3.21 \(\int \frac{1}{(b \coth ^2(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=65 \[ \frac{\coth (c+d x) \log (\cosh (c+d x))}{b d \sqrt{b \coth ^2(c+d x)}}-\frac{\tanh (c+d x)}{2 b d \sqrt{b \coth ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0356045, antiderivative size = 65, 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, 3475} \[ \frac{\coth (c+d x) \log (\cosh (c+d x))}{b d \sqrt{b \coth ^2(c+d x)}}-\frac{\tanh (c+d x)}{2 b d \sqrt{b \coth ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^2)^(-3/2),x]

[Out]

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

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 3475

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

Rubi steps

\begin{align*} \int \frac{1}{\left (b \coth ^2(c+d x)\right )^{3/2}} \, dx &=\frac{\coth (c+d x) \int \tanh ^3(c+d x) \, dx}{b \sqrt{b \coth ^2(c+d x)}}\\ &=-\frac{\tanh (c+d x)}{2 b d \sqrt{b \coth ^2(c+d x)}}+\frac{\coth (c+d x) \int \tanh (c+d x) \, dx}{b \sqrt{b \coth ^2(c+d x)}}\\ &=\frac{\coth (c+d x) \log (\cosh (c+d x))}{b d \sqrt{b \coth ^2(c+d x)}}-\frac{\tanh (c+d x)}{2 b d \sqrt{b \coth ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.143311, size = 48, normalized size = 0.74 \[ \frac{2 \coth (c+d x) \log (\cosh (c+d x))-\tanh (c+d x)}{2 b d \sqrt{b \coth ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x]^2)^(-3/2),x]

[Out]

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

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Maple [A]  time = 0.02, size = 79, normalized size = 1.2 \begin{align*} -{\frac{{\rm coth} \left (dx+c\right ) \left ( \ln \left ({\rm coth} \left (dx+c\right )+1 \right ) \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}-2\,\ln \left ({\rm coth} \left (dx+c\right ) \right ) \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}+\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}+1 \right ) }{2\,d} \left ( b \left ({\rm coth} \left (dx+c\right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(d*x+c)^2)^(3/2),x)

[Out]

-1/2/d*coth(d*x+c)*(ln(coth(d*x+c)+1)*coth(d*x+c)^2-2*ln(coth(d*x+c))*coth(d*x+c)^2+ln(coth(d*x+c)-1)*coth(d*x
+c)^2+1)/(b*coth(d*x+c)^2)^(3/2)

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Maxima [A]  time = 1.58899, size = 113, normalized size = 1.74 \begin{align*} -\frac{2 \, \sqrt{b} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + b^{2} e^{\left (-4 \, d x - 4 \, c\right )} + b^{2}\right )} d} - \frac{d x + c}{b^{\frac{3}{2}} d} - \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{\frac{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^2)^(3/2),x, algorithm="maxima")

[Out]

-2*sqrt(b)*e^(-2*d*x - 2*c)/((2*b^2*e^(-2*d*x - 2*c) + b^2*e^(-4*d*x - 4*c) + b^2)*d) - (d*x + c)/(b^(3/2)*d)
- log(e^(-2*d*x - 2*c) + 1)/(b^(3/2)*d)

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Fricas [B]  time = 1.96297, size = 2082, normalized size = 32.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^2)^(3/2),x, algorithm="fricas")

[Out]

(d*x*cosh(d*x + c)^4 - (d*x*e^(2*d*x + 2*c) - d*x)*sinh(d*x + c)^4 - 4*(d*x*cosh(d*x + c)*e^(2*d*x + 2*c) - d*
x*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(d*x - 1)*cosh(d*x + c)^2 + 2*(3*d*x*cosh(d*x + c)^2 + d*x - (3*d*x*cosh(
d*x + c)^2 + d*x - 1)*e^(2*d*x + 2*c) - 1)*sinh(d*x + c)^2 + d*x - (d*x*cosh(d*x + c)^4 + 2*(d*x - 1)*cosh(d*x
+ c)^2 + d*x)*e^(2*d*x + 2*c) + ((e^(2*d*x + 2*c) - 1)*sinh(d*x + c)^4 - cosh(d*x + c)^4 + 4*(cosh(d*x + c)*e
^(2*d*x + 2*c) - cosh(d*x + c))*sinh(d*x + c)^3 - 2*(3*cosh(d*x + c)^2 - (3*cosh(d*x + c)^2 + 1)*e^(2*d*x + 2*
c) + 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + (cosh(d*x + c)^4 + 2*cosh(d*x + c)^2 + 1)*e^(2*d*x + 2*c) - 4*(c
osh(d*x + c)^3 - (cosh(d*x + c)^3 + cosh(d*x + c))*e^(2*d*x + 2*c) + cosh(d*x + c))*sinh(d*x + c) - 1)*log(2*c
osh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(d*x*cosh(d*x + c)^3 + (d*x - 1)*cosh(d*x + c) - (d*x*cosh(d
*x + c)^3 + (d*x - 1)*cosh(d*x + c))*e^(2*d*x + 2*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^2*d*cosh(d*x + c)^4 + 2*b^2*d*cosh(d*x + c)^2 + (b^2*d
*e^(2*d*x + 2*c) + b^2*d)*sinh(d*x + c)^4 + 4*(b^2*d*cosh(d*x + c)*e^(2*d*x + 2*c) + b^2*d*cosh(d*x + c))*sinh
(d*x + c)^3 + b^2*d + 2*(3*b^2*d*cosh(d*x + c)^2 + b^2*d + (3*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(2*d*x + 2*c))*
sinh(d*x + c)^2 + (b^2*d*cosh(d*x + c)^4 + 2*b^2*d*cosh(d*x + c)^2 + b^2*d)*e^(2*d*x + 2*c) + 4*(b^2*d*cosh(d*
x + c)^3 + b^2*d*cosh(d*x + c) + (b^2*d*cosh(d*x + c)^3 + b^2*d*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \coth ^{2}{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)**2)**(3/2),x)

[Out]

Integral((b*coth(c + d*x)**2)**(-3/2), x)

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Giac [A]  time = 1.2425, size = 149, normalized size = 2.29 \begin{align*} -\frac{\frac{d x + c}{\sqrt{b} d \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} d \mathrm{sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )} - \frac{2 \, e^{\left (2 \, d x + 2 \, c\right )}}{\sqrt{b} d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2} \mathrm{sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^2)^(3/2),x, algorithm="giac")

[Out]

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