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

Optimal. Leaf size=174 \[ -\frac {i \text {Li}_2\left (\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}+\frac {i \text {Li}_2\left (\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}-\frac {1}{2} i x \log \left (1-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )+\frac {1}{2} i x \log \left (1-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+x \cot ^{-1}(d \coth (a+b x)+c) \]

[Out]

x*arccot(c+d*coth(b*x+a))-1/2*I*x*ln(1-(I-c-d)*exp(2*b*x+2*a)/(I-c+d))+1/2*I*x*ln(1-(I+c+d)*exp(2*b*x+2*a)/(I+
c-d))-1/4*I*polylog(2,(I-c-d)*exp(2*b*x+2*a)/(I-c+d))/b+1/4*I*polylog(2,(I+c+d)*exp(2*b*x+2*a)/(I+c-d))/b

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5194, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}+\frac {i \text {PolyLog}\left (2,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}-\frac {1}{2} i x \log \left (1-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )+\frac {1}{2} i x \log \left (1-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+x \cot ^{-1}(d \coth (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[c + d*Coth[a + b*x]],x]

[Out]

x*ArcCot[c + d*Coth[a + b*x]] - (I/2)*x*Log[1 - ((I - c - d)*E^(2*a + 2*b*x))/(I - c + d)] + (I/2)*x*Log[1 - (
(I + c + d)*E^(2*a + 2*b*x))/(I + c - d)] - ((I/4)*PolyLog[2, ((I - c - d)*E^(2*a + 2*b*x))/(I - c + d)])/b +
((I/4)*PolyLog[2, ((I + c + d)*E^(2*a + 2*b*x))/(I + c - d)])/b

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

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

Rule 5194

Int[ArcCot[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcCot[c + d*Coth[a + b*x]], x] + (Dis
t[I*b*(I - c - d), Int[(x*E^(2*a + 2*b*x))/(I - c + d - (I - c - d)*E^(2*a + 2*b*x)), x], x] - Dist[I*b*(I + c
+ d), Int[(x*E^(2*a + 2*b*x))/(I + c - d - (I + c + d)*E^(2*a + 2*b*x)), x], x]) /; FreeQ[{a, b, c, d}, x] &&
NeQ[(c - d)^2, -1]

Rubi steps

\begin {align*} \int \cot ^{-1}(c+d \coth (a+b x)) \, dx &=x \cot ^{-1}(c+d \coth (a+b x))+(b (1-i (c+d))) \int \frac {e^{2 a+2 b x} x}{i+c-d+(-i-c-d) e^{2 a+2 b x}} \, dx-(b (1+i (c+d))) \int \frac {e^{2 a+2 b x} x}{i-c+d+(-i+c+d) e^{2 a+2 b x}} \, dx\\ &=x \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{2} i x \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{2} i x \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {1}{2} i \int \log \left (1+\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx+\frac {1}{2} i \int \log \left (1+\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx\\ &=x \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{2} i x \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{2} i x \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {(-i-c-d) x}{i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}+\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {(-i+c+d) x}{i-c+d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=x \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{2} i x \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{2} i x \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i \text {Li}_2\left (\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i \text {Li}_2\left (\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.31, size = 287, normalized size = 1.65 \[ x \cot ^{-1}(d \coth (a+b x)+c)-\frac {d \text {Li}_2\left (\frac {\left (c^2+2 d c+d^2+1\right ) e^{2 (a+b x)}}{c^2-d^2+2 \sqrt {-d^2}+1}\right )-d \text {Li}_2\left (-\frac {\left (c^2+2 d c+d^2+1\right ) e^{2 (a+b x)}}{-c^2+d^2+2 \sqrt {-d^2}-1}\right )+2 d (a+b x) \log \left (1-\frac {\left ((c+d)^2+1\right ) e^{2 (a+b x)}}{c^2-d^2+2 \sqrt {-d^2}+1}\right )-2 d (a+b x) \log \left (\frac {\left ((c+d)^2+1\right ) e^{2 (a+b x)}}{-c^2+d^2+2 \sqrt {-d^2}-1}+1\right )+4 a \sqrt {-d^2} \tan ^{-1}\left (\frac {-\left (c^2+2 c d+d^2+1\right ) e^{2 (a+b x)}+c^2-d^2+1}{2 d}\right )}{4 b \sqrt {-d^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[c + d*Coth[a + b*x]],x]

[Out]

x*ArcCot[c + d*Coth[a + b*x]] - (4*a*Sqrt[-d^2]*ArcTan[(1 + c^2 - d^2 - (1 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)
))/(2*d)] + 2*d*(a + b*x)*Log[1 - ((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2])] - 2*d*(a +
b*x)*Log[1 + ((1 + (c + d)^2)*E^(2*(a + b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2])] + d*PolyLog[2, ((1 + c^2 + 2*
c*d + d^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2])] - d*PolyLog[2, -(((1 + c^2 + 2*c*d + d^2)*E^(2*(a
+ b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2]))])/(4*b*Sqrt[-d^2])

________________________________________________________________________________________

fricas [B]  time = 1.41, size = 839, normalized size = 4.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x*arctan(sinh(b*x + a)/(d*cosh(b*x + a) + c*sinh(b*x + a))) + I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(
b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + (c^2 - d^2 - 2*I*d + 1)*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/
(c^2 - 2*c*d + d^2 + 1))) + I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b
*x + a) - (c^2 - d^2 - 2*I*d + 1)*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))) - I*a*log(2*(c^2
+ 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + (c^2 - d^2 + 2*I*d + 1)*sqrt((4*c
^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))) - I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 +
2*c*d + d^2 + 1)*sinh(b*x + a) - (c^2 - d^2 + 2*I*d + 1)*sqrt((4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2
+ 1))) + (-I*b*x - I*a)*log(1/2*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sin
h(b*x + a)) + 1) + (-I*b*x - I*a)*log(-1/2*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x
+ a) + sinh(b*x + a)) + 1) + (I*b*x + I*a)*log(1/2*sqrt((4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*
(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b*x + I*a)*log(-1/2*sqrt((4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d +
d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - I*dilog(1/2*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d
^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) - I*dilog(-1/2*sqrt((4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 +
1))*(cosh(b*x + a) + sinh(b*x + a))) + I*dilog(1/2*sqrt((4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*
(cosh(b*x + a) + sinh(b*x + a))) + I*dilog(-1/2*sqrt((4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*(cos
h(b*x + a) + sinh(b*x + a))))/b

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arccot}\left (d \coth \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.58, size = 350, normalized size = 2.01 \[ -\frac {\mathrm {arccot}\left (c +d \coth \left (b x +a \right )\right ) \ln \left (d \coth \left (b x +a \right )-d \right )}{2 b}+\frac {\mathrm {arccot}\left (c +d \coth \left (b x +a \right )\right ) \ln \left (d \coth \left (b x +a \right )+d \right )}{2 b}+\frac {i \ln \left (d \coth \left (b x +a \right )-d \right ) \ln \left (\frac {-d \coth \left (b x +a \right )+i-c}{i-c -d}\right )}{4 b}-\frac {i \ln \left (d \coth \left (b x +a \right )-d \right ) \ln \left (\frac {i+d \coth \left (b x +a \right )+c}{i+c +d}\right )}{4 b}+\frac {i \dilog \left (\frac {-d \coth \left (b x +a \right )+i-c}{i-c -d}\right )}{4 b}-\frac {i \dilog \left (\frac {i+d \coth \left (b x +a \right )+c}{i+c +d}\right )}{4 b}-\frac {i \ln \left (d \coth \left (b x +a \right )+d \right ) \ln \left (\frac {-d \coth \left (b x +a \right )+i-c}{i-c +d}\right )}{4 b}+\frac {i \ln \left (d \coth \left (b x +a \right )+d \right ) \ln \left (\frac {i+d \coth \left (b x +a \right )+c}{i+c -d}\right )}{4 b}-\frac {i \dilog \left (\frac {-d \coth \left (b x +a \right )+i-c}{i-c +d}\right )}{4 b}+\frac {i \dilog \left (\frac {i+d \coth \left (b x +a \right )+c}{i+c -d}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*arccot(c+d*coth(b*x+a))*ln(d*coth(b*x+a)-d)+1/2/b*arccot(c+d*coth(b*x+a))*ln(d*coth(b*x+a)+d)+1/4*I/b*l
n(d*coth(b*x+a)-d)*ln((-d*coth(b*x+a)+I-c)/(I-c-d))-1/4*I/b*ln(d*coth(b*x+a)-d)*ln((I+d*coth(b*x+a)+c)/(I+c+d)
)+1/4*I/b*dilog((-d*coth(b*x+a)+I-c)/(I-c-d))-1/4*I/b*dilog((I+d*coth(b*x+a)+c)/(I+c+d))-1/4*I/b*ln(d*coth(b*x
+a)+d)*ln((-d*coth(b*x+a)+I-c)/(I-c+d))+1/4*I/b*ln(d*coth(b*x+a)+d)*ln((I+d*coth(b*x+a)+c)/(I+c-d))-1/4*I/b*di
log((-d*coth(b*x+a)+I-c)/(I-c+d))+1/4*I/b*dilog((I+d*coth(b*x+a)+c)/(I+c-d))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -4 \, b d \int \frac {x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} - 2 \, c d + d^{2} + {\left (c^{2} e^{\left (4 \, a\right )} + 2 \, c d e^{\left (4 \, a\right )} + d^{2} e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )} - 2 \, {\left (c^{2} e^{\left (2 \, a\right )} - d^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}\,{d x} + x \arctan \left (e^{\left (2 \, b x + 2 \, a\right )} - 1, {\left (c e^{\left (2 \, a\right )} + d e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} - c + d\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*d*integrate(x*e^(2*b*x + 2*a)/(c^2 - 2*c*d + d^2 + (c^2*e^(4*a) + 2*c*d*e^(4*a) + d^2*e^(4*a) + e^(4*a))*
e^(4*b*x) - 2*(c^2*e^(2*a) - d^2*e^(2*a) + e^(2*a))*e^(2*b*x) + 1), x) + x*arctan2(e^(2*b*x + 2*a) - 1, (c*e^(
2*a) + d*e^(2*a))*e^(2*b*x) - c + d)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {acot}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(c + d*coth(a + b*x)),x)

[Out]

int(acot(c + d*coth(a + b*x)), x)

________________________________________________________________________________________

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(acot(c+d*coth(b*x+a)),x)

[Out]

Timed out

________________________________________________________________________________________