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

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

[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

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Rubi [A]
time = 0.10, 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 = {6391, 2215, 2221, 2317, 2438} \begin {gather*} \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 \end {gather*}

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 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*} \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 \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 \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 [A]
time = 0.48, size = 85, normalized size = 0.90 \begin {gather*} x \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {2 b x \log \left (1-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+i \text {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (77 ) = 154\).
time = 0.94, size = 334, normalized size = 3.55

method	result	size
derivativedivides	\(-\frac {\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (i d -d \tan \left (b x +a \right )\right )}{2}-\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (-i d -d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (\frac {i \dilog \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{4 d}-\frac {i \dilog \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}-\frac {i \ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(334\)
default	\(-\frac {\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (i d -d \tan \left (b x +a \right )\right )}{2}-\frac {i \mathrm {arccoth}\left (1+i d -d \tan \left (b x +a \right )\right ) d \ln \left (-i d -d \tan \left (b x +a \right )\right )}{2}-\frac {d^{2} \left (\frac {i \dilog \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}-\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \ln \left (i d -d \tan \left (b x +a \right )\right )^{2}}{4 d}-\frac {i \dilog \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}-\frac {i \ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d -d \tan \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (-i d -d \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d -d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(334\)
risch	\(\text {Expression too large to display}\)	\(1650\)

Verification of antiderivative is not currently implemented for this CAS.

[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*(1/2*I/d*dilog(1+1/2*I*d-1/2*d*tan(b*x+a))+1/2*I/d*ln(I*d-d*tan(b*x+a))*ln(1+1/2*I*d-1/2*
d*tan(b*x+a))-1/4*I/d*ln(I*d-d*tan(b*x+a))^2-1/2*I/d*dilog(I*(-I*d-d*tan(b*x+a)-I*(2*I-2*d))/(2*I-2*d))-1/2*I/
d*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))+1/2*I/d*dilog(-1/2*I*(I*d-d*tan(b*x+a)
)/d)+1/2*I/d*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] 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 = 0.48, size = 260, normalized size = 2.77 \begin {gather*} -\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} \end {gather*}

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(-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

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Fricas [B] 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 = 0.38, size = 218, normalized size = 2.32 \begin {gather*} \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} \end {gather*}

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(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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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