\(\int \frac {1}{(a+b \text {csch}^2(c+d x))^{5/2}} \, dx\) [14]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{a^{5/2} d}+\frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}+\frac {(5 a-3 b) b \coth (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a-b+b \coth ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4213, 425, 541, 12, 385, 212} \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )}{a^{5/2} d}+\frac {b (5 a-3 b) \coth (c+d x)}{3 a^2 d (a-b)^2 \sqrt {a+b \coth ^2(c+d x)-b}}+\frac {b \coth (c+d x)}{3 a d (a-b) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}} \]

[In]

Int[(a + b*Csch[c + d*x]^2)^(-5/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a-b+b x^2\right )^{5/2}} \, dx,x,\coth (c+d x)\right )}{d} \\ & = \frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-3 a+b+2 b x^2}{\left (1-x^2\right ) \left (a-b+b x^2\right )^{3/2}} \, dx,x,\coth (c+d x)\right )}{3 a (a-b) d} \\ & = \frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}+\frac {(5 a-3 b) b \coth (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a-b+b \coth ^2(c+d x)}}+\frac {\text {Subst}\left (\int \frac {3 (a-b)^2}{\left (1-x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\coth (c+d x)\right )}{3 a^2 (a-b)^2 d} \\ & = \frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}+\frac {(5 a-3 b) b \coth (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a-b+b \coth ^2(c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\coth (c+d x)\right )}{a^2 d} \\ & = \frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}+\frac {(5 a-3 b) b \coth (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a-b+b \coth ^2(c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{a^2 d} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{a^{5/2} d}+\frac {b \coth (c+d x)}{3 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}+\frac {(5 a-3 b) b \coth (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a-b+b \coth ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \, dx=\frac {\text {csch}^5(c+d x) \left (-\frac {4 b \cosh (c+d x) (-a+2 b+a \cosh (2 (c+d x))) \left (3 a^2-7 a b+3 b^2+a (-3 a+2 b) \cosh (2 (c+d x))\right )}{3 a^2 (a-b)^2}+\frac {\sqrt {2} (-a+2 b+a \cosh (2 (c+d x)))^{5/2} \log \left (\sqrt {2} \sqrt {a} \cosh (c+d x)+\sqrt {-a+2 b+a \cosh (2 (c+d x))}\right )}{a^{5/2}}\right )}{8 d \left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \]

[In]

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

[Out]

(Csch[c + d*x]^5*((-4*b*Cosh[c + d*x]*(-a + 2*b + a*Cosh[2*(c + d*x)])*(3*a^2 - 7*a*b + 3*b^2 + a*(-3*a + 2*b)
*Cosh[2*(c + d*x)]))/(3*a^2*(a - b)^2) + (Sqrt[2]*(-a + 2*b + a*Cosh[2*(c + d*x)])^(5/2)*Log[Sqrt[2]*Sqrt[a]*C
osh[c + d*x] + Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]]])/a^(5/2)))/(8*d*(a + b*Csch[c + d*x]^2)^(5/2))

Maple [F]

\[\int \frac {1}{\left (a +b \operatorname {csch}\left (d x +c \right )^{2}\right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3577 vs. \(2 (121) = 242\).

Time = 0.61 (sec) , antiderivative size = 7831, normalized size of antiderivative = 58.01 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {sinh}\left (c+d\,x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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