\(\int \frac {1}{(b \coth ^m(c+d x))^{3/2}} \, dx\) [54]

Optimal result

Integrand size = 14, antiderivative size = 69 \[ \int \frac {1}{\left (b \coth ^m(c+d x)\right )^{3/2}} \, dx=\frac {2 \coth ^{1-m}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2-3 m),\frac {3 (2-m)}{4},\coth ^2(c+d x)\right )}{b d (2-3 m) \sqrt {b \coth ^m(c+d x)}} \]

[Out]

2*coth(d*x+c)^(1-m)*hypergeom([1, 1/2-3/4*m],[3/2-3/4*m],coth(d*x+c)^2)/b/d/(2-3*m)/(b*coth(d*x+c)^m)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3740, 3557, 371} \[ \int \frac {1}{\left (b \coth ^m(c+d x)\right )^{3/2}} \, dx=\frac {2 \coth ^{1-m}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2-3 m),\frac {3 (2-m)}{4},\coth ^2(c+d x)\right )}{b d (2-3 m) \sqrt {b \coth ^m(c+d x)}} \]

[In]

Int[(b*Coth[c + d*x]^m)^(-3/2),x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

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]

Rule 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, 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*} \text {integral}& = \frac {\coth ^{\frac {m}{2}}(c+d x) \int \coth ^{-\frac {3 m}{2}}(c+d x) \, dx}{b \sqrt {b \coth ^m(c+d x)}} \\ & = -\frac {\coth ^{\frac {m}{2}}(c+d x) \text {Subst}\left (\int \frac {x^{-3 m/2}}{-1+x^2} \, dx,x,\coth (c+d x)\right )}{b d \sqrt {b \coth ^m(c+d x)}} \\ & = \frac {2 \coth ^{1-m}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2-3 m),\frac {3 (2-m)}{4},\coth ^2(c+d x)\right )}{b d (2-3 m) \sqrt {b \coth ^m(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b \coth ^m(c+d x)\right )^{3/2}} \, dx=\frac {\coth (c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2-3 m),-\frac {3}{4} (-2+m),\coth ^2(c+d x)\right )}{d \left (1-\frac {3 m}{2}\right ) \left (b \coth ^m(c+d x)\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(b*coth(c + d*x)^m)^(3/2),x)

[Out]

int(1/(b*coth(c + d*x)^m)^(3/2), x)