3.1.7 \(\int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx\) [7]

3.1.7.1 Optimal result

Integrand size = 12, antiderivative size = 79 \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}} \]

output

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

3.1.7.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\frac {-2+3 \arctan \left (\sqrt [4]{\coth ^2(c+d x)}\right ) \coth ^2(c+d x)^{3/4}+3 \text {arctanh}\left (\sqrt [4]{\coth ^2(c+d x)}\right ) \coth ^2(c+d x)^{3/4}}{3 b d (b \coth (c+d x))^{3/2}} \]

input

Integrate[(b*Coth[c + d*x])^(-5/2),x]
 

output

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

3.1.7.3 Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3955, 3042, 3957, 25, 266, 756, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (-i b \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\int \frac {1}{\sqrt {b \coth (c+d x)}}dx}{b^2}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2}{3 b d (b \coth (c+d x))^{3/2}}+\frac {\int \frac {1}{\sqrt {-i b \tan \left (i c+i d x+\frac {\pi }{2}\right )}}dx}{b^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {\int -\frac {1}{\sqrt {b \coth (c+d x)} \left (b^2-b^2 \coth ^2(c+d x)\right )}d(b \coth (c+d x))}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {1}{\sqrt {b \coth (c+d x)} \left (b^2-b^2 \coth ^2(c+d x)\right )}d(b \coth (c+d x))}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 \int \frac {1}{b^2-b^4 \coth ^4(c+d x)}d\sqrt {b \coth (c+d x)}}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{b-b^2 \coth ^2(c+d x)}d\sqrt {b \coth (c+d x)}}{2 b}+\frac {\int \frac {1}{b^2 \coth ^2(c+d x)+b}d\sqrt {b \coth (c+d x)}}{2 b}\right )}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{b-b^2 \coth ^2(c+d x)}d\sqrt {b \coth (c+d x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}\right )}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (\frac {\arctan \left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}\right )}{b d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}}\)

input

Int[(b*Coth[c + d*x])^(-5/2),x]
 

output

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

3.1.7.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

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

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3955

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] 
)^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2   Int[(b*Tan[c + d*x])^(n + 2), x] 
, x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
 

3.1.7.4 Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}-\frac {2}{3 b d \left (b \coth \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(64\)
default	\(\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}-\frac {2}{3 b d \left (b \coth \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(64\)

input

int(1/(b*coth(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

output

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

3.1.7.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 689 vs. \(2 (63) = 126\).

Time = 0.28 (sec) , antiderivative size = 1428, normalized size of antiderivative = 18.08 \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(1/(b*coth(d*x+c))^(5/2),x, algorithm="fricas")
 

output

[-1/12*(6*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + 
c)^4 + 2*(3*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 2*cosh(d*x + c)^2 + 4*( 
cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 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)*sqr 
t(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*si 
nh(d*x + c) + b*sinh(d*x + c)^2 + b)) + 3*(cosh(d*x + c)^4 + 4*cosh(d*x + 
c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 + 1)*sinh(d*x 
+ c)^2 + 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x 
+ c) + 1)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x 
+ c) + 6*b*cosh(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)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2 
*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d*x 
 + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b*cosh(d*x + c)/sinh(d*x 
+ c)))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^ 
3*d*sinh(d*x + c)^4 + 2*b^3*d*cosh(d*x + c)^2 + b^3*d + 2*(3*b^3*d*cosh(d* 
x + c)^2 + b^3*d)*sinh(d*x + c)^2 + 4*(b^3*d*cosh(d*x + c)^3 + b^3*d*co...
 

3.1.7.6 Sympy [F]

\[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (b \coth {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(1/(b*coth(d*x+c))**(5/2),x)
 

output

Integral((b*coth(c + d*x))**(-5/2), x)
 

3.1.7.7 Maxima [F]

\[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \coth \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(b*coth(d*x+c))^(5/2),x, algorithm="maxima")
 

output

integrate((b*coth(d*x + c))^(-5/2), x)
 

3.1.7.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(b*coth(d*x+c))^(5/2),x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

3.1.7.9 Mupad [B] (verification not implemented)

Time = 2.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{5/2}\,d}-\frac {2}{3\,b\,d\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{3/2}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{5/2}\,d} \]

input

int(1/(b*coth(c + d*x))^(5/2),x)
 

output

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