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

Optimal. Leaf size=69 \[ \frac{2 \coth ^{1-m}(c+d x) \, _2F_1\left (1,\frac{1}{4} (2-3 m);\frac{3 (2-m)}{4};\coth ^2(c+d x)\right )}{b d (2-3 m) \sqrt{b \coth ^m(c+d x)}} \]

[Out]

(2*Coth[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])

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Rubi [A]  time = 0.0481878, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3659, 3476, 364} \[ \frac{2 \coth ^{1-m}(c+d x) \, _2F_1\left (1,\frac{1}{4} (2-3 m);\frac{3 (2-m)}{4};\coth ^2(c+d x)\right )}{b d (2-3 m) \sqrt{b \coth ^m(c+d x)}} \]

Antiderivative was successfully verified.

[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 3659

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]])

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (b \coth ^m(c+d x)\right )^{3/2}} \, dx &=\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) \operatorname{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) \, _2F_1\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]  time = 0.0661005, size = 58, normalized size = 0.84 \[ -\frac{2 \coth (c+d x) \, _2F_1\left (1,\frac{1}{4} (2-3 m);-\frac{3}{4} (m-2);\coth ^2(c+d x)\right )}{d (3 m-2) \left (b \coth ^m(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.168, size = 0, normalized size = 0. \begin{align*} \int \left ( b \left ({\rm coth} \left (dx+c\right ) \right ) ^{m} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{m}\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)^m)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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