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

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

[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

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Rubi [A]  time = 0.156441, antiderivative size = 94, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [B]  time = 2.82527, size = 723, normalized size = 7.69 \[ x \coth ^{-1}(d (-\tan (a+b x))+i d+1)-\frac{x \sec (a+b x) (\cos (b x)+i \sin (b x)) (\sin (b x)+i \cos (b x)) \left (-\text{PolyLog}\left (2,\frac{1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+(2+i d) \sin (a))\right )+\text{PolyLog}\left (2,\frac{\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+\text{PolyLog}(2,-\cos (2 b x)+i \sin (2 b x))+\log (1-i \tan (b x)) \log \left (\frac{(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )-\log (1+i \tan (b x)) \log \left (\frac{\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\tan (a+b x)-i) (i d \sin (a+b x)+(d-2 i) \cos (a+b x)) \left (-\frac{\sec ^2(b x) \log \left (\frac{\sec (b x) (-d \sin (a+b x)+(2+i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{\tan (b x)-i}+\frac{\sec ^2(b x) \log \left (1-\frac{1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+(2+i d) \sin (a))\right )}{\tan (b x)-i}+\frac{\sec ^2(b x) \log \left (\frac{(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d-2 i) \cos (a+b x))}{2 (d-i)}\right )}{\tan (b x)+i}+\frac{i \sec (b x) (d \cos (a)+(2+i d) \sin (a)) \log (1-i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}+\frac{\sec (b x) ((d-2 i) \sin (a)-i d \cos (a)) \log (1+i \tan (b x))}{i d \sin (a+b x)+(d-2 i) \cos (a+b x)}-(\tan (b x)-i) \log \left (1-\frac{\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d-i)}\right )+2 i b x (\tan (b x)+i)\right )} \]

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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Maple [B]  time = 0.178, size = 297, normalized size = 3.2 \begin{align*}{\frac{{\frac{i}{2}}{\rm arccoth} \left (1+id-d\tan \left ( bx+a \right ) \right )\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}}-{\frac{{\frac{i}{2}}{\rm arccoth} \left (1+id-d\tan \left ( bx+a \right ) \right )\ln \left ( id-d\tan \left ( bx+a \right ) \right ) }{b}}-{\frac{{\frac{i}{8}} \left ( \ln \left ( id-d\tan \left ( bx+a \right ) \right ) \right ) ^{2}}{b}}+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ( 1+{\frac{i}{2}}d-{\frac{d\tan \left ( bx+a \right ) }{2}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( id-d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ( 1+{\frac{i}{2}}d-{\frac{d\tan \left ( bx+a \right ) }{2}} \right ) }+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{{\frac{i}{2}} \left ( -id+d\tan \left ( bx+a \right ) \right ) }{d}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ({\frac{{\frac{i}{2}} \left ( -id+d\tan \left ( bx+a \right ) \right ) }{d}} \right ) }-{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{-2-id+d\tan \left ( bx+a \right ) }{-2\,id-2}} \right ) }-{\frac{{\frac{i}{4}}\ln \left ( id+d\tan \left ( bx+a \right ) \right ) }{b}\ln \left ({\frac{-2-id+d\tan \left ( bx+a \right ) }{-2\,id-2}} \right ) } \end{align*}

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(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)-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))

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Maxima [B]  time = 1.59644, size = 355, normalized size = 3.78 \begin{align*} -\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} \end{align*}

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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Fricas [B]  time = 1.66666, size = 610, normalized size = 6.49 \begin{align*} \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} \end{align*}

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

[Out]

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