\(\int \frac {\text {sech}(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\) [358]

Optimal result

Integrand size = 23, antiderivative size = 134 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {b \sinh (e+f x)}{3 a (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(5 a-2 b) b \sinh (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}} \] Output:

arctan((a-b)^(1/2)*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/ 
3*b*sinh(f*x+e)/a/(a-b)/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/3*(5*a-2*b)*b*sinh(f 
*x+e)/a^2/(a-b)^2/f/(a+b*sinh(f*x+e)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 8.35 (sec) , antiderivative size = 1331, normalized size of antiderivative = 9.93 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[Sech[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]
 

Output:

(Sech[e + f*x]*Tanh[e + f*x]*(1575*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a 
]] + (2100*b*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Sinh[e + f*x]^2)/a 
+ (840*b^2*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Sinh[e + f*x]^4)/a^2 
- (3150*(a - b)*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Tanh[e + f*x]^2) 
/a - (4200*(a - b)*b*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Sinh[e + f* 
x]^2*Tanh[e + f*x]^2)/a^2 - (1680*(a - b)*b^2*ArcSin[Sqrt[((a - b)*Tanh[e 
+ f*x]^2)/a]]*Sinh[e + f*x]^4*Tanh[e + f*x]^2)/a^3 + (1575*(a - b)^2*ArcSi 
n[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Tanh[e + f*x]^4)/a^2 + (2100*(a - b)^ 
2*b*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Sinh[e + f*x]^2*Tanh[e + f*x 
]^4)/a^3 + (840*(a - b)^2*b^2*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Si 
nh[e + f*x]^4*Tanh[e + f*x]^4)/a^4 + 2100*Sqrt[(Sech[e + f*x]^2*(a + b*Sin 
h[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(3/2) + (2800*b*Sinh[e + f 
*x]^2*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + 
 f*x]^2)/a)^(3/2))/a + (1120*b^2*Sinh[e + f*x]^4*Sqrt[(Sech[e + f*x]^2*(a 
+ b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(3/2))/a^2 + 96*Hyp 
ergeometric2F1[2, 2, 9/2, ((a - b)*Tanh[e + f*x]^2)/a]*Sqrt[(Sech[e + f*x] 
^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(7/2) + 24*Hy 
pergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Tanh[e + f*x]^2)/a]*Sqrt[(Se 
ch[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(7 
/2) + (168*b*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Tanh[e + f*x]^2)/a]*...
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3669, 316, 402, 27, 291, 216}

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.

\(\displaystyle \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (i e+i f x) \left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3669

\(\displaystyle \frac {\int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh (e+f x)}{f}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\int \frac {-2 b \sinh ^2(e+f x)+3 a-2 b}{\left (\sinh ^2(e+f x)+1\right ) \left (b \sinh ^2(e+f x)+a\right )^{3/2}}d\sinh (e+f x)}{3 a (a-b)}-\frac {b \sinh (e+f x)}{3 a (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2}{\left (\sinh ^2(e+f x)+1\right ) \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{a (a-b)}-\frac {b (5 a-2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{3 a (a-b)}-\frac {b \sinh (e+f x)}{3 a (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{a-b}-\frac {b (5 a-2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{3 a (a-b)}-\frac {b \sinh (e+f x)}{3 a (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{1-\frac {(b-a) \sinh ^2(e+f x)}{b \sinh ^2(e+f x)+a}}d\frac {\sinh (e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}}{a-b}-\frac {b (5 a-2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{3 a (a-b)}-\frac {b \sinh (e+f x)}{3 a (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {3 a \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{(a-b)^{3/2}}-\frac {b (5 a-2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{3 a (a-b)}-\frac {b \sinh (e+f x)}{3 a (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\)

Input:

Int[Sech[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(b*Sinh[e + f*x])/(a*(a - b)*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((3*a* 
ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]])/(a - b)^( 
3/2) - ((5*a - 2*b)*b*Sinh[e + f*x])/(a*(a - b)*Sqrt[a + b*Sinh[e + f*x]^2 
]))/(3*a*(a - b)))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 316

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

rule 402

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

rule 3042

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

rule 3669

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   S 
ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] 
/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.69 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {\operatorname {`\,int/indef0`\,}\left (\frac {-b^{2} \sinh \left (f x +e \right )^{4}-2 \sinh \left (f x +e \right )^{2} a b -a^{2}}{\left (-b^{4} \sinh \left (f x +e \right )^{10}+\left (-4 a \,b^{3}-b^{4}\right ) \sinh \left (f x +e \right )^{8}+\left (-6 a^{2} b^{2}-4 a \,b^{3}\right ) \sinh \left (f x +e \right )^{6}+\left (-4 a^{3} b -6 a^{2} b^{2}\right ) \sinh \left (f x +e \right )^{4}+\left (-a^{4}-4 a^{3} b \right ) \sinh \left (f x +e \right )^{2}-a^{4}\right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}}, \sinh \left (f x +e \right )\right )}{f}\)	\(169\)
risch	\(\text {Expression too large to display}\)	\(735304\)

Input:

int(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

`int/indef0`((-b^2*sinh(f*x+e)^4-2*sinh(f*x+e)^2*a*b-a^2)/(-b^4*sinh(f*x+e 
)^10+(-4*a*b^3-b^4)*sinh(f*x+e)^8+(-6*a^2*b^2-4*a*b^3)*sinh(f*x+e)^6+(-4*a 
^3*b-6*a^2*b^2)*sinh(f*x+e)^4+(-a^4-4*a^3*b)*sinh(f*x+e)^2-a^4)/(a+b*sinh( 
f*x+e)^2)^(1/2),sinh(f*x+e))/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2640 vs. \(2 (120) = 240\).

Time = 0.39 (sec) , antiderivative size = 5396, normalized size of antiderivative = 40.27 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\operatorname {sech}{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)**2)**(5/2),x)
 

Output:

Integral(sech(e + f*x)/(a + b*sinh(e + f*x)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\operatorname {sech}\left (f x + e\right )}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(sech(f*x + e)/(b*sinh(f*x + e)^2 + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\mathrm {cosh}\left (e+f\,x\right )\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

int(1/(cosh(e + f*x)*(a + b*sinh(e + f*x)^2)^(5/2)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(cosh(e + f*x)*(a + b*sinh(e + f*x)^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sinh \left (f x +e \right )^{2} b +a}\, \mathrm {sech}\left (f x +e \right )}{\sinh \left (f x +e \right )^{6} b^{3}+3 \sinh \left (f x +e \right )^{4} a \,b^{2}+3 \sinh \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sinh(e + f*x)**2*b + a)*sech(e + f*x))/(sinh(e + f*x)**6*b**3 + 
3*sinh(e + f*x)**4*a*b**2 + 3*sinh(e + f*x)**2*a**2*b + a**3),x)