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

Optimal. Leaf size=78 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}-\frac{2}{b d \sqrt{b \coth (c+d x)}} \]

[Out]

-(ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(3/2)*d)) + ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(3/2)*d) - 2/
(b*d*Sqrt[b*Coth[c + d*x]])

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Rubi [A]  time = 0.0512037, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3474, 3476, 329, 298, 203, 206} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}-\frac{2}{b d \sqrt{b \coth (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(-3/2),x]

[Out]

-(ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(3/2)*d)) + ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(3/2)*d) - 2/
(b*d*Sqrt[b*Coth[c + d*x]])

Rule 3474

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(b \coth (c+d x))^{3/2}} \, dx &=-\frac{2}{b d \sqrt{b \coth (c+d x)}}+\frac{\int \sqrt{b \coth (c+d x)} \, dx}{b^2}\\ &=-\frac{2}{b d \sqrt{b \coth (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{-b^2+x^2} \, dx,x,b \coth (c+d x)\right )}{b d}\\ &=-\frac{2}{b d \sqrt{b \coth (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{-b^2+x^4} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{b d}\\ &=-\frac{2}{b d \sqrt{b \coth (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{b d}-\frac{\operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{b d}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}-\frac{2}{b d \sqrt{b \coth (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0737085, size = 36, normalized size = 0.46 \[ -\frac{2 \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\coth ^2(c+d x)\right )}{b d \sqrt{b \coth (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(-3/2),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, Coth[c + d*x]^2])/(b*d*Sqrt[b*Coth[c + d*x]])

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Maple [A]  time = 0.013, size = 65, normalized size = 0.8 \begin{align*} -{\frac{1}{d}\arctan \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{3}{2}}}}+{\frac{1}{d}{\it Artanh} \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{3}{2}}}}-2\,{\frac{1}{bd\sqrt{b{\rm coth} \left (dx+c\right )}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/b^(3/2)/d+arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))/b^(3/2)/d-2/b/d/(b*cot
h(d*x+c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.71562, size = 2576, normalized size = 33.03 \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))^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*arctan((cosh(d*x + c
)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*
x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) + (cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh
(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*co
sh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(cosh(d*x + c)^2 + 2
*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - 2*b)/(cosh(
d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x +
c)^3 + sinh(d*x + c)^4)) + 8*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b*co
sh(d*x + c)/sinh(d*x + c)))/(b^2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*sinh(d*x + c) + b^2*d*sinh(d*x + c)
^2 + b^2*d), -1/4*(2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(b)*arctan(sq
rt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x +
c)^2 + b)) - (cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(b)*log(2*b*cosh(d*x
+ c)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x
+ c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*co
sh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c))*sq
rt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - b) + 8*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x
+ c)^2 - 1)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/(b^2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*sinh(d*x + c)
+ b^2*d*sinh(d*x + c)^2 + b^2*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \coth{\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))**(3/2),x)

[Out]

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

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Giac [B]  time = 1.33478, size = 277, normalized size = 3.55 \begin{align*} \frac{\frac{{\left (\pi + \log \left ({\left | b \right |}\right ) + 8\right )} \mathrm{sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{\sqrt{b} d} + \frac{4 \, \arctan \left (-\frac{\sqrt{b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt{b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt{b}}\right )}{\sqrt{b} d \mathrm{sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} - \frac{2 \, \log \left ({\left | -\sqrt{b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt{b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{\sqrt{b} d \mathrm{sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} - \frac{16}{{\left (\sqrt{b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt{b e^{\left (4 \, d x + 4 \, c\right )} - b} + \sqrt{b}\right )} d \mathrm{sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((pi + log(abs(b)) + 8)*sgn(e^(2*d*x + 2*c) - 1)/(sqrt(b)*d) + 4*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b
*e^(4*d*x + 4*c) - b))/sqrt(b))/(sqrt(b)*d*sgn(e^(2*d*x + 2*c) - 1)) - 2*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sq
rt(b*e^(4*d*x + 4*c) - b)))/(sqrt(b)*d*sgn(e^(2*d*x + 2*c) - 1)) - 16/((sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*
d*x + 4*c) - b) + sqrt(b))*d*sgn(e^(2*d*x + 2*c) - 1)))/b