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

Optimal result

Integrand size = 14, antiderivative size = 105 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=-\frac {2 \coth (c+d x)}{d \sqrt {b \coth ^3(c+d x)}}-\frac {\arctan \left (\sqrt {\coth (c+d x)}\right ) \coth ^{\frac {3}{2}}(c+d x)}{d \sqrt {b \coth ^3(c+d x)}}+\frac {\text {arctanh}\left (\sqrt {\coth (c+d x)}\right ) \coth ^{\frac {3}{2}}(c+d x)}{d \sqrt {b \coth ^3(c+d x)}} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3739, 3555, 3557, 335, 304, 209, 212} \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=-\frac {\coth ^{\frac {3}{2}}(c+d x) \arctan \left (\sqrt {\coth (c+d x)}\right )}{d \sqrt {b \coth ^3(c+d x)}}+\frac {\coth ^{\frac {3}{2}}(c+d x) \text {arctanh}\left (\sqrt {\coth (c+d x)}\right )}{d \sqrt {b \coth ^3(c+d x)}}-\frac {2 \coth (c+d x)}{d \sqrt {b \coth ^3(c+d x)}} \]

[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*} \text {integral}& = \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 {\arctan \left (\sqrt {\coth (c+d x)}\right ) \coth ^{\frac {3}{2}}(c+d x)}{d \sqrt {b \coth ^3(c+d x)}}+\frac {\text {arctanh}\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 [A] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=-\frac {\coth (c+d x) \left (2+\arctan \left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}-\text {arctanh}\left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}\right )}{d \sqrt {b \coth ^3(c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\coth \left (d x +c \right ) \left (-2 b^{\frac {5}{2}}+\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \coth \left (d x +c \right )^{3}}\, b^{\frac {5}{2}}}\)	\(91\)
default	\(\frac {\coth \left (d x +c \right ) \left (-2 b^{\frac {5}{2}}+\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \coth \left (d x +c \right )^{3}}\, b^{\frac {5}{2}}}\)	\(91\)

[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)+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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (91) = 182\).

Time = 0.30 (sec) , antiderivative size = 907, normalized size of antiderivative = 8.64 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^3}} \,d x \]

[In]

int(1/(b*coth(c + d*x)^3)^(1/2),x)

[Out]

int(1/(b*coth(c + d*x)^3)^(1/2), x)