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

Optimal result

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

[Out]

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*co
th(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3555, 3557, 335, 218, 212, 209} \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {2}{3 b d (b \coth (c+d x))^{3/2}} \]

[In]

Int[(b*Coth[c + d*x])^(-5/2),x]

[Out]

ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(5/2)*d) + ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(5/2)*d) - 2/(3*
b*d*(b*Coth[c + d*x])^(3/2))

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 218

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 b d (b \coth (c+d x))^{3/2}}+\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}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \coth (c+d x)\right )}{b d} \\ & = -\frac {2}{3 b d (b \coth (c+d x))^{3/2}}-\frac {2 \text {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b d} \\ & = -\frac {2}{3 b d (b \coth (c+d x))^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b^2 d}+\frac {\text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b^2 d} \\ & = \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}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(b*Coth[c + d*x])^(-5/2),x]

[Out]

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

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

[In]

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

[Out]

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*co
th(d*x+c))^(3/2)

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 \[ \int \frac {1}{(b \coth (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-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)*sin
h(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((cos
h(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)) + 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*sin
h(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))*s
inh(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*cosh(d*x + c))*sinh(d*x + c)), 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(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)) + 3*(cosh(d*x + c)^4 + 4*cos
h(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(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)*sqrt(b*cosh(d*x + c)/si
nh(d*x + c)) - b) - 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*c
osh(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*cosh(d*x + c))*sinh(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate(1/(b*coth(d*x+c))^(5/2),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 [B] (verification not implemented)

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

[In]

int(1/(b*coth(c + d*x))^(5/2),x)

[Out]

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)