3.1.31 \(\int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx\) [31]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3739, 3555, 3557, 335, 304, 209, 212} \begin {gather*} -\frac {\coth ^{\frac {3}{2}}(c+d x) \text {ArcTan}\left (\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) \tanh ^{-1}\left (\sqrt {\coth (c+d x)}\right )}{d \sqrt {b \coth ^3(c+d x)}} \end {gather*}

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 209

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

Rule 212

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

Rule 304

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 335

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 3555

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 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 3739

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
]])

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) \text {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 ) \text {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) \text {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) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 41, normalized size = 0.39 \begin {gather*} -\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)}} \end {gather*}

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 = 2.42, size = 92, normalized size = 0.88

method	result	size
derivativedivides	\(-\frac {\coth \left (d x +c \right ) \left (2 b^{\frac {5}{2}}+\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \left (\coth ^{3}\left (d x +c \right )\right )}\, b^{\frac {5}{2}}}\)	\(92\)
default	\(-\frac {\coth \left (d x +c \right ) \left (2 b^{\frac {5}{2}}+\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \left (\coth ^{3}\left (d x +c \right )\right )}\, b^{\frac {5}{2}}}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*coth(d*x+c)*(2*b^(5/2)+arctan((b*coth(d*x+c))^(1/2)/b^(1/2))*b^2*(b*coth(d*x+c))^(1/2)-arctanh((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (91) = 182\).
time = 0.40, size = 907, normalized size = 8.64 \begin {gather*} \left [-\frac {2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) + {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {-b} \log \left (-\frac {b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - 2 \, b}{\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4}}\right ) + 8 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, {\left (b d \cosh \left (d x + c\right )^{2} + 2 \, b d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b d \sinh \left (d x + c\right )^{2} + b d\right )}}, -\frac {2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) - {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {b} \log \left (2 \, b \cosh \left (d x + c\right )^{4} + 8 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 12 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 8 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b \sinh \left (d x + c\right )^{4} + 2 \, {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + {\left (6 \, \cosh \left (d x + c\right )^{2} - 1\right )} \sinh \left (d x + c\right )^{2} - \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cosh \left (d x + c\right )^{3} - \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - b\right ) + 8 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, {\left (b d \cosh \left (d x + c\right )^{2} + 2 \, b d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b d \sinh \left (d x + c\right )^{2} + b d\right )}}\right ] \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \coth ^{3}{\left (c + d x \right )}}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (91) = 182\).
time = 0.57, size = 279, normalized size = 2.66 \begin {gather*} \frac {\frac {2 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right )}{\sqrt {b} \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )} - \frac {\log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{\sqrt {b} \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )} - \frac {8}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} + \sqrt {b}\right )} \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}}{2 \, d} \end {gather*}

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]

1/2*(2*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))/(sqrt(b)*sgn(e^(6*d*x + 6*c) -
3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*d*x + 4*c) - 1)) - log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqr
t(b*e^(4*d*x + 4*c) - b)))/(sqrt(b)*sgn(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*
d*x + 4*c) - 1)) - 8/((sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) + sqrt(b))*sgn(e^(6*d*x + 6*c) -
3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*d*x + 4*c) - 1)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*coth(c + d*x)^3)^(1/2),x)

[Out]

int(1/(b*coth(c + d*x)^3)^(1/2), x)

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