3.31 \(\int \frac{1}{\sqrt{b \coth ^3(c+d x)}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\coth ^{\frac{3}{2}}(c+d x) \tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}}+\frac{\coth ^{\frac{3}{2}}(c+d x) \tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}} \]

[Out]

(-2*Coth[c + d*x])/(d*Sqrt[b*Coth[c + d*x]^3]) - (ArcTan[Sqrt[Coth[c + d*x]]]*Coth[c + d*x]^(3/2))/(d*Sqrt[b*C
oth[c + d*x]^3]) + (ArcTanh[Sqrt[Coth[c + d*x]]]*Coth[c + d*x]^(3/2))/(d*Sqrt[b*Coth[c + d*x]^3])

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Rubi [A]  time = 0.0480527, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3658, 3474, 3476, 329, 298, 203, 206} \[ -\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\coth ^{\frac{3}{2}}(c+d x) \tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}}+\frac{\coth ^{\frac{3}{2}}(c+d x) \tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Coth[c + d*x]^3],x]

[Out]

(-2*Coth[c + d*x])/(d*Sqrt[b*Coth[c + d*x]^3]) - (ArcTan[Sqrt[Coth[c + d*x]]]*Coth[c + d*x]^(3/2))/(d*Sqrt[b*C
oth[c + d*x]^3]) + (ArcTanh[Sqrt[Coth[c + d*x]]]*Coth[c + d*x]^(3/2))/(d*Sqrt[b*Coth[c + d*x]^3])

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, 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 3474

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 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 329

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 298

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 203

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{b \coth ^3(c+d x)}} \, dx &=\frac{\coth ^{\frac{3}{2}}(c+d x) \int \frac{1}{\coth ^{\frac{3}{2}}(c+d x)} \, dx}{\sqrt{b \coth ^3(c+d x)}}\\ &=-\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}+\frac{\coth ^{\frac{3}{2}}(c+d x) \int \sqrt{\coth (c+d x)} \, dx}{\sqrt{b \coth ^3(c+d x)}}\\ &=-\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\coth ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+x^2} \, dx,x,\coth (c+d x)\right )}{d \sqrt{b \coth ^3(c+d x)}}\\ &=-\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\left (2 \coth ^{\frac{3}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}}\\ &=-\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}+\frac{\coth ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\coth ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\coth (c+d x)}\right )}{d \sqrt{b \coth ^3(c+d x)}}\\ &=-\frac{2 \coth (c+d x)}{d \sqrt{b \coth ^3(c+d x)}}-\frac{\tan ^{-1}\left (\sqrt{\coth (c+d x)}\right ) \coth ^{\frac{3}{2}}(c+d x)}{d \sqrt{b \coth ^3(c+d x)}}+\frac{\tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right ) \coth ^{\frac{3}{2}}(c+d x)}{d \sqrt{b \coth ^3(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0294052, size = 41, normalized size = 0.39 \[ -\frac{2 \coth (c+d x) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\coth ^2(c+d x)\right )}{d \sqrt{b \coth ^3(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Coth[c + d*x]^3],x]

[Out]

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

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Maple [A]  time = 0.041, size = 91, normalized size = 0.9 \begin{align*}{\frac{{\rm coth} \left (dx+c\right )}{d} \left ( -2\,{b}^{5/2}+{\it Artanh} \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ){b}^{2}\sqrt{b{\rm coth} \left (dx+c\right )}-\arctan \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ){b}^{2}\sqrt{b{\rm coth} \left (dx+c\right )} \right ){\frac{1}{\sqrt{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*coth(d*x+c)*(-2*b^(5/2)+arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))*b^2*(b*coth(d*x+c))^(1/2)-arctan((b*coth(d*
x+c))^(1/2)/b^(1/2))*b^2*(b*coth(d*x+c))^(1/2))/(b*coth(d*x+c)^3)^(1/2)/b^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \coth \left (d x + c\right )^{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.38283, size = 2554, normalized size = 24.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*coth(d*x+c)^3)^(1/2),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*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(d*x + c)^2 + b
*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(sqrt(b)*sq
rt(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*cosh(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))*sqrt(b)*sq
rt(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*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + c) + b*d*sinh(
d*x + c)^2 + b*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \coth ^{3}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*coth(c + d*x)**3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError