\(\int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx\) [246]

Optimal result

Integrand size = 17, antiderivative size = 94 \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\frac {1}{2} i b x^2+x \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{2} x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i \operatorname {PolyLog}\left (2,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b} \]

[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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6391, 2215, 2221, 2317, 2438} \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\frac {i \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+x \coth ^{-1}(d (-\tan (a+b x))+i d+1)+\frac {1}{2} i b x^2 \]

[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 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 6391

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*} \text {integral}& = 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 \text {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 \operatorname {PolyLog}\left (2,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(723\) vs. \(2(94)=188\).

Time = 1.67 (sec) , antiderivative size = 723, normalized size of antiderivative = 7.69 \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=x \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {x \left (-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))+\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((-2 i+d) \cos (a+b x)+i d \sin (a+b x))}{2 (-i+d)}\right ) \log (1-i \tan (b x))-\log \left (\frac {\sec (b x) ((2+i d) \cos (a+b x)-d \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \log (1+i \tan (b x))+\operatorname {PolyLog}(2,-\cos (2 b x)+i \sin (2 b x))+\operatorname {PolyLog}\left (2,\frac {\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (-i+d)}\right )-\operatorname {PolyLog}\left (2,\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+(2+i d) \sin (a)) (-i+\tan (b x))\right )\right ) \sec (a+b x) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x))}{((-2 i+d) \cos (a+b x)+i d \sin (a+b x)) \left (\frac {i \log (1-i \tan (b x)) \sec (b x) (d \cos (a)+(2+i d) \sin (a))}{(-2 i+d) \cos (a+b x)+i d \sin (a+b x)}+\frac {\log (1+i \tan (b x)) \sec (b x) (-i d \cos (a)+(-2 i+d) \sin (a))}{(-2 i+d) \cos (a+b x)+i d \sin (a+b x)}-\frac {\log \left (\frac {\sec (b x) ((2+i d) \cos (a+b x)-d \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \sec ^2(b x)}{-i+\tan (b x)}+\frac {\log \left (1-\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+(2+i d) \sin (a)) (-i+\tan (b x))\right ) \sec ^2(b x)}{-i+\tan (b x)}-\log \left (1-\frac {\sec (b x) (d \cos (a)+(2+i d) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (-i+d)}\right ) (-i+\tan (b x))+\frac {\log \left (\frac {\sec (b x) (\cos (a)-i \sin (a)) ((-2 i+d) \cos (a+b x)+i d \sin (a+b x))}{2 (-i+d)}\right ) \sec ^2(b x)}{i+\tan (b x)}+2 i b x (i+\tan (b x))\right ) (-i+\tan (a+b x))} \]

[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]))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (77 ) = 154\).

Time = 2.50 (sec) , antiderivative size = 320, normalized size of antiderivative = 3.40

method	result	size
derivativedivides	\(-\frac {\frac {i \operatorname {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (i d -d \tan \left (b x +a \right )\right )}{2}-\frac {i \operatorname {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (-i d -d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (-\frac {i \left (\frac {\ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2}-\frac {\ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2}\right )}{d}+\frac {i \left (\frac {\operatorname {dilog}\left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}+\frac {\ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (2 i-2 d \right )\right )}{2 i-2 d}\right )}{2}-\frac {\ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (2 i-2 d \right )\right )}{2 i-2 d}\right )}{2}\right )}{d}\right )}{2}}{b d}\)	\(320\)
default	\(-\frac {\frac {i \operatorname {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (i d -d \tan \left (b x +a \right )\right )}{2}-\frac {i \operatorname {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (-i d -d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (-\frac {i \left (\frac {\ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{4}-\frac {\operatorname {dilog}\left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2}-\frac {\ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2}\right )}{d}+\frac {i \left (\frac {\operatorname {dilog}\left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}+\frac {\ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (2 i-2 d \right )\right )}{2 i-2 d}\right )}{2}-\frac {\ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (2 i-2 d \right )\right )}{2 i-2 d}\right )}{2}\right )}{d}\right )}{2}}{b d}\)	\(320\)
risch	\(\text {Expression too large to display}\)	\(1650\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (66) = 132\).

Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.32 \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\frac {i \, b^{2} x^{2} - b x \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) - i \, a^{2} - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) + a \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \]

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

Sympy [F]

\[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\int \operatorname {acoth}{\left (- d \tan {\left (a + b x \right )} + i d + 1 \right )}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (66) = 132\).

Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.77 \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=-\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right )}{d} - \frac {\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} + d {\left (-\frac {2 i \, {\left (\log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (-\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, {\left (d - i\right )}} + 1\right ) + {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d - 2 i}{2 \, {\left (d - i\right )}}\right )\right )}}{d} + \frac {2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d - 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (-\frac {1}{2} \, d \tan \left (b x + a\right ) + \frac {1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) + {\rm Li}_2\left (\frac {1}{2} \, d \tan \left (b x + a\right ) - \frac {1}{2} i \, d\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} + 8 \, {\left (b x + a\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{8 \, b} \]

[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(-1/2*(I*d*tan(b*x + a) + d - 2*I)/(d - I) + 1) + dilog(1/2*(I*d*tan(b*x + a) + d - 2*I)/(d - 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

Giac [F]

\[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\int { \operatorname {arcoth}\left (-d \tan \left (b x + a\right ) + i \, d + 1\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx=\int \mathrm {acoth}\left (1-d\,\mathrm {tan}\left (a+b\,x\right )+d\,1{}\mathrm {i}\right ) \,d x \]

[In]

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

[Out]

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