3.20 \(\int \frac {1}{\sqrt {b \coth ^2(c+d x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt {b \coth ^2(c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3658, 3475} \[ \frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt {b \coth ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Coth[c + d*x]^2],x]

[Out]

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

Rule 3475

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

Rule 3658

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*} \int \frac {1}{\sqrt {b \coth ^2(c+d x)}} \, dx &=\frac {\coth (c+d x) \int \tanh (c+d x) \, dx}{\sqrt {b \coth ^2(c+d x)}}\\ &=\frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt {b \coth ^2(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 31, normalized size = 1.00 \[ \frac {\coth (c+d x) \log (\cosh (c+d x))}{d \sqrt {b \coth ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Coth[c + d*x]^2],x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 128, normalized size = 4.13 \[ -\frac {{\left (d x e^{\left (2 \, d x + 2 \, c\right )} - d x - {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )\right )} \sqrt {\frac {b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}}}{b d e^{\left (2 \, d x + 2 \, c\right )} + b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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*cosh(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))/(b*d*e^(2*d*x + 2*c)
+ b*d)

________________________________________________________________________________________

giac [B]  time = 0.15, size = 60, normalized size = 1.94 \[ -\frac {\frac {d x + c}{\sqrt {b} \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )} - \frac {\log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{\sqrt {b} \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.16, size = 52, normalized size = 1.68 \[ -\frac {\coth \left (d x +c \right ) \left (\ln \left (\coth \left (d x +c \right )-1\right )+\ln \left (\coth \left (d x +c \right )+1\right )-2 \ln \left (\coth \left (d x +c \right )\right )\right )}{2 d \sqrt {b \left (\coth ^{2}\left (d x +c \right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 34, normalized size = 1.10 \[ -\frac {d x + c}{\sqrt {b} d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{\sqrt {b} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.26, size = 30, normalized size = 0.97 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {b}\,\mathrm {coth}\left (c+d\,x\right )}{\sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^2}}\right )}{\sqrt {b}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \coth ^{2}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________