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

Optimal. Leaf size=174 \[ \frac{i \text{PolyLog}\left (2,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}-\frac{i \text{PolyLog}\left (2,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}+\frac{1}{2} i x \log \left (1+\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )-\frac{1}{2} i x \log \left (1+\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+x \tan ^{-1}(d \tanh (a+b x)+c) \]

[Out]

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

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Rubi [A]  time = 0.230635, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {5191, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}-\frac{i \text{PolyLog}\left (2,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}+\frac{1}{2} i x \log \left (1+\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )-\frac{1}{2} i x \log \left (1+\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+x \tan ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[c + d*Tanh[a + b*x]],x]

[Out]

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

Rule 5191

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

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]

Rubi steps

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

Mathematica [A]  time = 3.90743, size = 288, normalized size = 1.66 \[ \frac{d \text{PolyLog}\left (2,-\frac{\left (c^2+2 c d+d^2+1\right ) e^{2 (a+b x)}}{c^2-d^2+2 \sqrt{-d^2}+1}\right )-d \text{PolyLog}\left (2,\frac{\left (c^2+2 c d+d^2+1\right ) e^{2 (a+b x)}}{-c^2+d^2+2 \sqrt{-d^2}-1}\right )-2 d (a+b x) \log \left (\frac{2 \left ((c+d)^2+1\right ) e^{2 (a+b x)}}{2 c^2-2 d^2-4 \sqrt{-d^2}+2}+1\right )+2 d (a+b x) \log \left (\frac{\left ((c+d)^2+1\right ) e^{2 (a+b x)}}{c^2-d^2+2 \sqrt{-d^2}+1}+1\right )+4 a \sqrt{-d^2} \tan ^{-1}\left (\frac{\left (c^2+2 c d+d^2+1\right ) e^{2 (a+b x)}+c^2-d^2+1}{2 d}\right )}{4 b \sqrt{-d^2}}+x \tan ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[c + d*Tanh[a + b*x]],x]

[Out]

x*ArcTan[c + d*Tanh[a + b*x]] + (4*a*Sqrt[-d^2]*ArcTan[(1 + c^2 - d^2 + (1 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)
))/(2*d)] - 2*d*(a + b*x)*Log[1 + (2*(1 + (c + d)^2)*E^(2*(a + b*x)))/(2 + 2*c^2 - 2*d^2 - 4*Sqrt[-d^2])] + 2*
d*(a + b*x)*Log[1 + ((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2])] + d*PolyLog[2, -(((1 + c
^2 + 2*c*d + d^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2]))] - d*PolyLog[2, ((1 + c^2 + 2*c*d + d^2)*E^
(2*(a + b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2])])/(4*b*Sqrt[-d^2])

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Maple [B]  time = 0.085, size = 350, normalized size = 2. \begin{align*}{\frac{\arctan \left ( c+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{2\,b}}-{\frac{\arctan \left ( c+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{2\,b}}-{\frac{{\frac{i}{4}}\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{b}\ln \left ({\frac{-d\tanh \left ( bx+a \right ) +i-c}{i-c-d}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c+i}{i+c+d}} \right ) }-{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{-d\tanh \left ( bx+a \right ) +i-c}{i-c-d}} \right ) }+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+i}{i+c+d}} \right ) }+{\frac{{\frac{i}{4}}\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{b}\ln \left ({\frac{-d\tanh \left ( bx+a \right ) +i-c}{i-c+d}} \right ) }-{\frac{{\frac{i}{4}}\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c+i}{i+c-d}} \right ) }+{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{-d\tanh \left ( bx+a \right ) +i-c}{i-c+d}} \right ) }-{\frac{{\frac{i}{4}}}{b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+i}{i+c-d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(c+d*tanh(b*x+a)),x)

[Out]

1/2/b*arctan(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)+d)-1/2/b*arctan(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)-d)-1/4*I/b*ln
(d*tanh(b*x+a)-d)*ln((-d*tanh(b*x+a)+I-c)/(I-c-d))+1/4*I/b*ln(d*tanh(b*x+a)-d)*ln((d*tanh(b*x+a)+c+I)/(I+c+d))
-1/4*I/b*dilog((-d*tanh(b*x+a)+I-c)/(I-c-d))+1/4*I/b*dilog((d*tanh(b*x+a)+c+I)/(I+c+d))+1/4*I/b*ln(d*tanh(b*x+
a)+d)*ln((-d*tanh(b*x+a)+I-c)/(I-c+d))-1/4*I/b*ln(d*tanh(b*x+a)+d)*ln((d*tanh(b*x+a)+c+I)/(I+c-d))+1/4*I/b*dil
og((-d*tanh(b*x+a)+I-c)/(I-c+d))-1/4*I/b*dilog((d*tanh(b*x+a)+c+I)/(I+c-d))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -4 \, b d \int \frac{x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} - 2 \, c d + d^{2} +{\left (c^{2} e^{\left (4 \, a\right )} + 2 \, c d e^{\left (4 \, a\right )} + d^{2} e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )} + 2 \,{\left (c^{2} e^{\left (2 \, a\right )} - d^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}\,{d x} + x \arctan \left (\frac{{\left (c e^{\left (2 \, a\right )} + d e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + c - d}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+d*tanh(b*x+a)),x, algorithm="maxima")

[Out]

-4*b*d*integrate(x*e^(2*b*x + 2*a)/(c^2 - 2*c*d + d^2 + (c^2*e^(4*a) + 2*c*d*e^(4*a) + d^2*e^(4*a) + e^(4*a))*
e^(4*b*x) + 2*(c^2*e^(2*a) - d^2*e^(2*a) + e^(2*a))*e^(2*b*x) + 1), x) + x*arctan(((c*e^(2*a) + d*e^(2*a))*e^(
2*b*x) + c - d)/(e^(2*b*x + 2*a) + 1))

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Fricas [B]  time = 5.62637, size = 2288, normalized size = 13.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+d*tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/2*(2*b*x*arctan((c*cosh(b*x + a) + d*sinh(b*x + a))/cosh(b*x + a)) - I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(
b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + (c^2 - d^2 - 2*I*d + 1)*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)
/(c^2 - 2*c*d + d^2 + 1))) - I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(
b*x + a) - (c^2 - d^2 - 2*I*d + 1)*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))) + I*a*log(2*(c^
2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + (c^2 - d^2 + 2*I*d + 1)*sqrt(-(
4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))) + I*a*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^
2 + 2*c*d + d^2 + 1)*sinh(b*x + a) - (c^2 - d^2 + 2*I*d + 1)*sqrt(-(4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d +
d^2 + 1))) + (I*b*x + I*a)*log(1/2*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) +
sinh(b*x + a)) + 1) + (I*b*x + I*a)*log(-1/2*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c*d + d^2 + 1))*(cosh
(b*x + a) + sinh(b*x + a)) + 1) + (-I*b*x - I*a)*log(1/2*sqrt(-(4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^2
+ 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b*x - I*a)*log(-1/2*sqrt(-(4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 -
2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + I*dilog(1/2*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 -
2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) + I*dilog(-1/2*sqrt(-(4*c^2 - 4*d^2 + 8*I*d + 4)/(c^2 - 2*c
*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) - I*dilog(1/2*sqrt(-(4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d +
d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) - I*dilog(-1/2*sqrt(-(4*c^2 - 4*d^2 - 8*I*d + 4)/(c^2 - 2*c*d + d^
2 + 1))*(cosh(b*x + a) + sinh(b*x + a))))/b

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(c+d*tanh(b*x+a)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arctan \left (d \tanh \left (b x + a\right ) + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+d*tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(arctan(d*tanh(b*x + a) + c), x)