3.2.4 \(\int \text {ArcTan}(c+(i+c) \coth (a+b x)) \, dx\) [104]

Optimal. Leaf size=79 \[ -\frac {1}{2} i b x^2+x \text {ArcTan}(c+(i+c) \coth (a+b x))+\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+\frac {i \text {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b} \]

[Out]

-1/2*I*b*x^2+x*arctan(c+(I+c)*coth(b*x+a))+1/2*I*x*ln(1-I*c*exp(2*b*x+2*a))+1/4*I*polylog(2,I*c*exp(2*b*x+2*a)
)/b

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Rubi [A]
time = 0.08, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5297, 2215, 2221, 2317, 2438} \begin {gather*} x \text {ArcTan}(c+(c+i) \coth (a+b x))+\frac {i \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}+\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )-\frac {1}{2} i b x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[c + (I + c)*Coth[a + b*x]],x]

[Out]

(-1/2*I)*b*x^2 + x*ArcTan[c + (I + c)*Coth[a + b*x]] + (I/2)*x*Log[1 - I*c*E^(2*a + 2*b*x)] + ((I/4)*PolyLog[2
, I*c*E^(2*a + 2*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 5297

Int[ArcTan[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcTan[c + d*Coth[a + b*x]], x] - Dist
[b, Int[x/(c - d - c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c - d)^2, -1]

Rubi steps

\begin {align*} \int \tan ^{-1}(c+(i+c) \coth (a+b x)) \, dx &=x \tan ^{-1}(c+(i+c) \coth (a+b x))-b \int \frac {x}{-i-c e^{2 a+2 b x}} \, dx\\ &=-\frac {1}{2} i b x^2+x \tan ^{-1}(c+(i+c) \coth (a+b x))-(i b c) \int \frac {e^{2 a+2 b x} x}{-i-c e^{2 a+2 b x}} \, dx\\ &=-\frac {1}{2} i b x^2+x \tan ^{-1}(c+(i+c) \coth (a+b x))+\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )-\frac {1}{2} i \int \log \left (1-i c e^{2 a+2 b x}\right ) \, dx\\ &=-\frac {1}{2} i b x^2+x \tan ^{-1}(c+(i+c) \coth (a+b x))+\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )-\frac {i \text {Subst}\left (\int \frac {\log (1-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=-\frac {1}{2} i b x^2+x \tan ^{-1}(c+(i+c) \coth (a+b x))+\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+\frac {i \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 71, normalized size = 0.90 \begin {gather*} x \text {ArcTan}(c+(i+c) \coth (a+b x))+\frac {i \left (2 b x \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-\text {PolyLog}\left (2,-\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[c + (I + c)*Coth[a + b*x]],x]

[Out]

x*ArcTan[c + (I + c)*Coth[a + b*x]] + ((I/4)*(2*b*x*Log[1 + I/(c*E^(2*(a + b*x)))] - PolyLog[2, (-I)/(c*E^(2*(
a + b*x)))]))/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. 597 vs. \(2 (65 ) = 130\).
time = 0.31, size = 598, normalized size = 7.57

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{2 i+2 c}-\frac {2 i \arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right ) c}{2 i+2 c}-\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right ) c^{2}}{2 i+2 c}-\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2 i+2 c}+\frac {2 i \arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right ) c}{2 i+2 c}+\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right ) c^{2}}{2 i+2 c}-\left (i+c \right )^{2} \left (\frac {i \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )^{2}}{8 i+8 c}+\frac {i \ln \left (-\frac {i \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right ) \ln \left (-\frac {i \left (i-c -\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right )}{4 i+4 c}-\frac {i \ln \left (-\frac {i \left (i-c -\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{4 \left (i+c \right )}+\frac {i \dilog \left (-\frac {i \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right )}{4 i+4 c}+\frac {i \ln \left (-\frac {i-c -\left (i+c \right ) \coth \left (b x +a \right )}{2 c}\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{4 i+4 c}+\frac {i \dilog \left (-\frac {i-c -\left (i+c \right ) \coth \left (b x +a \right )}{2 c}\right )}{4 i+4 c}-\frac {i \ln \left (\frac {-i-c -\left (i+c \right ) \coth \left (b x +a \right )}{-2 i-2 c}\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{4 \left (i+c \right )}-\frac {i \dilog \left (\frac {-i-c -\left (i+c \right ) \coth \left (b x +a \right )}{-2 i-2 c}\right )}{4 \left (i+c \right )}\right )}{b \left (i+c \right )}\)	\(598\)
default	\(\frac {\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{2 i+2 c}-\frac {2 i \arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right ) c}{2 i+2 c}-\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right ) c^{2}}{2 i+2 c}-\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2 i+2 c}+\frac {2 i \arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right ) c}{2 i+2 c}+\frac {\arctan \left (c +\left (i+c \right ) \coth \left (b x +a \right )\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right ) c^{2}}{2 i+2 c}-\left (i+c \right )^{2} \left (\frac {i \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )^{2}}{8 i+8 c}+\frac {i \ln \left (-\frac {i \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right ) \ln \left (-\frac {i \left (i-c -\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right )}{4 i+4 c}-\frac {i \ln \left (-\frac {i \left (i-c -\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right ) \ln \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{4 \left (i+c \right )}+\frac {i \dilog \left (-\frac {i \left (i+c +\left (i+c \right ) \coth \left (b x +a \right )\right )}{2}\right )}{4 i+4 c}+\frac {i \ln \left (-\frac {i-c -\left (i+c \right ) \coth \left (b x +a \right )}{2 c}\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{4 i+4 c}+\frac {i \dilog \left (-\frac {i-c -\left (i+c \right ) \coth \left (b x +a \right )}{2 c}\right )}{4 i+4 c}-\frac {i \ln \left (\frac {-i-c -\left (i+c \right ) \coth \left (b x +a \right )}{-2 i-2 c}\right ) \ln \left (c -\left (i+c \right ) \coth \left (b x +a \right )+i\right )}{4 \left (i+c \right )}-\frac {i \dilog \left (\frac {-i-c -\left (i+c \right ) \coth \left (b x +a \right )}{-2 i-2 c}\right )}{4 \left (i+c \right )}\right )}{b \left (i+c \right )}\)	\(598\)
risch	\(\text {Expression too large to display}\)	\(1221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(c+(I+c)*coth(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/b/(I+c)*(arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)-2*I*arctan(c+(I+c)*coth(b*x+a))/(2*
I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)*c-arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(c-(I+c)*coth(b*x+a)+I)*c^2-arctan(
c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(I+c+(I+c)*coth(b*x+a))+2*I*arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(I+c+(I+c
)*coth(b*x+a))*c+arctan(c+(I+c)*coth(b*x+a))/(2*I+2*c)*ln(I+c+(I+c)*coth(b*x+a))*c^2-(I+c)^2*(1/8*I/(I+c)*ln(I
+c+(I+c)*coth(b*x+a))^2+1/4*I/(I+c)*ln(-1/2*I*(I+c+(I+c)*coth(b*x+a)))*ln(-1/2*I*(I-c-(I+c)*coth(b*x+a)))-1/4*
I/(I+c)*ln(-1/2*I*(I-c-(I+c)*coth(b*x+a)))*ln(I+c+(I+c)*coth(b*x+a))+1/4*I/(I+c)*dilog(-1/2*I*(I+c+(I+c)*coth(
b*x+a)))+1/4*I/(I+c)*ln(-1/2*(I-c-(I+c)*coth(b*x+a))/c)*ln(c-(I+c)*coth(b*x+a)+I)+1/4*I/(I+c)*dilog(-1/2*(I-c-
(I+c)*coth(b*x+a))/c)-1/4*I/(I+c)*ln((-I-c-(I+c)*coth(b*x+a))/(-2*I-2*c))*ln(c-(I+c)*coth(b*x+a)+I)-1/4*I/(I+c
)*dilog((-I-c-(I+c)*coth(b*x+a))/(-2*I-2*c))))

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Maxima [A]
time = 1.16, size = 80, normalized size = 1.01 \begin {gather*} 2 \, b {\left (c + i\right )} {\left (\frac {2 \, x^{2}}{2 i \, c - 2} - \frac {2 \, b x \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right )}{-2 \, b^{2} {\left (-i \, c + 1\right )}}\right )} + x \arctan \left ({\left (c + i\right )} \coth \left (b x + a\right ) + c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+(I+c)*coth(b*x+a)),x, algorithm="maxima")

[Out]

2*b*(c + I)*(2*x^2/(2*I*c - 2) - (2*b*x*log(-I*c*e^(2*b*x + 2*a) + 1) + dilog(I*c*e^(2*b*x + 2*a)))/(b^2*(2*I*
c - 2))) + x*arctan((c + I)*coth(b*x + a) + c)

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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. 187 vs. \(2 (58) = 116\).
time = 2.42, size = 187, normalized size = 2.37 \begin {gather*} \frac {-i \, b^{2} x^{2} + i \, b x \log \left (-\frac {{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} + i}\right ) + i \, a^{2} + {\left (i \, b x + i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b x + i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) - i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+(I+c)*coth(b*x+a)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(c+(I+c)*coth(b*x+a)),x)

[Out]

Exception raised: CoercionFailed >> Cannot convert _t0**2*exp(2*a) - 1 of type <class 'sympy.core.add.Add'> to
QQ_I[b,_t0,exp(a)]

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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(arctan(c+(I+c)*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(arctan((c + I)*coth(b*x + a) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(c + coth(a + b*x)*(c + 1i)),x)

[Out]

int(atan(c + coth(a + b*x)*(c + 1i)), x)

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