 a+2 b x}\right )}{8 b^2}\\ \end{align*}

Mathematica [A] time = 0.115977, size = 102, normalized size = 0.88 \[ \frac{i \left (-2 b x \text{PolyLog}\left (2,\frac{i e^{-2 (a+b x)}}{c}\right )-\text{PolyLog}\left (3,\frac{i e^{-2 (a+b x)}}{c}\right )+2 b^2 x^2 \log \left (1-\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{8 b^2}+\frac{1}{2} x^2 \cot ^{-1}(c+(c-i) \coth (a+b x)) \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*ArcCot[c - (I - c)*Coth[a + b*x]],x]

[Out]

(x^2*ArcCot[c + (-I + c)*Coth[a + b*x]])/2 + ((I/8)*(2*b^2*x^2*Log[1 - I/(c*E^(2*(a + b*x)))] - 2*b*x*PolyLog[

2, I/(c*E^(2*(a + b*x)))] - PolyLog[3, I/(c*E^(2*(a + b*x)))]))/b^2

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Maple [C] time = 8.506, size = 1535, normalized size = 13.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(c-(I-c)*coth(b*x+a)),x)

[Out]

1/2*Pi*x^2-1/8*Pi*x^2*csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^2-1/3/b^2/(I-c)*a^3+1/6*b*x^3/(I-c)-1/

2/b/(I-c)*x*a^2-1/8*Pi*x^2*csgn((-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))^3+1/2*I/b*ln(1+I*

c*exp(2*b*x+2*a))*x*a-1/2*I/b*a*ln(1+I*exp(b*x+a)*(I*c)^(1/2))*x-1/2*I/b*a*ln(1-I*exp(b*x+a)*(I*c)^(1/2))*x-1/

8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^2-1/8*Pi*x^2*csg

n((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^3+1/4*I/b^2*ln(1+I*c*exp(2*b*x+2*a))*a^2+1/4*I/b^2*polylog(2,-I

*c*exp(2*b*x+2*a))*a-1/2*I/b^2*a^2*ln(1+I*exp(b*x+a)*(I*c)^(1/2))-1/2*I/b^2*a^2*ln(1-I*exp(b*x+a)*(I*c)^(1/2))

-1/2*I/b^2*a*dilog(1+I*exp(b*x+a)*(I*c)^(1/2))-1/2*I/b^2*a*dilog(1-I*exp(b*x+a)*(I*c)^(1/2))-1/8*I*polylog(3,-

I*c*exp(2*b*x+2*a))/b^2-1/8*Pi*x^2*csgn((-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))^2-1/4*I*x

^2*ln(-2*exp(2*b*x+2*a)*c+2*I)-1/3*I/b^2*c/(I-c)*a^3+1/6*I*b*c/(I-c)*x^3-1/2*I/b*c/(I-c)*x*a^2+1/2*I/b^2*c*a^2

/(I-c)*ln(exp(b*x+a))+1/4*I*x*polylog(2,-I*c*exp(2*b*x+2*a))/b+1/8*Pi*x^2*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*

b*x+2*a)*c))*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))^2-1/8*Pi*x^2*csgn(I*(2*exp(2*

b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))*csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^2+1/8*Pi*x^2*csgn(I*(-2*

I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))*csgn((-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*

b*x+2*a)-1))^2+1/8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^2+1/4

*I*x^2*ln(2*I*exp(2*b*x+2*a)-2*exp(2*b*x+2*a)*c)-1/8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(-2*I*exp(2*b*x+

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

b*x+2*a))-1/8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I))*csgn(I*(2*exp(2*b*x+2*a)*c-2*

I)/(exp(2*b*x+2*a)-1))+1/8*Pi*x^2*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))*csgn((-2

*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))-1/8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*

x+2*a)-1))*csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))+1/2/b^2*a^2/(I-c)*ln(exp(b*x+a))+1/8*Pi*x^2*csgn(

I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)-1))^3-1/8*Pi*x^2*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(e

xp(2*b*x+2*a)-1))^3+1/8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)-1))*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c))*csg

n(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))

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Maxima [A] time = 5.90224, size = 143, normalized size = 1.23 \begin{align*}{\left (\frac{2 \, x^{3}}{3 i \, c + 3} - \frac{2 \, b^{2} x^{2} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) -{\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{3}{\left (-i \, c - 1\right )}}\right )} b{\left (c - i\right )} + \frac{1}{2} \, x^{2} \operatorname{arccot}\left ({\left (c - i\right )} \coth \left (b x + a\right ) + c\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c-(I-c)*coth(b*x+a)),x, algorithm="maxima")

[Out]

(2*x^3/(3*I*c + 3) - (2*b^2*x^2*log(I*c*e^(2*b*x + 2*a) + 1) + 2*b*x*dilog(-I*c*e^(2*b*x + 2*a)) - polylog(3,

-I*c*e^(2*b*x + 2*a)))/(b^3*(2*I*c + 2)))*b*(c - I) + 1/2*x^2*arccot((c - I)*coth(b*x + a) + c)

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Fricas [C] time = 2.26993, size = 717, normalized size = 6.18 \begin{align*} \frac{-2 i \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (\frac{{\left (c - i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} - i}\right ) - 2 i \, a^{3} + 6 i \, b x{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b x{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (b x + a\right )} + i \, \sqrt{-4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (b x + a\right )} - i \, \sqrt{-4 i \, c}}{2 \, c}\right ) +{\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) +{\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 6 i \,{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) - 6 i \,{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c-(I-c)*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(-2*I*b^3*x^3 + 3*I*b^2*x^2*log((c - I)*e^(2*b*x + 2*a)/(c*e^(2*b*x + 2*a) - I)) - 2*I*a^3 + 6*I*b*x*dilo

g(1/2*sqrt(-4*I*c)*e^(b*x + a)) + 6*I*b*x*dilog(-1/2*sqrt(-4*I*c)*e^(b*x + a)) + 3*I*a^2*log(1/2*(2*c*e^(b*x +

a) + I*sqrt(-4*I*c))/c) + 3*I*a^2*log(1/2*(2*c*e^(b*x + a) - I*sqrt(-4*I*c))/c) + (3*I*b^2*x^2 - 3*I*a^2)*log

(1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) + (3*I*b^2*x^2 - 3*I*a^2)*log(-1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) - 6*I*poly

log(3, 1/2*sqrt(-4*I*c)*e^(b*x + a)) - 6*I*polylog(3, -1/2*sqrt(-4*I*c)*e^(b*x + 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*acot(c-(I-c)*coth(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{arccot}\left ({\left (c - i\right )} \coth \left (b x + a\right ) + c\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c-(I-c)*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arccot((c - I)*coth(b*x + a) + c), x)

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