\(\int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx\) [320]

Optimal result

Integrand size = 25, antiderivative size = 144 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\frac {a (3 a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{8 (a-b)^{3/2} f}+\frac {(3 a-2 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{8 (a-b) f}+\frac {\text {sech}^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{4 f} \] Output:

1/8*a*(3*a-4*b)*arctan((a-b)^(1/2)*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))/ 
(a-b)^(3/2)/f+1/8*(3*a-2*b)*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x 
+e)/(a-b)/f+1/4*sech(f*x+e)^3*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 10.05 (sec) , antiderivative size = 684, normalized size of antiderivative = 4.75 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=-\frac {\text {sech}^3(e+f x) \left (1+\frac {b \sinh ^2(e+f x)}{a}\right ) \tanh (e+f x) \left (-15 a \arcsin \left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right )-10 b \arcsin \left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right ) \sinh ^2(e+f x)-30 a \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}-20 b \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}-32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}-32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}+32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{7/2}+32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{7/2}+15 a \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}+10 b \sinh ^2(e+f x) \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}\right )}{40 f \sqrt {a+b \sinh ^2(e+f x)} \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}} \] Input:

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

Output:

-1/40*(Sech[e + f*x]^3*(1 + (b*Sinh[e + f*x]^2)/a)*Tanh[e + f*x]*(-15*a*Ar 
cSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]] - 10*b*ArcSin[Sqrt[((a - b)*Tanh[e 
 + f*x]^2)/a]]*Sinh[e + f*x]^2 - 30*a*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e 
+ f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(3/2) - 20*b*Sinh[e + f*x]^2*S 
qrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2 
)/a)^(3/2) - 32*a*Hypergeometric2F1[2, 4, 7/2, ((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)^(5/2) - 32*b*Hypergeometric2F1[2, 4, 7/2, ((a - b)*Tanh[e + f*x]^2 
)/a]*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)^(5/2) + 32*a*Hypergeometric2F1[2, 4, 7/2, ((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) + 32*b*Hypergeometric2F1[2, 4, 7/2, (( 
a - b)*Tanh[e + f*x]^2)/a]*Sinh[e + f*x]^2*Sqrt[(Sech[e + f*x]^2*(a + b*Si 
nh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(7/2) + 15*a*Sqrt[((a - b 
)*Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2)*Tanh[e + f*x]^2)/a^2] + 10*b*Sin 
h[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]))/(f*Sqrt[a + 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))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3669, 296, 292, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^(3/2))/(4*(a - b)*(1 + Sinh[e + f* 
x]^2)^2) + ((3*a - 4*b)*((a*ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b* 
Sinh[e + f*x]^2]])/(2*Sqrt[a - b]) + (Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f* 
x]^2])/(2*(1 + Sinh[e + f*x]^2))))/(4*(a - b)))/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 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

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 = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.24

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

Input:

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

Output:

`int/indef0`(1/cosh(f*x+e)^6*(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. 1805 vs. \(2 (128) = 256\).

Time = 0.28 (sec) , antiderivative size = 3727, normalized size of antiderivative = 25.88 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(sech(f*x+e)^5*(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(sech(f*x+e)**5*(a+b*sinh(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(sech(f*x+e)^5*(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2122 vs. \(2 (128) = 256\).

Time = 0.46 (sec) , antiderivative size = 2122, normalized size of antiderivative = 14.74 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(sech(f*x+e)^5*(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

1/4*((3*a^2 - 4*a*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*f*e^(4*e) - b*f*e^(4*e))*sqrt(a - b)) - 2*(3*(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))^7*a^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))^7*a*b + 21*(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))^6*a^2*sqrt(b) - 60*(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))^6*a*b^(3/2) + 32*(sqr 
t(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))^6*b^(5/2) + 44*(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))^5*a^3 - 253 
*(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))^5*a^2*b + 252*(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))^5*a*b^2 
 - 64*(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))^5*b^3 - 292*(sqrt(b)*e^(2*f*x + 2*e) - sqr 
t(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^4*a^ 
3*sqrt(b) + 317*(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))^4*a^2*b^(3/2) - 28*(sqrt(b)*e...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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