3.3.42 \(\int \coth ^{-1}(1-i d+d \tan (a+b x)) \, dx\) [242]

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

[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 = 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 = {6391, 2215, 2221, 2317, 2438} \begin {gather*} \frac {i \text {Li}_2\left (-\left ((1-i d) 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 = 0.42, size = 84, 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. 320 vs. \(2 (76 ) = 152\).
time = 0.91, size = 321, normalized size = 3.45

method	result	size
derivativedivides	\(\frac {-\frac {i \mathrm {arccoth}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (-i d +d \tan \left (b x +a \right )\right )}{2}+\frac {i \mathrm {arccoth}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (i d +d \tan \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (-\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right )^{2}}{4 d}+\frac {i \dilog \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \dilog \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 d +2 i\right )\right )}{2 d +2 i}\right )}{2 d}-\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 d +2 i\right )\right )}{2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(321\)
default	\(\frac {-\frac {i \mathrm {arccoth}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (-i d +d \tan \left (b x +a \right )\right )}{2}+\frac {i \mathrm {arccoth}\left (1-i d +d \tan \left (b x +a \right )\right ) d \ln \left (i d +d \tan \left (b x +a \right )\right )}{2}+\frac {d^{2} \left (-\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right )^{2}}{4 d}+\frac {i \dilog \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}+\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{2 d}-\frac {i \dilog \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 d +2 i\right )\right )}{2 d +2 i}\right )}{2 d}-\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (i d +d \tan \left (b x +a \right )-i \left (2 d +2 i\right )\right )}{2 d +2 i}\right )}{2 d}+\frac {i \dilog \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}+\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{2 d}\right )}{2}}{b d}\)	\(321\)
risch	\(\text {Expression too large to display}\)	\(1618\)

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*(-1/4*I/d*ln(-I*d+d*tan(b*x+a))^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))-1/2*I/d*dilog(I*(I*d+d*tan(b*x+a)-I*(2*d+2*I))/(2*d+2*I))-1/2*
I/d*ln(I*d+d*tan(b*x+a))*ln(I*(I*d+d*tan(b*x+a)-I*(2*d+2*I))/(2*d+2*I))+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. 262 vs. \(2 (65) = 130\).
time = 0.47, size = 262, normalized size = 2.82 \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. 217 vs. \(2 (65) = 130\).
time = 0.39, size = 217, normalized size = 2.33 \begin {gather*} \frac {i \, b^{2} x^{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 ) - 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 + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/d) - 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 (d\,\mathrm {tan}\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*tan(a + b*x) - d*1i + 1),x)

[Out]

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

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