3.246 \(\int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx\)

Optimal. Leaf size=94 \[ \frac {i \text {Li}_2\left (-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\tan (a+b x))+i d+1)+\frac {1}{2} i b x^2 \]

[Out]

1/2*I*b*x^2+x*arccoth(1+I*d-d*tan(b*x+a))-1/2*x*ln(1+(1+I*d)*exp(2*I*a+2*I*b*x))+1/4*I*polylog(2,-(1+I*d)*exp(
2*I*a+2*I*b*x))/b

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Rubi [A]  time = 0.16, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6256, 2184, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,-(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\tan (a+b x))+i d+1)+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[1 + I*d - d*Tan[a + b*x]],x]

[Out]

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

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 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 6256

Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCoth[c + d*Tan[a + b*x]], x] + Dist
[I*b, Int[x/(c + I*d + c*E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c + I*d)^2, 1]

Rubi steps

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

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Mathematica [B]  time = 3.24, size = 723, normalized size = 7.69 \[ x \coth ^{-1}(d (-\tan (a+b x))+i d+1)-\frac {x \sec (a+b x) (\cos (b x)+i \sin (b x)) (\sin (b x)+i \cos (b x)) \left (-\text {Li}_2\left (\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+(i d+2) \sin (a)) (\tan (b x)-i)\right )+\text {Li}_2\left (\frac {\sec (b x) (d \cos (a)+(i d+2) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+\log (1-i \tan (b x)) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )-\log (1+i \tan (b x)) \log \left (\frac {\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )+\text {Li}_2(i \sin (2 b x)-\cos (2 b x))-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\tan (a+b x)-i) (i d \sin (a+b x)+(d-2 i) \cos (a+b x)) \left (-\frac {\sec ^2(b x) \log \left (\frac {\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{\tan (b x)-i}+\frac {\sec ^2(b x) \log \left (1-\frac {1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+(2+i d) \sin (a))\right )}{\tan (b x)-i}+\frac {\sec ^2(b x) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )}{\tan (b x)+i}+\frac {i \sec (b x) (d \cos (a)+(2+i d) \sin (a)) \log (1-i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}+\frac {\sec (b x) ((d-2 i) \sin (a)-i d \cos (a)) \log (1+i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}-(\tan (b x)-i) \log \left (1-\frac {\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+2 i b x (\tan (b x)+i)\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[1 + I*d - d*Tan[a + b*x]],x]

[Out]

x*ArcCoth[1 + I*d - d*Tan[a + b*x]] - (x*((-2*I)*b*x*Log[2*Cos[b*x]*(Cos[b*x] - I*Sin[b*x])] + Log[(Sec[b*x]*(
Cos[a] - I*Sin[a])*((-2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]))/(2*(-I + d))]*Log[1 - I*Tan[b*x]] - Log[(Sec[
b*x]*((2 + I*d)*Cos[a + b*x] - d*Sin[a + b*x]))/(2*Cos[a] - (2*I)*Sin[a])]*Log[1 + I*Tan[b*x]] + PolyLog[2, -C
os[2*b*x] + I*Sin[2*b*x]] + PolyLog[2, (Sec[b*x]*(d*Cos[a] + (2 + I*d)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x])
)/(2*(-I + d))] - PolyLog[2, ((Cos[a] + I*Sin[a])*(d*Cos[a] + (2 + I*d)*Sin[a])*(-I + Tan[b*x]))/2])*Sec[a + b
*x]*(Cos[b*x] + I*Sin[b*x])*(I*Cos[b*x] + Sin[b*x]))/(((-2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x])*((I*Log[1 -
I*Tan[b*x]]*Sec[b*x]*(d*Cos[a] + (2 + I*d)*Sin[a]))/((-2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]) + (Log[1 + I
*Tan[b*x]]*Sec[b*x]*((-I)*d*Cos[a] + (-2*I + d)*Sin[a]))/((-2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]) - (Log[(
Sec[b*x]*((2 + I*d)*Cos[a + b*x] - d*Sin[a + b*x]))/(2*Cos[a] - (2*I)*Sin[a])]*Sec[b*x]^2)/(-I + Tan[b*x]) + (
Log[1 - ((Cos[a] + I*Sin[a])*(d*Cos[a] + (2 + I*d)*Sin[a])*(-I + Tan[b*x]))/2]*Sec[b*x]^2)/(-I + Tan[b*x]) - L
og[1 - (Sec[b*x]*(d*Cos[a] + (2 + I*d)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x]))/(2*(-I + d))]*(-I + Tan[b*x])
+ (Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((-2*I + d)*Cos[a + b*x] + I*d*Sin[a + b*x]))/(2*(-I + d))]*Sec[b*x]^2)/(
I + Tan[b*x]) + (2*I)*b*x*(I + Tan[b*x]))*(-I + Tan[a + b*x]))

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fricas [B]  time = 0.48, size = 222, normalized size = 2.36 \[ \frac {i \, b^{2} x^{2} - b x \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) - i \, a^{2} - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac {{\left (2 \, d - 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, d - 2 i}\right ) + a \log \left (\frac {{\left (2 \, d - 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, d - 2 i}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (-d \tan \left (b x + a\right ) + i \, d + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(-d*tan(b*x + a) + I*d + 1), x)

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maple [B]  time = 0.51, size = 297, normalized size = 3.16 \[ \frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) \ln \left (i d +d \tan \left (b x +a \right )\right )}{2 b}-\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) \ln \left (i d -d \tan \left (b x +a \right )\right )}{2 b}-\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{8 b}+\frac {i \dilog \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b}+\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b}-\frac {i \dilog \left (\frac {-2-i d +d \tan \left (b x +a \right )}{-2 i d -2}\right )}{4 b}-\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {-2-i d +d \tan \left (b x +a \right )}{-2 i d -2}\right )}{4 b}+\frac {i \dilog \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b}+\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1+I*d-d*tan(b*x+a)),x)

[Out]

1/2*I/b*arccoth(1+I*d-d*tan(b*x+a))*ln(I*d+d*tan(b*x+a))-1/2*I/b*arccoth(1+I*d-d*tan(b*x+a))*ln(I*d-d*tan(b*x+
a))-1/8*I/b*ln(I*d-d*tan(b*x+a))^2+1/4*I/b*dilog(1+1/2*I*d-1/2*d*tan(b*x+a))+1/4*I/b*ln(I*d-d*tan(b*x+a))*ln(1
+1/2*I*d-1/2*d*tan(b*x+a))-1/4*I/b*dilog((-2-I*d+d*tan(b*x+a))/(-2*I*d-2))-1/4*I/b*ln(I*d+d*tan(b*x+a))*ln((-2
-I*d+d*tan(b*x+a))/(-2*I*d-2))+1/4*I/b*dilog(1/2*I*(-I*d+d*tan(b*x+a))/d)+1/4*I/b*ln(I*d+d*tan(b*x+a))*ln(1/2*
I*(-I*d+d*tan(b*x+a))/d)

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maxima [B]  time = 0.42, size = 263, normalized size = 2.80 \[ -\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right )}{d} - \frac {\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} + d {\left (-\frac {2 i \, {\left (\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (-\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, d - 2 i} + 1\right ) + {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, d - 2 i}\right )\right )}}{d} + \frac {2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (-\frac {1}{2} \, d \tan \left (b x + a\right ) + \frac {1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) + {\rm Li}_2\left (\frac {1}{2} \, d \tan \left (b x + a\right ) - \frac {1}{2} i \, d\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} + 8 \, {\left (b x + a\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="maxima")

[Out]

-1/8*(4*(b*x + a)*d*(log(d*tan(b*x + a) - I*d - 2)/d - log(tan(b*x + a) - I)/d) + d*(-2*I*(log(d*tan(b*x + a)
- I*d - 2)*log(-(I*d*tan(b*x + a) + d - 2*I)/(2*d - 2*I) + 1) + dilog((I*d*tan(b*x + a) + d - 2*I)/(2*d - 2*I)
))/d + (2*I*log(d*tan(b*x + a) - I*d - 2)*log(tan(b*x + a) - I) - I*log(tan(b*x + a) - I)^2)/d - 2*I*(log(-1/2
*d*tan(b*x + a) + 1/2*I*d + 1)*log(tan(b*x + a) - I) + dilog(1/2*d*tan(b*x + a) - 1/2*I*d))/d + 2*I*(log(tan(b
*x + a) - I)*log(-1/2*I*tan(b*x + a) + 1/2) + dilog(1/2*I*tan(b*x + a) + 1/2))/d) + 8*(b*x + a)*arccoth(d*tan(
b*x + a) - I*d - 1))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(d*1i - d*tan(a + b*x) + 1),x)

[Out]

int(acoth(d*1i - d*tan(a + b*x) + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {acoth}{\left (- d \tan {\left (a + b x \right )} + i d + 1 \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(1+I*d-d*tan(b*x+a)),x)

[Out]

Integral(acoth(-d*tan(a + b*x) + I*d + 1), x)

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