3.3.59 \(\int \coth ^{-1}(1+i d+d \cot (a+b x)) \, dx\) [259]

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

[Out]

1/2*I*b*x^2+x*arccoth(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 = 0.10, 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 = {6393, 2215, 2221, 2317, 2438} \begin {gather*} \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 \coth ^{-1}(d \cot (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*Cot[a + b*x]],x]

[Out]

(I/2)*b*x^2 + x*ArcCoth[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 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 6393

Int[ArcCoth[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcCoth[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 \coth ^{-1}(1+i d+d \cot (a+b x)) \, dx &=x \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 {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 \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 [A]
time = 0.33, size = 83, normalized size = 0.89 \begin {gather*} x \coth ^{-1}(1+i d+d \cot (a+b x))-\frac {2 b x \log \left (1+\frac {e^{-2 i (a+b x)}}{-1-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*Cot[a + b*x]],x]

[Out]

x*ArcCoth[1 + I*d + d*Cot[a + b*x]] - (2*b*x*Log[1 + 1/((-1 - 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. 320 vs. \(2 (76 ) = 152\).
time = 1.25, size = 321, normalized size = 3.45

method	result	size
derivativedivides	\(\frac {-\frac {i \mathrm {arccoth}\left (1+i d +d \cot \left (b x +a \right )\right ) d \ln \left (i d +d \cot \left (b x +a \right )\right )}{2}+\frac {i \mathrm {arccoth}\left (1+i d +d \cot \left (b x +a \right )\right ) d \ln \left (-i d +d \cot \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (-\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right )^{2}}{4 d}+\frac {i \dilog \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \dilog \left (\frac {i \left (-i d +d \cot \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}-\frac {i \ln \left (-i d +d \cot \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \cot \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (-\frac {i \left (i d +d \cot \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (-i d +d \cot \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d +d \cot \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(321\)
default	\(\frac {-\frac {i \mathrm {arccoth}\left (1+i d +d \cot \left (b x +a \right )\right ) d \ln \left (i d +d \cot \left (b x +a \right )\right )}{2}+\frac {i \mathrm {arccoth}\left (1+i d +d \cot \left (b x +a \right )\right ) d \ln \left (-i d +d \cot \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (-\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right )^{2}}{4 d}+\frac {i \dilog \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (i d +d \cot \left (b x +a \right )\right ) \ln \left (1+\frac {i d}{2}+\frac {d \cot \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \dilog \left (\frac {i \left (-i d +d \cot \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}-\frac {i \ln \left (-i d +d \cot \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \cot \left (b x +a \right )-i \left (-2 d +2 i\right )\right )}{-2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (-\frac {i \left (i d +d \cot \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (-i d +d \cot \left (b x +a \right )\right ) \ln \left (-\frac {i \left (i d +d \cot \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(321\)
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*cot(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/b/d*(-1/2*I*arccoth(1+I*d+d*cot(b*x+a))*d*ln(I*d+d*cot(b*x+a))+1/2*I*arccoth(1+I*d+d*cot(b*x+a))*d*ln(-I*d+d
*cot(b*x+a))+1/2*d^2*(-1/4*I/d*ln(I*d+d*cot(b*x+a))^2+1/2*I/d*dilog(1+1/2*I*d+1/2*d*cot(b*x+a))+1/2*I/d*ln(I*d
+d*cot(b*x+a))*ln(1+1/2*I*d+1/2*d*cot(b*x+a))-1/2*I/d*dilog(I*(-I*d+d*cot(b*x+a)-I*(2*I-2*d))/(2*I-2*d))-1/2*I
/d*ln(-I*d+d*cot(b*x+a))*ln(I*(-I*d+d*cot(b*x+a)-I*(2*I-2*d))/(2*I-2*d))+1/2*I/d*dilog(-1/2*I*(I*d+d*cot(b*x+a
))/d)+1/2*I/d*ln(-I*d+d*cot(b*x+a))*ln(-1/2*I*(I*d+d*cot(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. 286 vs. \(2 (66) = 132\).
time = 0.47, size = 286, normalized size = 3.08 \begin {gather*} -\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 {arcoth}\left (i \, d + \frac {d}{\tan \left (b x + a\right )} + 1\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(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)*t
an(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)*arccoth(I*d + d/tan(b*x + a) + 1))/b

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Fricas [A]
time = 0.36, size = 121, normalized size = 1.30 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {acoth}{\left (d \cot {\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*cot(b*x+a)),x)

[Out]

Integral(acoth(d*cot(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*cot(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(d*cot(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 (d\,\mathrm {cot}\left (a+b\,x\right )+1+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*cot(a + b*x) + 1),x)

[Out]

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

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