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

Optimal. Leaf size=150 \[ \frac {\text {Li}_2\left (-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {\text {Li}_2\left (-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{2} x \log \left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac {1}{2} x \log \left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+c) \]

[Out]

x*arctanh(c+d*tanh(b*x+a))+1/2*x*ln(1+(1-c-d)*exp(2*b*x+2*a)/(1-c+d))-1/2*x*ln(1+(1+c+d)*exp(2*b*x+2*a)/(1+c-d
))+1/4*polylog(2,-(1-c-d)*exp(2*b*x+2*a)/(1-c+d))/b-1/4*polylog(2,-(1+c+d)*exp(2*b*x+2*a)/(1+c-d))/b

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Rubi [A]  time = 0.22, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 = {6235, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {\text {PolyLog}\left (2,-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{2} x \log \left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac {1}{2} x \log \left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[c + d*Tanh[a + b*x]],x]

[Out]

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

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 6235

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTanh[c + d*Tanh[a + b*x]], x] + (D
ist[b*(1 - c - d), Int[(x*E^(2*a + 2*b*x))/(1 - c + d + (1 - c - d)*E^(2*a + 2*b*x)), x], x] - Dist[b*(1 + c +
d), Int[(x*E^(2*a + 2*b*x))/(1 + c - d + (1 + c + d)*E^(2*a + 2*b*x)), x], x]) /; FreeQ[{a, b, c, d}, x] && N
eQ[(c - d)^2, 1]

Rubi steps

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

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Mathematica [A]  time = 6.81, size = 131, normalized size = 0.87 \[ \frac {\text {Li}_2\left (-\frac {(c+d-1) e^{2 (a+b x)}}{c-d-1}\right )-\text {Li}_2\left (-\frac {(c+d+1) e^{2 (a+b x)}}{c-d+1}\right )+2 b x \left (\log \left (\frac {(c+d-1) e^{2 (a+b x)}}{c-d-1}+1\right )-\log \left (\frac {(c+d+1) e^{2 (a+b x)}}{c-d+1}+1\right )\right )}{4 b}+x \tanh ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.94, size = 552, normalized size = 3.68 \[ \frac {b x \log \left (-\frac {{\left (c + 1\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (c - 1\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (c + d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d + 1\right )} \sinh \left (b x + a\right ) + 2 \, {\left (c - d + 1\right )} \sqrt {-\frac {c + d + 1}{c - d + 1}}\right ) + a \log \left (2 \, {\left (c + d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d + 1\right )} \sinh \left (b x + a\right ) - 2 \, {\left (c - d + 1\right )} \sqrt {-\frac {c + d + 1}{c - d + 1}}\right ) - a \log \left (2 \, {\left (c + d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d - 1\right )} \sinh \left (b x + a\right ) + 2 \, {\left (c - d - 1\right )} \sqrt {-\frac {c + d - 1}{c - d - 1}}\right ) - a \log \left (2 \, {\left (c + d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d - 1\right )} \sinh \left (b x + a\right ) - 2 \, {\left (c - d - 1\right )} \sqrt {-\frac {c + d - 1}{c - d - 1}}\right ) - {\left (b x + a\right )} \log \left (\sqrt {-\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\sqrt {-\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (b x + a\right )} \log \left (\sqrt {-\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (b x + a\right )} \log \left (-\sqrt {-\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\sqrt {-\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\sqrt {-\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\rm Li}_2\left (\sqrt {-\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\rm Li}_2\left (-\sqrt {-\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x*log(-((c + 1)*cosh(b*x + a) + d*sinh(b*x + a))/((c - 1)*cosh(b*x + a) + d*sinh(b*x + a))) + a*log(2*(
c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) + 2*(c - d + 1)*sqrt(-(c + d + 1)/(c - d + 1))) + a*log
(2*(c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) - 2*(c - d + 1)*sqrt(-(c + d + 1)/(c - d + 1))) - a
*log(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) + 2*(c - d - 1)*sqrt(-(c + d - 1)/(c - d - 1)))
- a*log(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) - 2*(c - d - 1)*sqrt(-(c + d - 1)/(c - d -
1))) - (b*x + a)*log(sqrt(-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b*x + a)*log(-sqrt
(-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (b*x + a)*log(sqrt(-(c + d - 1)/(c - d - 1))
*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (b*x + a)*log(-sqrt(-(c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*
x + a)) + 1) - dilog(sqrt(-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) - dilog(-sqrt(-(c + d + 1
)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + dilog(sqrt(-(c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b
*x + a))) + dilog(-sqrt(-(c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.30, size = 306, normalized size = 2.04 \[ -\frac {\arctanh \left (c +d \tanh \left (b x +a \right )\right ) \ln \left (d \tanh \left (b x +a \right )-d \right )}{2 b}+\frac {\arctanh \left (c +d \tanh \left (b x +a \right )\right ) \ln \left (d \tanh \left (b x +a \right )+d \right )}{2 b}+\frac {\dilog \left (\frac {d \tanh \left (b x +a \right )+c -1}{c -d -1}\right )}{4 b}+\frac {\ln \left (d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {d \tanh \left (b x +a \right )+c -1}{c -d -1}\right )}{4 b}-\frac {\dilog \left (\frac {d \tanh \left (b x +a \right )+c +1}{1+c -d}\right )}{4 b}-\frac {\ln \left (d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {d \tanh \left (b x +a \right )+c +1}{1+c -d}\right )}{4 b}+\frac {\dilog \left (\frac {d \tanh \left (b x +a \right )+c +1}{1+c +d}\right )}{4 b}+\frac {\ln \left (d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {d \tanh \left (b x +a \right )+c +1}{1+c +d}\right )}{4 b}-\frac {\dilog \left (\frac {d \tanh \left (b x +a \right )+c -1}{c +d -1}\right )}{4 b}-\frac {\ln \left (d \tanh \left (b x +a \right )-d \right ) \ln \left (\frac {d \tanh \left (b x +a \right )+c -1}{c +d -1}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*arctanh(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)-d)+1/2/b*arctanh(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)+d)+1/4/b*d
ilog((d*tanh(b*x+a)+c-1)/(c-d-1))+1/4/b*ln(d*tanh(b*x+a)+d)*ln((d*tanh(b*x+a)+c-1)/(c-d-1))-1/4/b*dilog((d*tan
h(b*x+a)+c+1)/(1+c-d))-1/4/b*ln(d*tanh(b*x+a)+d)*ln((d*tanh(b*x+a)+c+1)/(1+c-d))+1/4/b*dilog((d*tanh(b*x+a)+c+
1)/(1+c+d))+1/4/b*ln(d*tanh(b*x+a)-d)*ln((d*tanh(b*x+a)+c+1)/(1+c+d))-1/4/b*dilog((d*tanh(b*x+a)+c-1)/(c+d-1))
-1/4/b*ln(d*tanh(b*x+a)-d)*ln((d*tanh(b*x+a)+c-1)/(c+d-1))

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maxima [A]  time = 0.69, size = 142, normalized size = 0.95 \[ -\frac {1}{4} \, b d {\left (\frac {2 \, b x \log \left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right )}{b^{2} d} - \frac {2 \, b x \log \left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right )}{b^{2} d}\right )} + x \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*d*((2*b*x*log((c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1) + 1) + dilog(-(c + d + 1)*e^(2*b*x + 2*a)/(c - d
+ 1)))/(b^2*d) - (2*b*x*log((c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1) + 1) + dilog(-(c + d - 1)*e^(2*b*x + 2*a)/
(c - d - 1)))/(b^2*d)) + x*arctanh(d*tanh(b*x + a) + c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(c + d*tanh(a + b*x)),x)

[Out]

int(atanh(c + d*tanh(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(c + d*tanh(a + b*x)), x)

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