3.340 \(\int \tanh ^{-1}(1+i d+d \cot (a+b x)) \, dx\)

Optimal. Leaf size=93 \[ \frac {i \text {Li}_2\left ((i d+1) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )+x \tanh ^{-1}(d \cot (a+b x)+i d+1)+\frac {1}{2} i b x^2 \]

[Out]

1/2*I*b*x^2+x*arctanh(1+I*d+d*cot(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 = 0.15, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6257, 2184, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )+x \tanh ^{-1}(d \cot (a+b x)+i d+1)+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[1 + I*d + d*Cot[a + b*x]],x]

[Out]

(I/2)*b*x^2 + x*ArcTanh[1 + I*d + d*Cot[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 6257

Int[ArcTanh[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcTanh[c + d*Cot[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 \tanh ^{-1}(1+i d+d \cot (a+b x)) \, dx &=x \tanh ^{-1}(1+i d+d \cot (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 \tanh ^{-1}(1+i d+d \cot (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 \tanh ^{-1}(1+i d+d \cot (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 \tanh ^{-1}(1+i d+d \cot (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 \tanh ^{-1}(1+i d+d \cot (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 = 39.25, size = 709, normalized size = 7.62 \[ \frac {x \csc ^2(a+b x) (\cos (b x)-i \sin (b x)) (\cos (b x)+i \sin (b x)) \left (i \text {Li}_2\left (\frac {(\cos (a)-i \sin (a)) ((-i d-2) \cos (a)+d \sin (a)) (\tan (b x)+i)}{2 (d-i)}\right )-i \text {Li}_2\left (\frac {1}{2} \sec (b x) ((i d+2) \cos (a)-d \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )+i \log (1-i \tan (b x)) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (d \cos (a+b x)+(2+i d) \sin (a+b x))}{2 (d-i)}\right )-i \log (1+i \tan (b x)) \log \left (\frac {1}{2} (\sin (a)-i \cos (a)) \sec (b x) (d \cos (a+b x)+(2+i d) \sin (a+b x))\right )+i \text {Li}_2(i \sin (2 b x)-\cos (2 b x))+2 b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\cot (a+b x)+i) (d \cot (a+b x)+i d+2) \left (\frac {(d-2 i) \cos (a+b x) (\log (1-i \tan (b x))-\log (1+i \tan (b x)))}{d \cos (a+b x)+(2+i d) \sin (a+b x)}+\frac {d \sin (a+b x) (\log (1-i \tan (b x))-\log (1+i \tan (b x)))}{(d-2 i) \sin (a+b x)-i d \cos (a+b x)}+\log \left (1+\frac {1}{2} \sec (b x) (d \sin (a)+(-2-i d) \cos (a)) (\cos (a+b x)+i \sin (a+b x))\right )-\log \left (\frac {1}{2} (\sin (a)-i \cos (a)) \sec (b x) (d \cos (a+b x)+(2+i d) \sin (a+b x))\right )-i \tan (b x) \log \left (1+\frac {1}{2} \sec (b x) (d \sin (a)+(-2-i d) \cos (a)) (\cos (a+b x)+i \sin (a+b x))\right )+i \tan (b x) \log \left (\frac {1}{2} (\sin (a)-i \cos (a)) \sec (b x) (d \cos (a+b x)+(2+i d) \sin (a+b x))\right )+2 i b x+2 b x \tan (b x)-i \tan (b x) \log (1-i \tan (b x))+i \tan (b x) \log (1+i \tan (b x))\right )}+x \tanh ^{-1}(d \cot (a+b x)+i d+1) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[1 + I*d + d*Cot[a + b*x]],x]

[Out]

x*ArcTanh[1 + I*d + d*Cot[a + b*x]] + (x*Csc[a + b*x]^2*(2*b*x*Log[2*Cos[b*x]*(Cos[b*x] - I*Sin[b*x])] + I*Log
[(Sec[b*x]*(Cos[a] - I*Sin[a])*(d*Cos[a + b*x] + (2 + I*d)*Sin[a + b*x]))/(2*(-I + d))]*Log[1 - I*Tan[b*x]] -
I*Log[(Sec[b*x]*((-I)*Cos[a] + Sin[a])*(d*Cos[a + b*x] + (2 + I*d)*Sin[a + b*x]))/2]*Log[1 + I*Tan[b*x]] + I*P
olyLog[2, -Cos[2*b*x] + I*Sin[2*b*x]] - I*PolyLog[2, (Sec[b*x]*((2 + I*d)*Cos[a] - d*Sin[a])*(Cos[a + b*x] + I
*Sin[a + b*x]))/2] + I*PolyLog[2, ((Cos[a] - I*Sin[a])*((-2 - I*d)*Cos[a] + d*Sin[a])*(I + Tan[b*x]))/(2*(-I +
d))])*(Cos[b*x] - I*Sin[b*x])*(Cos[b*x] + I*Sin[b*x]))/((I + Cot[a + b*x])*(2 + I*d + d*Cot[a + b*x])*((2*I)*
b*x + Log[1 + (Sec[b*x]*((-2 - I*d)*Cos[a] + d*Sin[a])*(Cos[a + b*x] + I*Sin[a + b*x]))/2] - Log[(Sec[b*x]*((-
I)*Cos[a] + Sin[a])*(d*Cos[a + b*x] + (2 + I*d)*Sin[a + b*x]))/2] + ((-2*I + d)*Cos[a + b*x]*(Log[1 - I*Tan[b*
x]] - Log[1 + I*Tan[b*x]]))/(d*Cos[a + b*x] + (2 + I*d)*Sin[a + b*x]) + (d*(Log[1 - I*Tan[b*x]] - Log[1 + I*Ta
n[b*x]])*Sin[a + b*x])/((-I)*d*Cos[a + b*x] + (-2*I + d)*Sin[a + b*x]) + 2*b*x*Tan[b*x] - I*Log[1 + (Sec[b*x]*
((-2 - I*d)*Cos[a] + d*Sin[a])*(Cos[a + b*x] + I*Sin[a + b*x]))/2]*Tan[b*x] + I*Log[(Sec[b*x]*((-I)*Cos[a] + S
in[a])*(d*Cos[a + b*x] + (2 + I*d)*Sin[a + b*x]))/2]*Tan[b*x] - I*Log[1 - I*Tan[b*x]]*Tan[b*x] + I*Log[1 + I*T
an[b*x]]*Tan[b*x]))

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fricas [A]  time = 0.45, size = 122, normalized size = 1.31 \[ \frac {2 i \, b^{2} x^{2} + 2 \, b x \log \left (-\frac {{\left ({\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) - 2 i \, a^{2} - 2 \, {\left (b x + a\right )} \log \left ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) + 2 \, a \log \left (\frac {{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i}{d - i}\right ) + i \, {\rm Li}_2\left (-{\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="giac")

[Out]

integrate(arctanh(d*cot(b*x + a) + I*d + 1), x)

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maple [B]  time = 0.61, size = 299, normalized size = 3.22 \[ \frac {i \arctanh \left (1+i d +d \cot \left (b x +a \right )\right ) \ln \left (i d -d \cot \left (b x +a \right )\right )}{2 b}-\frac {i \arctanh \left (1+i d +d \cot \left (b x +a \right )\right ) \ln \left (i d +d \cot \left (b x +a \right )\right )}{2 b}+\frac {i \dilog \left (\frac {i \left (-i d -d \cot \left (b x +a \right )\right )}{2 d}\right )}{4 b}+\frac {i \ln \left (i d -d \cot \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \cot \left (b x +a \right )\right )}{2 d}\right )}{4 b}-\frac {i \dilog \left (\frac {-2-i d -d \cot \left (b x +a \right )}{-2 i d -2}\right )}{4 b}-\frac {i \ln \left (i d -d \cot \left (b x +a \right )\right ) \ln \left (\frac {-2-i d -d \cot \left (b x +a \right )}{-2 i d -2}\right )}{4 b}-\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right )^{2}}{8 b}+\frac {i \dilog \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{4 b}+\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(1+I*d+d*cot(b*x+a)),x)

[Out]

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

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maxima [B]  time = 0.49, size = 288, normalized size = 3.10 \[ -\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left ({\left (i \, d + 2\right )} \tan \left (b x + a\right ) + d\right )}{d} - \frac {\log \left (i \, \tan \left (b x + a\right ) + 1\right )}{d}\right )} - d {\left (\frac {2 i \, {\left (\log \left ({\left (i \, d + 2\right )} \tan \left (b x + a\right ) + d\right ) \log \left (\frac {{\left (d - 2 i\right )} \tan \left (b x + a\right ) - i \, d}{2 i \, d + 2} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (d - 2 i\right )} \tan \left (b x + a\right ) - i \, d}{2 i \, d + 2}\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\frac {1}{2} \, {\left (d - 2 i\right )} \tan \left (b x + a\right ) - \frac {1}{2} i \, d\right ) \log \left (i \, \tan \left (b x + a\right ) + 1\right ) + {\rm Li}_2\left (-\frac {1}{2} \, {\left (d - 2 i\right )} \tan \left (b x + a\right ) + \frac {1}{2} i \, d + 1\right )\right )}}{d} - \frac {2 i \, \log \left ({\left (i \, d + 2\right )} \tan \left (b x + a\right ) + d\right ) \log \left (i \, \tan \left (b x + a\right ) + 1\right ) - i \, \log \left (i \, \tan \left (b x + a\right ) + 1\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (i \, \tan \left (b x + a\right ) + 1\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} - 8 \, {\left (b x + a\right )} \operatorname {artanh}\left (i \, d + \frac {d}{\tan \left (b x + a\right )} + 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1+I*d+d*cot(b*x+a)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(d*1i + d*cot(a + b*x) + 1),x)

[Out]

int(atanh(d*1i + d*cot(a + b*x) + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(1+I*d+d*cot(b*x+a)),x)

[Out]

Integral(atanh(d*cot(a + b*x) + I*d + 1), x)

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