3.42 \(\int \frac{1}{\sqrt{b \coth ^4(c+d x)}} \, dx\)

Optimal. Leaf size=50 \[ \frac{x \coth ^2(c+d x)}{\sqrt{b \coth ^4(c+d x)}}-\frac{\coth (c+d x)}{d \sqrt{b \coth ^4(c+d x)}} \]

[Out]

-(Coth[c + d*x]/(d*Sqrt[b*Coth[c + d*x]^4])) + (x*Coth[c + d*x]^2)/Sqrt[b*Coth[c + d*x]^4]

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Rubi [A]  time = 0.0226638, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ \frac{x \coth ^2(c+d x)}{\sqrt{b \coth ^4(c+d x)}}-\frac{\coth (c+d x)}{d \sqrt{b \coth ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Coth[c + d*x]^4],x]

[Out]

-(Coth[c + d*x]/(d*Sqrt[b*Coth[c + d*x]^4])) + (x*Coth[c + d*x]^2)/Sqrt[b*Coth[c + d*x]^4]

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
]])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{b \coth ^4(c+d x)}} \, dx &=\frac{\coth ^2(c+d x) \int \tanh ^2(c+d x) \, dx}{\sqrt{b \coth ^4(c+d x)}}\\ &=-\frac{\coth (c+d x)}{d \sqrt{b \coth ^4(c+d x)}}+\frac{\coth ^2(c+d x) \int 1 \, dx}{\sqrt{b \coth ^4(c+d x)}}\\ &=-\frac{\coth (c+d x)}{d \sqrt{b \coth ^4(c+d x)}}+\frac{x \coth ^2(c+d x)}{\sqrt{b \coth ^4(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.072736, size = 40, normalized size = 0.8 \[ \frac{\coth (c+d x) \left (\tanh ^{-1}(\tanh (c+d x)) \coth (c+d x)-1\right )}{d \sqrt{b \coth ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Coth[c + d*x]^4],x]

[Out]

(Coth[c + d*x]*(-1 + ArcTanh[Tanh[c + d*x]]*Coth[c + d*x]))/(d*Sqrt[b*Coth[c + d*x]^4])

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Maple [A]  time = 0.037, size = 59, normalized size = 1.2 \begin{align*}{\frac{{\rm coth} \left (dx+c\right ) \left ( \ln \left ({\rm coth} \left (dx+c\right )+1 \right ){\rm coth} \left (dx+c\right )-\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){\rm coth} \left (dx+c\right )-2 \right ) }{2\,d}{\frac{1}{\sqrt{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(d*x+c)^4)^(1/2),x)

[Out]

1/2/d*coth(d*x+c)*(ln(coth(d*x+c)+1)*coth(d*x+c)-ln(coth(d*x+c)-1)*coth(d*x+c)-2)/(b*coth(d*x+c)^4)^(1/2)

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Maxima [A]  time = 1.73557, size = 49, normalized size = 0.98 \begin{align*} \frac{d x + c}{\sqrt{b} d} - \frac{2 \, \sqrt{b}}{{\left (b e^{\left (-2 \, d x - 2 \, c\right )} + b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.19262, size = 1079, normalized size = 21.58 \begin{align*} \frac{{\left (d x \cosh \left (d x + c\right )^{2} +{\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x\right )} \sinh \left (d x + c\right )^{2} + d x +{\left (d x \cosh \left (d x + c\right )^{2} + d x + 2\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \,{\left (d x \cosh \left (d x + c\right )^{2} + d x + 2\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d x \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2\right )} \sqrt{\frac{b e^{\left (8 \, d x + 8 \, c\right )} + 4 \, b e^{\left (6 \, d x + 6 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (8 \, d x + 8 \, c\right )} - 4 \, e^{\left (6 \, d x + 6 \, c\right )} + 6 \, e^{\left (4 \, d x + 4 \, c\right )} - 4 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}}}{b d \cosh \left (d x + c\right )^{2} +{\left (b d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d e^{\left (2 \, d x + 2 \, c\right )} + b d\right )} \sinh \left (d x + c\right )^{2} + b d +{\left (b d \cosh \left (d x + c\right )^{2} + b d\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \,{\left (b d \cosh \left (d x + c\right )^{2} + b d\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (b d \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + b d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

(d*x*cosh(d*x + c)^2 + (d*x*e^(4*d*x + 4*c) - 2*d*x*e^(2*d*x + 2*c) + d*x)*sinh(d*x + c)^2 + d*x + (d*x*cosh(d
*x + c)^2 + d*x + 2)*e^(4*d*x + 4*c) - 2*(d*x*cosh(d*x + c)^2 + d*x + 2)*e^(2*d*x + 2*c) + 2*(d*x*cosh(d*x + c
)*e^(4*d*x + 4*c) - 2*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + d*x*cosh(d*x + c))*sinh(d*x + c) + 2)*sqrt((b*e^(8*d
*x + 8*c) + 4*b*e^(6*d*x + 6*c) + 6*b*e^(4*d*x + 4*c) + 4*b*e^(2*d*x + 2*c) + b)/(e^(8*d*x + 8*c) - 4*e^(6*d*x
+ 6*c) + 6*e^(4*d*x + 4*c) - 4*e^(2*d*x + 2*c) + 1))/(b*d*cosh(d*x + c)^2 + (b*d*e^(4*d*x + 4*c) + 2*b*d*e^(2
*d*x + 2*c) + b*d)*sinh(d*x + c)^2 + b*d + (b*d*cosh(d*x + c)^2 + b*d)*e^(4*d*x + 4*c) + 2*(b*d*cosh(d*x + c)^
2 + b*d)*e^(2*d*x + 2*c) + 2*(b*d*cosh(d*x + c)*e^(4*d*x + 4*c) + 2*b*d*cosh(d*x + c)*e^(2*d*x + 2*c) + b*d*co
sh(d*x + c))*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \coth ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)**4)**(1/2),x)

[Out]

Integral(1/sqrt(b*coth(c + d*x)**4), x)

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Giac [A]  time = 1.16942, size = 46, normalized size = 0.92 \begin{align*} \frac{d x + c}{\sqrt{b} d} + \frac{2}{\sqrt{b} d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

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