\(\int \sqrt {b \coth ^2(c+d x)} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=\frac {\sqrt {b \coth ^2(c+d x)} \log (\sinh (c+d x)) \tanh (c+d x)}{d} \]

[Out]

ln(sinh(d*x+c))*(b*coth(d*x+c)^2)^(1/2)*tanh(d*x+c)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3739, 3556} \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=\frac {\tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \log (\sinh (c+d x))}{d} \]

[In]

Int[Sqrt[b*Coth[c + d*x]^2],x]

[Out]

(Sqrt[b*Coth[c + d*x]^2]*Log[Sinh[c + d*x]]*Tanh[c + d*x])/d

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=\frac {\sqrt {b \coth ^2(c+d x)} (\log (\cosh (c+d x))+\log (\tanh (c+d x))) \tanh (c+d x)}{d} \]

[In]

Integrate[Sqrt[b*Coth[c + d*x]^2],x]

[Out]

(Sqrt[b*Coth[c + d*x]^2]*(Log[Cosh[c + d*x]] + Log[Tanh[c + d*x]])*Tanh[c + d*x])/d

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(-\frac {\sqrt {\coth \left (d x +c \right )^{2} b}\, \left (\ln \left (\coth \left (d x +c \right )-1\right )+\ln \left (\coth \left (d x +c \right )+1\right )\right )}{2 d \coth \left (d x +c \right )}\)	\(45\)
default	\(-\frac {\sqrt {\coth \left (d x +c \right )^{2} b}\, \left (\ln \left (\coth \left (d x +c \right )-1\right )+\ln \left (\coth \left (d x +c \right )+1\right )\right )}{2 d \coth \left (d x +c \right )}\)	\(45\)
risch	\(\frac {\sqrt {\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} b}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}-1\right ) x}{{\mathrm e}^{2 d x +2 c}+1}-\frac {2 \sqrt {\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} b}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (d x +c \right )}{\left ({\mathrm e}^{2 d x +2 c}+1\right ) d}+\frac {\sqrt {\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} b}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}-1\right ) \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\left ({\mathrm e}^{2 d x +2 c}+1\right ) d}\)	\(192\)

[In]

int((coth(d*x+c)^2*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (29) = 58\).

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 4.03 \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=-\frac {{\left (d x e^{\left (2 \, d x + 2 \, c\right )} - d x - {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )\right )} \sqrt {\frac {b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}}}{d e^{\left (2 \, d x + 2 \, c\right )} + d} \]

[In]

integrate((b*coth(d*x+c)^2)^(1/2),x, algorithm="fricas")

[Out]

-(d*x*e^(2*d*x + 2*c) - d*x - (e^(2*d*x + 2*c) - 1)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))*sqrt
((b*e^(4*d*x + 4*c) + 2*b*e^(2*d*x + 2*c) + b)/(e^(4*d*x + 4*c) - 2*e^(2*d*x + 2*c) + 1))/(d*e^(2*d*x + 2*c) +
d)

Sympy [F]

\[ \int \sqrt {b \coth ^2(c+d x)} \, dx=\int \sqrt {b \coth ^{2}{\left (c + d x \right )}}\, dx \]

[In]

integrate((b*coth(d*x+c)**2)**(1/2),x)

[Out]

Integral(sqrt(b*coth(c + d*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=-\frac {{\left (d x + c\right )} \sqrt {b}}{d} - \frac {\sqrt {b} \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {\sqrt {b} \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} \]

[In]

integrate((b*coth(d*x+c)^2)^(1/2),x, algorithm="maxima")

[Out]

-(d*x + c)*sqrt(b)/d - sqrt(b)*log(e^(-d*x - c) + 1)/d - sqrt(b)*log(e^(-d*x - c) - 1)/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=-\frac {{\left ({\left (d x + c\right )} \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) - \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )\right )} \sqrt {b}}{d} \]

[In]

integrate((b*coth(d*x+c)^2)^(1/2),x, algorithm="giac")

[Out]

-((d*x + c)*sgn(e^(4*d*x + 4*c) - 1) - log(abs(e^(2*d*x + 2*c) - 1))*sgn(e^(4*d*x + 4*c) - 1))*sqrt(b)/d

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \coth ^2(c+d x)} \, dx=\int \sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^2} \,d x \]

[In]

int((b*coth(c + d*x)^2)^(1/2),x)

[Out]

int((b*coth(c + d*x)^2)^(1/2), x)