\(\int \frac {\text {sech}^3(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\) [349]

Optimal result

Integrand size = 25, antiderivative size = 142 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 (a-b)^{5/2} f}+\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}} \] Output:

1/2*(a-4*b)*arctan((a-b)^(1/2)*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))/(a-b 
)^(5/2)/f+1/2*b*(a+2*b)*sinh(f*x+e)/a/(a-b)^2/f/(a+b*sinh(f*x+e)^2)^(1/2)+ 
1/2*sech(f*x+e)*tanh(f*x+e)/(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 = 3.84 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.63 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {sech}^5(e+f x) \left (16 (a-b) \, _3F_2\left (2,2,3;1,\frac {9}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^2+16 (a-b) \operatorname {Hypergeometric2F1}\left (2,3,\frac {9}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (4 a^2+7 a b \sinh ^2(e+f x)+3 b^2 \sinh ^4(e+f x)\right )+7 a \cosh ^2(e+f x) \operatorname {Hypergeometric2F1}\left (1,2,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \left (15 a^2+20 a b \sinh ^2(e+f x)+8 b^2 \sinh ^4(e+f x)\right )\right ) \tanh (e+f x)}{105 a^4 f \sqrt {a+b \sinh ^2(e+f x)}} \] Input:

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

Output:

(Sech[e + f*x]^5*(16*(a - b)*HypergeometricPFQ[{2, 2, 3}, {1, 9/2}, ((a - 
b)*Tanh[e + f*x]^2)/a]*Sinh[e + f*x]^2*(a + b*Sinh[e + f*x]^2)^2 + 16*(a - 
 b)*Hypergeometric2F1[2, 3, 9/2, ((a - b)*Tanh[e + f*x]^2)/a]*Sinh[e + f*x 
]^2*(4*a^2 + 7*a*b*Sinh[e + f*x]^2 + 3*b^2*Sinh[e + f*x]^4) + 7*a*Cosh[e + 
 f*x]^2*Hypergeometric2F1[1, 2, 7/2, ((a - b)*Tanh[e + f*x]^2)/a]*(15*a^2 
+ 20*a*b*Sinh[e + f*x]^2 + 8*b^2*Sinh[e + f*x]^4))*Tanh[e + f*x])/(105*a^4 
*f*Sqrt[a + b*Sinh[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3669, 316, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {a (a-4 b)}{\left (\sinh ^2(e+f x)+1\right ) \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{a (a-b)}+\frac {b (a+2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{2 (a-b)}+\frac {\sinh (e+f x)}{2 (a-b) \left (\sinh ^2(e+f x)+1\right ) \sqrt {a+b \sinh ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {(a-4 b) \int \frac {1}{\left (\sinh ^2(e+f x)+1\right ) \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{a-b}+\frac {b (a+2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{2 (a-b)}+\frac {\sinh (e+f x)}{2 (a-b) \left (\sinh ^2(e+f x)+1\right ) \sqrt {a+b \sinh ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {(a-4 b) \int \frac {1}{1-\frac {(b-a) \sinh ^2(e+f x)}{b \sinh ^2(e+f x)+a}}d\frac {\sinh (e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}}{a-b}+\frac {b (a+2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{2 (a-b)}+\frac {\sinh (e+f x)}{2 (a-b) \left (\sinh ^2(e+f x)+1\right ) \sqrt {a+b \sinh ^2(e+f x)}}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {(a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{(a-b)^{3/2}}+\frac {b (a+2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{2 (a-b)}+\frac {\sinh (e+f x)}{2 (a-b) \left (\sinh ^2(e+f x)+1\right ) \sqrt {a+b \sinh ^2(e+f x)}}}{f}\)

Input:

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

Output:

(Sinh[e + f*x]/(2*(a - b)*(1 + Sinh[e + f*x]^2)*Sqrt[a + b*Sinh[e + f*x]^2 
]) + (((a - 4*b)*ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f* 
x]^2]])/(a - b)^(3/2) + (b*(a + 2*b)*Sinh[e + f*x])/(a*(a - b)*Sqrt[a + b* 
Sinh[e + f*x]^2]))/(2*(a - b)))/f
 

Defintions of rubi rules used

rule 25

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

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 = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67

\[\frac {\operatorname {`\,int/indef0`\,}\left (-\frac {\sqrt {a +b \sinh \left (f x +e \right )^{2}}\, \cosh \left (f x +e \right )^{2}}{-b^{2} \cosh \left (f x +e \right )^{10}+\left (-2 a b +2 b^{2}\right ) \cosh \left (f x +e \right )^{8}+\left (-a^{2}+2 a b -b^{2}\right ) \cosh \left (f x +e \right )^{6}}, \sinh \left (f x +e \right )\right )}{f}\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2364 vs. \(2 (126) = 252\).

Time = 0.37 (sec) , antiderivative size = 4845, normalized size of antiderivative = 34.12 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sech(f*x + e)^3/(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. 1031 vs. \(2 (126) = 252\).

Time = 0.51 (sec) , antiderivative size = 1031, normalized size of antiderivative = 7.26 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(((a^3*b^2*f*e^(6*e) - 3*a^2*b^3*f*e^(6*e) + 3*a*b^4*f*e^(6*e) - b^5*f*e^( 
6*e))*e^(2*f*x)/(a^6*f^2*e^(10*e) - 5*a^5*b*f^2*e^(10*e) + 10*a^4*b^2*f^2* 
e^(10*e) - 10*a^3*b^3*f^2*e^(10*e) + 5*a^2*b^4*f^2*e^(10*e) - a*b^5*f^2*e^ 
(10*e)) - (a^3*b^2*f*e^(4*e) - 3*a^2*b^3*f*e^(4*e) + 3*a*b^4*f*e^(4*e) - b 
^5*f*e^(4*e))/(a^6*f^2*e^(10*e) - 5*a^5*b*f^2*e^(10*e) + 10*a^4*b^2*f^2*e^ 
(10*e) - 10*a^3*b^3*f^2*e^(10*e) + 5*a^2*b^4*f^2*e^(10*e) - a*b^5*f^2*e^(1 
0*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) + (a - 4*b)*arctan(-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^2*f*e^(6*e) - 2*a*b*f*e^(6*e) + b^2*f*e^(6*e))*sqrt(a - b)) - 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))^3*a - 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))^3*b - 5*(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))^2*a*sqrt(b) + 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))^2*b^(3/2) - 4 
*(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))*a^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))*a*b + 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*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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