\(\int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx\) [6]

Optimal result

Integrand size = 12, antiderivative size = 78 \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2}{b d \sqrt {b \coth (c+d x)}} \]

[Out]

-arctan((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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3555, 3557, 335, 304, 209, 212} \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2}{b d \sqrt {b \coth (c+d x)}} \]

[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 209

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

Rule 212

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

Rule 304

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 335

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 3555

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 3557

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]

Rubi steps \begin{align*} \text {integral}& = -\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 {\text {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 \text {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 {\text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b d}-\frac {\text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b d} \\ & = -\frac {\arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {\text {arctanh}\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 [A] (verified)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\frac {-2-\arctan \left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}+\text {arctanh}\left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}}{b d \sqrt {b \coth (c+d x)}} \]

[In]

Integrate[(b*Coth[c + d*x])^(-3/2),x]

[Out]

(-2 - ArcTan[(Coth[c + d*x]^2)^(1/4)]*(Coth[c + d*x]^2)^(1/4) + ArcTanh[(Coth[c + d*x]^2)^(1/4)]*(Coth[c + d*x
]^2)^(1/4))/(b*d*Sqrt[b*Coth[c + d*x]])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}-\frac {2}{b d \sqrt {b \coth \left (d x +c \right )}}\)	\(65\)
default	\(-\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}-\frac {2}{b d \sqrt {b \coth \left (d x +c \right )}}\)	\(65\)

[In]

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

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (64) = 128\).

Time = 0.28 (sec) , antiderivative size = 923, normalized size of antiderivative = 11.83 \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

\[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (b \coth {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\int { \frac {1}{\left (b \coth \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 2.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {2}{b\,d\,\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}} \]

[In]

int(1/(b*coth(c + d*x))^(3/2),x)

[Out]

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