Integrand size = 25, antiderivative size = 142 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 (a-b)^{5/2} f}+\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}} \]
1/2*(a-4*b)*arctan(sinh(f*x+e)*(a-b)^(1/2)/(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)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.34 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.63 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {sech}^5(e+f x) \left (16 (a-b) \, _3F_2\left (2,2,3;1,\frac {9}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^2+16 (a-b) \operatorname {Hypergeometric2F1}\left (2,3,\frac {9}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (4 a^2+7 a b \sinh ^2(e+f x)+3 b^2 \sinh ^4(e+f x)\right )+7 a \cosh ^2(e+f x) \operatorname {Hypergeometric2F1}\left (1,2,\frac {7}{2},\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \left (15 a^2+20 a b \sinh ^2(e+f x)+8 b^2 \sinh ^4(e+f x)\right )\right ) \tanh (e+f x)}{105 a^4 f \sqrt {a+b \sinh ^2(e+f x)}} \]
(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])
Time = 0.35 (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 |
| \(\displaystyle \frac {\frac {\frac {(a-4 b) \arctan \left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{(a-b)^{3/2}}+\frac {b (a+2 b) \sinh (e+f x)}{a (a-b) \sqrt {a+b \sinh ^2(e+f x)}}}{2 (a-b)}+\frac {\sinh (e+f x)}{2 (a-b) \left (\sinh ^2(e+f x)+1\right ) \sqrt {a+b \sinh ^2(e+f x)}}}{f}\) |
(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
3.4.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] :> 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]
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]
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 = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67
\[\frac {\operatorname {`\,int/indef0`\,}\left (-\frac {\sqrt {a +b \sinh \left (f x +e \right )^{2}}\, \cosh \left (f x +e \right )^{2}}{-b^{2} \cosh \left (f x +e \right )^{10}+\left (-2 a b +2 b^{2}\right ) \cosh \left (f x +e \right )^{8}+\left (-a^{2}+2 a b -b^{2}\right ) \cosh \left (f x +e \right )^{6}}, \sinh \left (f x +e \right )\right )}{f}\]
`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
Leaf count of result is larger than twice the leaf count of optimal. 2364 vs. \(2 (126) = 252\).
Time = 0.49 (sec) , antiderivative size = 4845, normalized size of antiderivative = 34.12 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(((a^2*b - 4*a*b^2)*cosh(f*x + e)^8 + 8*(a^2*b - 4*a*b^2)*cosh(f*x + e)*sinh(f*x + e)^7 + (a^2*b - 4*a*b^2)*sinh(f*x + e)^8 + 4*(a^3 - 4*a^2*b) *cosh(f*x + e)^6 + 4*(a^3 - 4*a^2*b + 7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^2) *sinh(f*x + e)^6 + 8*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^3 + 3*(a^3 - 4*a^2 *b)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f *x + e)^4 + 2*(35*(a^2*b - 4*a*b^2)*cosh(f*x + e)^4 + 4*a^3 - 17*a^2*b + 4 *a*b^2 + 30*(a^3 - 4*a^2*b)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 8*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^5 + 10*(a^3 - 4*a^2*b)*cosh(f*x + e)^3 + (4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + a^2*b - 4*a*b^2 + 4 *(a^3 - 4*a^2*b)*cosh(f*x + e)^2 + 4*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^6 + 15*(a^3 - 4*a^2*b)*cosh(f*x + e)^4 + a^3 - 4*a^2*b + 3*(4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*((a^2*b - 4*a*b^2)*cosh(f *x + e)^7 + 3*(a^3 - 4*a^2*b)*cosh(f*x + e)^5 + (4*a^3 - 17*a^2*b + 4*a*b^ 2)*cosh(f*x + e)^3 + (a^3 - 4*a^2*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)*co...
\[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\operatorname {sech}^{3}{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\operatorname {sech}\left (f x + e\right )^{3}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 984 vs. \(2 (126) = 252\).
Time = 0.73 (sec) , antiderivative size = 984, normalized size of antiderivative = 6.93 \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]
(((a^3*b^2*e^(6*e) - 3*a^2*b^3*e^(6*e) + 3*a*b^4*e^(6*e) - b^5*e^(6*e))*e^ (2*f*x)/(a^6*e^(10*e) - 5*a^5*b*e^(10*e) + 10*a^4*b^2*e^(10*e) - 10*a^3*b^ 3*e^(10*e) + 5*a^2*b^4*e^(10*e) - a*b^5*e^(10*e)) - (a^3*b^2*e^(4*e) - 3*a ^2*b^3*e^(4*e) + 3*a*b^4*e^(4*e) - b^5*e^(4*e))/(a^6*e^(10*e) - 5*a^5*b*e^ (10*e) + 10*a^4*b^2*e^(10*e) - 10*a^3*b^3*e^(10*e) + 5*a^2*b^4*e^(10*e) - a*b^5*e^(10*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*e^(6*e) - 2*a*b*e^(6*e) + b^2*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*(sq rt(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*(sq rt(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))*b^2 - 4*a^2*sqrt(b) + 5*a*b^(3/2) - 2*b^(5/2))/(...
Timed out. \[ \int \frac {\text {sech}^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\mathrm {cosh}\left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]