3.1.9 \(\int (a+b \text {csch}^2(c+d x))^{5/2} \, dx\) [9]

3.1.9.1 Optimal result

Integrand size = 16, antiderivative size = 174 \[ \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx=\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{8 d}-\frac {(7 a-3 b) b \coth (c+d x) \sqrt {a-b+b \coth ^2(c+d x)}}{8 d}-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d} \]

output

a^(5/2)*arctanh(coth(d*x+c)*a^(1/2)/(a-b+b*coth(d*x+c)^2)^(1/2))/d-1/4*b*c 
oth(d*x+c)*(a-b+b*coth(d*x+c)^2)^(3/2)/d-1/8*(15*a^2-10*a*b+3*b^2)*arctanh 
(coth(d*x+c)*b^(1/2)/(a-b+b*coth(d*x+c)^2)^(1/2))*b^(1/2)/d-1/8*(7*a-3*b)* 
b*coth(d*x+c)*(a-b+b*coth(d*x+c)^2)^(1/2)/d
 

3.1.9.2 Mathematica [A] (verified)

Time = 3.39 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.33 \[ \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx=\frac {\left (a+b \text {csch}^2(c+d x)\right )^{5/2} \left (-2 \sqrt {2} \sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \cosh (c+d x)}{\sqrt {-a+2 b+a \cosh (2 (c+d x))}}\right )+b \sqrt {-a+2 b+a \cosh (2 (c+d x))} (9 a-7 b+(-9 a+3 b) \cosh (2 (c+d x))) \coth (c+d x) \text {csch}^3(c+d x)+16 \sqrt {2} a^{5/2} \log \left (\sqrt {2} \sqrt {a} \cosh (c+d x)+\sqrt {-a+2 b+a \cosh (2 (c+d x))}\right )\right ) \sinh ^5(c+d x)}{4 d (-a+2 b+a \cosh (2 (c+d x)))^{5/2}} \]

input

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

output

((a + b*Csch[c + d*x]^2)^(5/2)*(-2*Sqrt[2]*Sqrt[b]*(15*a^2 - 10*a*b + 3*b^ 
2)*ArcTanh[(Sqrt[2]*Sqrt[b]*Cosh[c + d*x])/Sqrt[-a + 2*b + a*Cosh[2*(c + d 
*x)]]] + b*Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]]*(9*a - 7*b + (-9*a + 3*b)* 
Cosh[2*(c + d*x)])*Coth[c + d*x]*Csch[c + d*x]^3 + 16*Sqrt[2]*a^(5/2)*Log[ 
Sqrt[2]*Sqrt[a]*Cosh[c + d*x] + Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]]])*Sin 
h[c + d*x]^5)/(4*d*(-a + 2*b + a*Cosh[2*(c + d*x)])^(5/2))
 

3.1.9.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {3042, 4616, 318, 25, 403, 25, 398, 224, 219, 291, 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.

\(\displaystyle \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

\(\displaystyle \frac {\int \frac {\left (b \coth ^2(c+d x)+a-b\right )^{5/2}}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d}\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {-\frac {1}{4} \int -\frac {\sqrt {b \coth ^2(c+d x)+a-b} \left ((7 a-3 b) b \coth ^2(c+d x)+(4 a-3 b) (a-b)\right )}{1-\coth ^2(c+d x)}d\coth (c+d x)-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sqrt {b \coth ^2(c+d x)+a-b} \left ((7 a-3 b) b \coth ^2(c+d x)+(4 a-3 b) (a-b)\right )}{1-\coth ^2(c+d x)}d\coth (c+d x)-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {1}{2} \int -\frac {b \left (15 a^2-10 b a+3 b^2\right ) \coth ^2(c+d x)+(a-b) \left (8 a^2-7 b a+3 b^2\right )}{\left (1-\coth ^2(c+d x)\right ) \sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 a^2-10 b a+3 b^2\right ) \coth ^2(c+d x)+(a-b) \left (8 a^2-7 b a+3 b^2\right )}{\left (1-\coth ^2(c+d x)\right ) \sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (1-\coth ^2(c+d x)\right ) \sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)-b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{\sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)\right )-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (1-\coth ^2(c+d x)\right ) \sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)-b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{1-\frac {b \coth ^2(c+d x)}{b \coth ^2(c+d x)+a-b}}d\frac {\coth (c+d x)}{\sqrt {b \coth ^2(c+d x)+a-b}}\right )-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (1-\coth ^2(c+d x)\right ) \sqrt {b \coth ^2(c+d x)+a-b}}d\coth (c+d x)-\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )\right )-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{1-\frac {a \coth ^2(c+d x)}{b \coth ^2(c+d x)+a-b}}d\frac {\coth (c+d x)}{\sqrt {b \coth ^2(c+d x)+a-b}}-\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )\right )-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )-\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )\right )-\frac {1}{2} b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}\right )-\frac {1}{4} b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{d}\)

input

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

output

(-1/4*(b*Coth[c + d*x]*(a - b + b*Coth[c + d*x]^2)^(3/2)) + ((8*a^(5/2)*Ar 
cTanh[(Sqrt[a]*Coth[c + d*x])/Sqrt[a - b + b*Coth[c + d*x]^2]] - Sqrt[b]*( 
15*a^2 - 10*a*b + 3*b^2)*ArcTanh[(Sqrt[b]*Coth[c + d*x])/Sqrt[a - b + b*Co 
th[c + d*x]^2]])/2 - ((7*a - 3*b)*b*Coth[c + d*x]*Sqrt[a - b + b*Coth[c + 
d*x]^2])/2)/4)/d
 

3.1.9.3.1 Defintions of rubi rules used

rule 25

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

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[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]
 

3.1.9.4 Maple [F]

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

input

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

output

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

3.1.9.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2759 vs. \(2 (152) = 304\).

Time = 0.73 (sec) , antiderivative size = 12590, normalized size of antiderivative = 72.36 \[ \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.1.9.6 Sympy [F]

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

input

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

output

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

3.1.9.7 Maxima [F]

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

input

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

output

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

3.1.9.8 Giac [F(-2)]

Exception generated. \[ \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

3.1.9.9 Mupad [F(-1)]

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

input

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

output

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