\(\int (a+b \coth ^{-1}(c+d x)) \, dx\) [105]

Optimal result

Integrand size = 10, antiderivative size = 40 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a x+\frac {b (c+d x) \coth ^{-1}(c+d x)}{d}+\frac {b \log \left (1-(c+d x)^2\right )}{2 d} \]

[Out]

a*x+b*(d*x+c)*arccoth(d*x+c)/d+1/2*b*ln(1-(d*x+c)^2)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6239, 6022, 266} \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a x+\frac {b \log \left (1-(c+d x)^2\right )}{2 d}+\frac {b (c+d x) \coth ^{-1}(c+d x)}{d} \]

[In]

Int[a + b*ArcCoth[c + d*x],x]

[Out]

a*x + (b*(c + d*x)*ArcCoth[c + d*x])/d + (b*Log[1 - (c + d*x)^2])/(2*d)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 6022

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6239

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCoth[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = a x+b \int \coth ^{-1}(c+d x) \, dx \\ & = a x+\frac {b \text {Subst}\left (\int \coth ^{-1}(x) \, dx,x,c+d x\right )}{d} \\ & = a x+\frac {b (c+d x) \coth ^{-1}(c+d x)}{d}-\frac {b \text {Subst}\left (\int \frac {x}{1-x^2} \, dx,x,c+d x\right )}{d} \\ & = a x+\frac {b (c+d x) \coth ^{-1}(c+d x)}{d}+\frac {b \log \left (1-(c+d x)^2\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a x+b x \coth ^{-1}(c+d x)+\frac {b (-((-1+c) \log (1-c-d x))+(1+c) \log (1+c+d x))}{2 d} \]

[In]

Integrate[a + b*ArcCoth[c + d*x],x]

[Out]

a*x + b*x*ArcCoth[c + d*x] + (b*(-((-1 + c)*Log[1 - c - d*x]) + (1 + c)*Log[1 + c + d*x]))/(2*d)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88

method	result	size
default	\(a x +\frac {b \left (\left (d x +c \right ) \operatorname {arccoth}\left (d x +c \right )+\frac {\ln \left (\left (d x +c \right )^{2}-1\right )}{2}\right )}{d}\)	\(35\)
parts	\(a x +\frac {b \left (\left (d x +c \right ) \operatorname {arccoth}\left (d x +c \right )+\frac {\ln \left (\left (d x +c \right )^{2}-1\right )}{2}\right )}{d}\)	\(35\)
derivativedivides	\(\frac {\left (d x +c \right ) a +b \left (\left (d x +c \right ) \operatorname {arccoth}\left (d x +c \right )+\frac {\ln \left (\left (d x +c \right )^{2}-1\right )}{2}\right )}{d}\)	\(40\)
parallelrisch	\(-\frac {b \left (-x \,\operatorname {arccoth}\left (d x +c \right ) d^{2}-\operatorname {arccoth}\left (d x +c \right ) c d -\ln \left (d x +c -1\right ) d -\operatorname {arccoth}\left (d x +c \right ) d \right )}{d^{2}}+a x\)	\(53\)
risch	\(a x +\frac {b x \ln \left (d x +c +1\right )}{2}-\frac {b x \ln \left (d x +c -1\right )}{2}-\frac {b \ln \left (d x +c -1\right ) c}{2 d}+\frac {b \ln \left (-d x -c -1\right ) c}{2 d}+\frac {b \ln \left (d x +c -1\right )}{2 d}+\frac {b \ln \left (-d x -c -1\right )}{2 d}\)	\(87\)

[In]

int(a+b*arccoth(d*x+c),x,method=_RETURNVERBOSE)

[Out]

a*x+b/d*((d*x+c)*arccoth(d*x+c)+1/2*ln((d*x+c)^2-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.50 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\frac {b d x \log \left (\frac {d x + c + 1}{d x + c - 1}\right ) + 2 \, a d x + {\left (b c + b\right )} \log \left (d x + c + 1\right ) - {\left (b c - b\right )} \log \left (d x + c - 1\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(b*d*x*log((d*x + c + 1)/(d*x + c - 1)) + 2*a*d*x + (b*c + b)*log(d*x + c + 1) - (b*c - b)*log(d*x + c - 1
))/d

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a x + b \left (\begin {cases} \frac {c \operatorname {acoth}{\left (c + d x \right )}}{d} + x \operatorname {acoth}{\left (c + d x \right )} + \frac {\log {\left (c + d x + 1 \right )}}{d} - \frac {\operatorname {acoth}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \operatorname {acoth}{\left (c \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(a+b*acoth(d*x+c),x)

[Out]

a*x + b*Piecewise((c*acoth(c + d*x)/d + x*acoth(c + d*x) + log(c + d*x + 1)/d - acoth(c + d*x)/d, Ne(d, 0)), (
x*acoth(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b}{2 \, d} \]

[In]

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

[Out]

a*x + 1/2*(2*(d*x + c)*arccoth(d*x + c) + log(-(d*x + c)^2 + 1))*b/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (38) = 76\).

Time = 0.26 (sec) , antiderivative size = 202, normalized size of antiderivative = 5.05 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\frac {1}{2} \, {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} b {\left (\frac {\log \left (\frac {{\left | d x + c + 1 \right |}}{{\left | d x + c - 1 \right |}}\right )}{d^{2}} - \frac {\log \left ({\left | \frac {d x + c + 1}{d x + c - 1} - 1 \right |}\right )}{d^{2}} + \frac {\log \left (-\frac {\frac {1}{c - \frac {{\left (\frac {{\left (d x + c + 1\right )} {\left (c - 1\right )}}{d x + c - 1} - c - 1\right )} d}{\frac {{\left (d x + c + 1\right )} d}{d x + c - 1} - d}} + 1}{\frac {1}{c - \frac {{\left (\frac {{\left (d x + c + 1\right )} {\left (c - 1\right )}}{d x + c - 1} - c - 1\right )} d}{\frac {{\left (d x + c + 1\right )} d}{d x + c - 1} - d}} - 1}\right )}{d^{2} {\left (\frac {d x + c + 1}{d x + c - 1} - 1\right )}}\right )} + a x \]

[In]

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

[Out]

1/2*((c + 1)*d - (c - 1)*d)*b*(log(abs(d*x + c + 1)/abs(d*x + c - 1))/d^2 - log(abs((d*x + c + 1)/(d*x + c - 1
) - 1))/d^2 + log(-(1/(c - ((d*x + c + 1)*(c - 1)/(d*x + c - 1) - c - 1)*d/((d*x + c + 1)*d/(d*x + c - 1) - d)
) + 1)/(1/(c - ((d*x + c + 1)*(c - 1)/(d*x + c - 1) - c - 1)*d/((d*x + c + 1)*d/(d*x + c - 1) - d)) - 1))/(d^2
*((d*x + c + 1)/(d*x + c - 1) - 1))) + a*x

Mupad [B] (verification not implemented)

Time = 4.56 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=a\,x+\frac {\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2}+b\,c\,\mathrm {acoth}\left (c+d\,x\right )}{d}+b\,x\,\mathrm {acoth}\left (c+d\,x\right ) \]

[In]

int(a + b*acoth(c + d*x),x)

[Out]

a*x + ((b*log(c^2 + d^2*x^2 + 2*c*d*x - 1))/2 + b*c*acoth(c + d*x))/d + b*x*acoth(c + d*x)