\(\int \sqrt {b \coth ^m(c+d x)} \, dx\) [52]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 54 \[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\frac {2 \coth (c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{4},\frac {6+m}{4},\coth ^2(c+d x)\right )}{d (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, 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} \[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\frac {2 \coth (c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{4},\frac {m+6}{4},\coth ^2(c+d x)\right )}{d (m+2)} \]

[In]

Int[Sqrt[b*Coth[c + d*x]^m],x]

[Out]

(2*Coth[c + d*x]*Sqrt[b*Coth[c + d*x]^m]*Hypergeometric2F1[1, (2 + m)/4, (6 + m)/4, Coth[c + d*x]^2])/(d*(2 +
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}& = \left (\coth ^{-\frac {m}{2}}(c+d x) \sqrt {b \coth ^m(c+d x)}\right ) \int \coth ^{\frac {m}{2}}(c+d x) \, dx \\ & = -\frac {\left (\coth ^{-\frac {m}{2}}(c+d x) \sqrt {b \coth ^m(c+d x)}\right ) \text {Subst}\left (\int \frac {x^{m/2}}{-1+x^2} \, dx,x,\coth (c+d x)\right )}{d} \\ & = \frac {2 \coth (c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{4},\frac {6+m}{4},\coth ^2(c+d x)\right )}{d (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\frac {2 \coth (c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{4},\frac {6+m}{4},\coth ^2(c+d x)\right )}{d (2+m)} \]

[In]

Integrate[Sqrt[b*Coth[c + d*x]^m],x]

[Out]

(2*Coth[c + d*x]*Sqrt[b*Coth[c + d*x]^m]*Hypergeometric2F1[1, (2 + m)/4, (6 + m)/4, Coth[c + d*x]^2])/(d*(2 +
m))

Maple [F]

\[\int \sqrt {b \coth \left (d x +c \right )^{m}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\int \sqrt {b \coth ^{m}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(b*coth(c + d*x)**m), x)

Maxima [F]

\[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\int { \sqrt {b \coth \left (d x + c\right )^{m}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(b*coth(d*x + c)^m), x)

Giac [F]

\[ \int \sqrt {b \coth ^m(c+d x)} \, dx=\int { \sqrt {b \coth \left (d x + c\right )^{m}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(b*coth(d*x + c)^m), x)

Mupad [F(-1)]

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

[In]

int((b*coth(c + d*x)^m)^(1/2),x)

[Out]

int((b*coth(c + d*x)^m)^(1/2), x)