Integrand size = 25, antiderivative size = 151 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\frac {a (3 a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{8 (a-b)^{3/2} f}+\frac {(3 a-4 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{8 (a-b) f}+\frac {\text {sech}^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh (e+f x)}{4 (a-b) f} \]
1/8*a*(3*a-4*b)*arctan(sinh(f*x+e)*(a-b)^(1/2)/(a+b*sinh(f*x+e)^2)^(1/2))/ (a-b)^(3/2)/f+1/4*sech(f*x+e)^3*(a+b*sinh(f*x+e)^2)^(3/2)*tanh(f*x+e)/(a-b )/f+1/8*(3*a-4*b)*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/(a-b)/ f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.46 (sec) , antiderivative size = 684, normalized size of antiderivative = 4.53 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=-\frac {\text {sech}^3(e+f x) \left (1+\frac {b \sinh ^2(e+f x)}{a}\right ) \tanh (e+f x) \left (-15 a \arcsin \left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right )-10 b \arcsin \left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right ) \sinh ^2(e+f x)-30 a \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}-20 b \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}-32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}-32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}+32 a \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{7/2}+32 b \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{7/2}+15 a \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}+10 b \sinh ^2(e+f x) \sqrt {\frac {(a-b) \text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \tanh ^2(e+f x)}{a^2}}\right )}{40 f \sqrt {a+b \sinh ^2(e+f x)} \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}} \]
-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))
Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02, 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 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \left (\frac {a \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 \sqrt {a-b}}+\frac {\sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{2 \left (\sinh ^2(e+f x)+1\right )}\right )}{4 (a-b)}+\frac {\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{4 (a-b) \left (\sinh ^2(e+f x)+1\right )^2}}{f}\) |
((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
3.4.56.3.1 Defintions of rubi rules used
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])
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.23
\[\frac {\operatorname {`\,int/indef0`\,}\left (\frac {\sqrt {a +b \sinh \left (f x +e \right )^{2}}}{\cosh \left (f x +e \right )^{6}}, \sinh \left (f x +e \right )\right )}{f}\]
Leaf count of result is larger than twice the leaf count of optimal. 1805 vs. \(2 (135) = 270\).
Time = 0.43 (sec) , antiderivative size = 3727, normalized size of antiderivative = 24.68 \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\text {Too large to display} \]
[-1/16*(((3*a^2 - 4*a*b)*cosh(f*x + e)^8 + 8*(3*a^2 - 4*a*b)*cosh(f*x + e) *sinh(f*x + e)^7 + (3*a^2 - 4*a*b)*sinh(f*x + e)^8 + 4*(3*a^2 - 4*a*b)*cos h(f*x + e)^6 + 4*(7*(3*a^2 - 4*a*b)*cosh(f*x + e)^2 + 3*a^2 - 4*a*b)*sinh( f*x + e)^6 + 8*(7*(3*a^2 - 4*a*b)*cosh(f*x + e)^3 + 3*(3*a^2 - 4*a*b)*cosh (f*x + e))*sinh(f*x + e)^5 + 6*(3*a^2 - 4*a*b)*cosh(f*x + e)^4 + 2*(35*(3* a^2 - 4*a*b)*cosh(f*x + e)^4 + 30*(3*a^2 - 4*a*b)*cosh(f*x + e)^2 + 9*a^2 - 12*a*b)*sinh(f*x + e)^4 + 8*(7*(3*a^2 - 4*a*b)*cosh(f*x + e)^5 + 10*(3*a ^2 - 4*a*b)*cosh(f*x + e)^3 + 3*(3*a^2 - 4*a*b)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(3*a^2 - 4*a*b)*cosh(f*x + e)^2 + 4*(7*(3*a^2 - 4*a*b)*cosh(f*x + e)^6 + 15*(3*a^2 - 4*a*b)*cosh(f*x + e)^4 + 9*(3*a^2 - 4*a*b)*cosh(f*x + e)^2 + 3*a^2 - 4*a*b)*sinh(f*x + e)^2 + 3*a^2 - 4*a*b + 8*((3*a^2 - 4*a*b) *cosh(f*x + e)^7 + 3*(3*a^2 - 4*a*b)*cosh(f*x + e)^5 + 3*(3*a^2 - 4*a*b)*c osh(f*x + e)^3 + (3*a^2 - 4*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*sinh(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 ...
\[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\int \sqrt {a + b \sinh ^{2}{\left (e + f x \right )}} \operatorname {sech}^{5}{\left (e + f x \right )}\, dx \]
\[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\int { \sqrt {b \sinh \left (f x + e\right )^{2} + a} \operatorname {sech}\left (f x + e\right )^{5} \,d x } \]
Exception generated. \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \text {sech}^5(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}}{{\mathrm {cosh}\left (e+f\,x\right )}^5} \,d x \]