\(\int \frac {\text {sech}(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\) [348]

Optimal result

Integrand size = 23, antiderivative size = 85 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \sinh (e+f x)}{a (a-b) f \sqrt {a+b \sinh ^2(e+f x)}} \] Output:

arctan((a-b)^(1/2)*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-b* 
sinh(f*x+e)/a/(a-b)/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 = 3.42 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.99 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {sech}(e+f x) \tanh (e+f x) \left (\frac {4 (a-b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{15 a^2}+\frac {\text {csch}^2(e+f x) \left (3 a+2 b \sinh ^2(e+f x)\right ) \left (-\arcsin \left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right ) \left (a+b \sinh ^2(e+f x)\right )+a \cosh ^2(e+f x) \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}\right )}{a (a-b) \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}}\right )}{a f \sqrt {a+b \sinh ^2(e+f x)}} \] Input:

Integrate[Sech[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]
 

Output:

(Sech[e + f*x]*Tanh[e + f*x]*((4*(a - b)*Hypergeometric2F1[2, 2, 7/2, ((a 
- b)*Tanh[e + f*x]^2)/a]*(a + b*Sinh[e + f*x]^2)*Tanh[e + f*x]^2)/(15*a^2) 
 + (Csch[e + f*x]^2*(3*a + 2*b*Sinh[e + f*x]^2)*(-(ArcSin[Sqrt[((a - b)*Ta 
nh[e + f*x]^2)/a]]*(a + b*Sinh[e + f*x]^2)) + a*Cosh[e + f*x]^2*Sqrt[((a - 
 b)*Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2)*Tanh[e + f*x]^2)/a^2]))/(a*(a 
- b)*Sqrt[((a - b)*Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2)*Tanh[e + f*x]^2 
)/a^2])))/(a*f*Sqrt[a + b*Sinh[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3669, 296, 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.

\(\displaystyle \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (i e+i f x) \left (a-b \sin (i e+i f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[Sech[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

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 296

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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[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 = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\operatorname {`\,int/indef0`\,}\left (\frac {-a -b \sinh \left (f x +e \right )^{2}}{\left (-b^{2} \sinh \left (f x +e \right )^{6}+\left (-2 a b -b^{2}\right ) \sinh \left (f x +e \right )^{4}+\left (-a^{2}-2 a b \right ) \sinh \left (f x +e \right )^{2}-a^{2}\right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}}, \sinh \left (f x +e \right )\right )}{f}\)	\(101\)

Input:

int(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

`int/indef0`((-a-b*sinh(f*x+e)^2)/(-b^2*sinh(f*x+e)^6+(-2*a*b-b^2)*sinh(f* 
x+e)^4+(-a^2-2*a*b)*sinh(f*x+e)^2-a^2)/(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. 800 vs. \(2 (77) = 154\).

Time = 0.19 (sec) , antiderivative size = 1717, normalized size of antiderivative = 20.20 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/2*((a*b*cosh(f*x + e)^4 + 4*a*b*cosh(f*x + e)*sinh(f*x + e)^3 + a*b*sin 
h(f*x + e)^4 + 2*(2*a^2 - a*b)*cosh(f*x + e)^2 + 2*(3*a*b*cosh(f*x + e)^2 
+ 2*a^2 - a*b)*sinh(f*x + e)^2 + a*b + 4*(a*b*cosh(f*x + e)^3 + (2*a^2 - a 
*b)*cosh(f*x + e))*sinh(f*x + e))*sqrt(-a + b)*log(((a - 2*b)*cosh(f*x + e 
)^4 + 4*(a - 2*b)*cosh(f*x + e)*sinh(f*x + e)^3 + (a - 2*b)*sinh(f*x + e)^ 
4 - 2*(3*a - 2*b)*cosh(f*x + e)^2 + 2*(3*(a - 2*b)*cosh(f*x + e)^2 - 3*a + 
 2*b)*sinh(f*x + e)^2 + 2*sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh( 
f*x + e) + sinh(f*x + e)^2 - 1)*sqrt(-a + b)*sqrt((b*cosh(f*x + e)^2 + b*s 
inh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) 
 + sinh(f*x + e)^2)) + 4*((a - 2*b)*cosh(f*x + e)^3 - (3*a - 2*b)*cosh(f*x 
 + e))*sinh(f*x + e) + a - 2*b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f* 
x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e)^2 + 2 
*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1)) 
 - 2*sqrt(2)*((a*b - b^2)*cosh(f*x + e)^2 + 2*(a*b - b^2)*cosh(f*x + e)*si 
nh(f*x + e) + (a*b - b^2)*sinh(f*x + e)^2 - a*b + b^2)*sqrt((b*cosh(f*x + 
e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sin 
h(f*x + e) + sinh(f*x + e)^2)))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + 
e)^4 + 4*(a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^ 
3*b - 2*a^2*b^2 + a*b^3)*f*sinh(f*x + e)^4 + 2*(2*a^4 - 5*a^3*b + 4*a^2*b^ 
2 - a*b^3)*f*cosh(f*x + e)^2 + 2*(3*(a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(...
 

Sympy [F]

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

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)**2)**(3/2),x)
 

Output:

Integral(sech(e + f*x)/(a + b*sinh(e + f*x)**2)**(3/2), x)
 

Maxima [F]

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

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(sech(f*x + e)/(b*sinh(f*x + e)^2 + a)^(3/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (77) = 154\).

Time = 0.26 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.80 \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=-{\left (\frac {\frac {{\left (a^{2} b f e^{\left (4 \, e\right )} - 2 \, a b^{2} f e^{\left (4 \, e\right )} + b^{3} f e^{\left (4 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a^{4} f^{2} e^{\left (6 \, e\right )} - 3 \, a^{3} b f^{2} e^{\left (6 \, e\right )} + 3 \, a^{2} b^{2} f^{2} e^{\left (6 \, e\right )} - a b^{3} f^{2} e^{\left (6 \, e\right )}} - \frac {a^{2} b f e^{\left (2 \, e\right )} - 2 \, a b^{2} f e^{\left (2 \, e\right )} + b^{3} f e^{\left (2 \, e\right )}}{a^{4} f^{2} e^{\left (6 \, e\right )} - 3 \, a^{3} b f^{2} e^{\left (6 \, e\right )} + 3 \, a^{2} b^{2} f^{2} e^{\left (6 \, e\right )} - a b^{3} f^{2} e^{\left (6 \, e\right )}}}{\sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}} - \frac {2 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right )}{{\left (a f e^{\left (4 \, e\right )} - b f e^{\left (4 \, e\right )}\right )} \sqrt {a - b}}\right )} e^{\left (4 \, e\right )} \] Input:

integrate(sech(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

-(((a^2*b*f*e^(4*e) - 2*a*b^2*f*e^(4*e) + b^3*f*e^(4*e))*e^(2*f*x)/(a^4*f^ 
2*e^(6*e) - 3*a^3*b*f^2*e^(6*e) + 3*a^2*b^2*f^2*e^(6*e) - a*b^3*f^2*e^(6*e 
)) - (a^2*b*f*e^(2*e) - 2*a*b^2*f*e^(2*e) + b^3*f*e^(2*e))/(a^4*f^2*e^(6*e 
) - 3*a^3*b*f^2*e^(6*e) + 3*a^2*b^2*f^2*e^(6*e) - a*b^3*f^2*e^(6*e)))/sqrt 
(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b) - 2*ar 
ctan(-1/2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x 
 + 2*e) - 2*b*e^(2*f*x + 2*e) + b) + sqrt(b))/sqrt(a - b))/((a*f*e^(4*e) - 
 b*f*e^(4*e))*sqrt(a - b)))*e^(4*e)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\mathrm {cosh}\left (e+f\,x\right )\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(1/(cosh(e + f*x)*(a + b*sinh(e + f*x)^2)^(3/2)),x)
 

Output:

int(1/(cosh(e + f*x)*(a + b*sinh(e + f*x)^2)^(3/2)), x)
 

Reduce [F]

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

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

Output:

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