3.323 \(\int \tanh ^{-1}(1-i d+d \tan (a+b x)) \, dx\)

Optimal. Leaf size=93 \[ \frac {i \text {Li}_2\left (-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+x \tanh ^{-1}(d \tan (a+b x)-i d+1)+\frac {1}{2} i b x^2 \]

[Out]

1/2*I*b*x^2+x*arctanh(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 = 0.15, 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 = {6255, 2184, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} x \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+x \tanh ^{-1}(d \tan (a+b x)-i d+1)+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[1 - I*d + d*Tan[a + b*x]],x]

[Out]

(I/2)*b*x^2 + x*ArcTanh[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 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 6255

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTanh[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 \tanh ^{-1}(1-i d+d \tan (a+b x)) \, dx &=x \tanh ^{-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 \tanh ^{-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 \tanh ^{-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 \tanh ^{-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 {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 \tanh ^{-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 [B]  time = 14.37, size = 766, normalized size = 8.24 \[ \frac {x \sec ^2(a+b x) (\cos (b x)+i \sin (b x)) (\sin (b x)+i \cos (b x)) (d \sin (a+b x)+(2-i d) \cos (a+b x)) \left (\text {Li}_2\left (-\frac {1}{2} (\cos (a)+i \sin (a)) (d \cos (a)+i (d+2 i) \sin (a)) (\tan (b x)-i)\right )-\text {Li}_2\left (\frac {\sec (b x) (d \cos (a)+i (d+2 i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 (d+i)}\right )-\log (1-i \tan (b x)) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )+\log (1+i \tan (b x)) \log \left (\frac {\sec (b x) (d \sin (a+b x)+(2-i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )-\text {Li}_2(i \sin (2 b x)-\cos (2 b x))+2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))\right )}{(\tan (a+b x)-i) (d \tan (a+b x)-i d+2) (i d \sin (a+b x)+(d+2 i) \cos (a+b x)) \left (\frac {\sec ^2(b x) \log \left (\frac {\sec (b x) (d \sin (a+b x)+(2-i d) \cos (a+b x))}{2 \cos (a)-2 i \sin (a)}\right )}{\tan (b x)-i}-\frac {\sec ^2(b x) \log \left (1+\frac {1}{2} (\cos (a)+i \sin (a)) (\tan (b x)-i) (d \cos (a)+i (d+2 i) \sin (a))\right )}{\tan (b x)-i}-\frac {\sec ^2(b x) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )}{\tan (b x)+i}+\frac {i \sec (b x) (d \cos (a)+i (d+2 i) \sin (a)) \log (1+i \tan (b x))}{i d \sin (a+b x)+(d+2 i) \cos (a+b x)}+\frac {\sec (b x) ((d+2 i) \sin (a)-i d \cos (a)) \log (1-i \tan (b x))}{i d \sin (a+b x)+(d+2 i) \cos (a+b x)}+(\tan (b x)-i) \log \left (\frac {(\cos (a)-i \sin (a)) \sec (b x) (i d \sin (a+b x)+(d+2 i) \cos (a+b x))}{2 (d+i)}\right )+2 b x (1-i \tan (b x))\right )}+x \tanh ^{-1}(d \tan (a+b x)-i d+1) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x*ArcTanh[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]*(C
os[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, -Cos[
2*b*x] + I*Sin[2*b*x]] - PolyLog[2, (Sec[b*x]*(d*Cos[a] + I*(2*I + d)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x]))
/(2*(I + d))] + PolyLog[2, -1/2*((Cos[a] + I*Sin[a])*(d*Cos[a] + I*(2*I + d)*Sin[a])*(-I + Tan[b*x]))])*Sec[a
+ b*x]^2*(Cos[b*x] + I*Sin[b*x])*(I*Cos[b*x] + Sin[b*x])*((2 - I*d)*Cos[a + b*x] + d*Sin[a + 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] + I*(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]) + 2*b*x*(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])]*Sec[b*x]^2)/(-I + Tan[b*x]) - (Log[1 + ((Cos[a] + I*Sin[a])*(d*Cos[a] +
I*(2*I + d)*Sin[a])*(-I + Tan[b*x]))/2]*Sec[b*x]^2)/(-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))]*(-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]))*(-I + Tan[a + b*x])*(2 - I*d
+ d*Tan[a + b*x]))

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fricas [B]  time = 1.01, size = 222, normalized size = 2.39 \[ \frac {i \, b^{2} x^{2} + b x \log \left (-\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) - i \, a^{2} - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) + a \log \left (\frac {{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, d - 4}}{2 \, d + 2 i}\right ) + a \log \left (\frac {{\left (2 \, d + 2 i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, d - 4}}{2 \, d + 2 i}\right ) + i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(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(((2*d
+ 2*I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(2*d + 2*I)) + a*log(((2*d + 2*I)*e^(I*b*x + I*a) - I*sqrt(4*I*d
- 4))/(2*d + 2*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.48, size = 292, normalized size = 3.14 \[ \frac {i \arctanh \left (1-i d +d \tan \left (b x +a \right )\right ) \ln \left (i d +d \tan \left (b x +a \right )\right )}{2 b}-\frac {i \arctanh \left (1-i d +d \tan \left (b x +a \right )\right ) \ln \left (-i d +d \tan \left (b x +a \right )\right )}{2 b}+\frac {i \dilog \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b}+\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {i \left (-i d +d \tan \left (b x +a \right )\right )}{2 d}\right )}{4 b}-\frac {i \dilog \left (\frac {2-i d +d \tan \left (b x +a \right )}{-2 i d +2}\right )}{4 b}-\frac {i \ln \left (i d +d \tan \left (b x +a \right )\right ) \ln \left (\frac {2-i d +d \tan \left (b x +a \right )}{-2 i d +2}\right )}{4 b}-\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right )^{2}}{8 b}+\frac {i \dilog \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b}+\frac {i \ln \left (-i d +d \tan \left (b x +a \right )\right ) \ln \left (1-\frac {i d}{2}+\frac {d \tan \left (b x +a \right )}{2}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(1-I*d+d*tan(b*x+a)),x)

[Out]

1/2*I/b*arctanh(1-I*d+d*tan(b*x+a))*ln(I*d+d*tan(b*x+a))-1/2*I/b*arctanh(1-I*d+d*tan(b*x+a))*ln(-I*d+d*tan(b*x
+a))+1/4*I/b*dilog(1/2*I*(-I*d+d*tan(b*x+a))/d)+1/4*I/b*ln(I*d+d*tan(b*x+a))*ln(1/2*I*(-I*d+d*tan(b*x+a))/d)-1
/4*I/b*dilog((2-I*d+d*tan(b*x+a))/(-2*I*d+2))-1/4*I/b*ln(I*d+d*tan(b*x+a))*ln((2-I*d+d*tan(b*x+a))/(-2*I*d+2))
-1/8*I/b*ln(-I*d+d*tan(b*x+a))^2+1/4*I/b*dilog(1-1/2*I*d+1/2*d*tan(b*x+a))+1/4*I/b*ln(-I*d+d*tan(b*x+a))*ln(1-
1/2*I*d+1/2*d*tan(b*x+a))

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maxima [B]  time = 0.43, size = 263, normalized size = 2.83 \[ -\frac {4 \, {\left (b x + a\right )} d {\left (\frac {\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right )}{d} - \frac {\log \left (\tan \left (b x + a\right ) - i\right )}{d}\right )} + d {\left (-\frac {2 i \, {\left (\log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (-\frac {i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, d + 2 i} + 1\right ) + {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d + 2 i}{2 \, d + 2 i}\right )\right )}}{d} + \frac {2 i \, \log \left (d \tan \left (b x + a\right ) - i \, d + 2\right ) \log \left (\tan \left (b x + a\right ) - i\right ) - i \, \log \left (\tan \left (b x + a\right ) - i\right )^{2}}{d} - \frac {2 i \, {\left (\log \left (\frac {1}{2} \, d \tan \left (b x + a\right ) - \frac {1}{2} i \, d + 1\right ) \log \left (\tan \left (b x + a\right ) - i\right ) + {\rm Li}_2\left (-\frac {1}{2} \, d \tan \left (b x + a\right ) + \frac {1}{2} i \, d\right )\right )}}{d} + \frac {2 i \, {\left (\log \left (\tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{d}\right )} - 8 \, {\left (b x + a\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {atanh}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {atanh}{\left (d \tan {\left (a + b x \right )} - i d + 1 \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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