\(\int (b \coth ^m(c+d x))^{3/2} \, dx\) [51]

Optimal result

Integrand size = 14, antiderivative size = 63 \[ \int \left (b \coth ^m(c+d x)\right )^{3/2} \, dx=\frac {2 b \coth ^{1+m}(c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2+3 m),\frac {3 (2+m)}{4},\coth ^2(c+d x)\right )}{d (2+3 m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, 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} \[ \int \left (b \coth ^m(c+d x)\right )^{3/2} \, dx=\frac {2 b \coth ^{m+1}(c+d x) \sqrt {b \coth ^m(c+d x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3 m+2),\frac {3 (m+2)}{4},\coth ^2(c+d x)\right )}{d (3 m+2)} \]

[In]

Int[(b*Coth[c + d*x]^m)^(3/2),x]

[Out]

(2*b*Coth[c + d*x]^(1 + m)*Sqrt[b*Coth[c + d*x]^m]*Hypergeometric2F1[1, (2 + 3*m)/4, (3*(2 + m))/4, Coth[c + d
*x]^2])/(d*(2 + 3*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 (b \coth ^{-\frac {m}{2}}(c+d x) \sqrt {b \coth ^m(c+d x)}\right ) \int \coth ^{\frac {3 m}{2}}(c+d x) \, dx \\ & = -\frac {\left (b \coth ^{-\frac {m}{2}}(c+d x) \sqrt {b \coth ^m(c+d x)}\right ) \text {Subst}\left (\int \frac {x^{3 m/2}}{-1+x^2} \, dx,x,\coth (c+d x)\right )}{d} \\ & = \frac {2 b \coth ^{1+m}(c+d x) \sqrt {b \coth ^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 )}{d (2+3 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \left (b \coth ^m(c+d x)\right )^{3/2} \, dx=\frac {\coth (c+d x) \left (b \coth ^m(c+d x)\right )^{3/2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2+3 m),\frac {3 (2+m)}{4},\coth ^2(c+d x)\right )}{d \left (1+\frac {3 m}{2}\right )} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

integrate((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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[In]

int((b*coth(c + d*x)^m)^(3/2),x)

[Out]

int((b*coth(c + d*x)^m)^(3/2), x)