Integrand size = 23, antiderivative size = 157 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {(a-5 b) \arctan (\sinh (c+d x))}{2 (a-b)^3 d}+\frac {(5 a-b) b^{3/2} \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^3 d}+\frac {b (a+b) \sinh (c+d x)}{2 a (a-b)^2 d \left (a+b \sinh ^2(c+d x)\right )}+\frac {\text {sech}(c+d x) \tanh (c+d x)}{2 (a-b) d \left (a+b \sinh ^2(c+d x)\right )} \]
1/2*(a-5*b)*arctan(sinh(d*x+c))/(a-b)^3/d+1/2*(5*a-b)*b^(3/2)*arctan(sinh( d*x+c)*b^(1/2)/a^(1/2))/a^(3/2)/(a-b)^3/d+1/2*b*(a+b)*sinh(d*x+c)/a/(a-b)^ 2/d/(a+b*sinh(d*x+c)^2)+1/2*sech(d*x+c)*tanh(d*x+c)/(a-b)/d/(a+b*sinh(d*x+ c)^2)
Time = 0.69 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.46 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {2 \sqrt {a} (a-b) b^2 \sinh (c+d x)+(2 a-b) \left (b^{3/2} (-5 a+b) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+2 a^{3/2} (a-5 b) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+a^{3/2} (a-b) \text {sech}(c+d x) \tanh (c+d x)\right )+b \cosh (2 (c+d x)) \left (b^{3/2} (-5 a+b) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+2 a^{3/2} (a-5 b) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+a^{3/2} (a-b) \text {sech}(c+d x) \tanh (c+d x)\right )}{2 a^{3/2} (a-b)^3 d (2 a-b+b \cosh (2 (c+d x)))} \]
(2*Sqrt[a]*(a - b)*b^2*Sinh[c + d*x] + (2*a - b)*(b^(3/2)*(-5*a + b)*ArcTa n[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 2*a^(3/2)*(a - 5*b)*ArcTan[Tanh[(c + d*x)/2]] + a^(3/2)*(a - b)*Sech[c + d*x]*Tanh[c + d*x]) + b*Cosh[2*(c + d* x)]*(b^(3/2)*(-5*a + b)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 2*a^(3/2 )*(a - 5*b)*ArcTan[Tanh[(c + d*x)/2]] + a^(3/2)*(a - b)*Sech[c + d*x]*Tanh [c + d*x]))/(2*a^(3/2)*(a - b)^3*d*(2*a - b + b*Cosh[2*(c + d*x)]))
Time = 0.37 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3669, 316, 25, 402, 27, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (i c+i d x)^3 \left (a-b \sin (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {1}{\left (\sinh ^2(c+d x)+1\right )^2 \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}-\frac {\int -\frac {3 b \sinh ^2(c+d x)+a-2 b}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{2 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b \sinh ^2(c+d x)+a-2 b}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (a^2-4 b a+b^2+b (a+b) \sinh ^2(c+d x)\right )}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{2 a (a-b)}+\frac {b (a+b) \sinh (c+d x)}{a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^2-4 b a+b^2+b (a+b) \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{a (a-b)}+\frac {b (a+b) \sinh (c+d x)}{a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a-b) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}+\frac {a (a-5 b) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{a-b}}{a (a-b)}+\frac {b (a+b) \sinh (c+d x)}{a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a-b) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}+\frac {a (a-5 b) \arctan (\sinh (c+d x))}{a-b}}{a (a-b)}+\frac {b (a+b) \sinh (c+d x)}{a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^{3/2} (5 a-b) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}+\frac {a (a-5 b) \arctan (\sinh (c+d x))}{a-b}}{a (a-b)}+\frac {b (a+b) \sinh (c+d x)}{a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
(Sinh[c + d*x]/(2*(a - b)*(1 + Sinh[c + d*x]^2)*(a + b*Sinh[c + d*x]^2)) + (((a*(a - 5*b)*ArcTan[Sinh[c + d*x]])/(a - b) + ((5*a - b)*b^(3/2)*ArcTan [(Sqrt[b]*Sinh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(a*(a - b)) + (b*(a + b)*Sinh[c + d*x])/(a*(a - b)*(a + b*Sinh[c + d*x]^2)))/(2*(a - b)))/d
3.4.37.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(367\) vs. \(2(141)=282\).
Time = 0.18 (sec) , antiderivative size = 368, normalized size of antiderivative = 2.34
\[\frac {\frac {\frac {2 \left (\left (-\frac {a}{2}+\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {a}{2}-\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{2}}+\left (a -5 b \right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{3}}+\frac {2 b^{2} \left (\frac {-\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}+\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}+\frac {\left (5 a -b \right ) \left (-\frac {\left (a +\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (-a +\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2}\right )}{\left (a -b \right )^{3}}}{d}\]
1/d*(2/(a-b)^3*(((-1/2*a+1/2*b)*tanh(1/2*d*x+1/2*c)^3+(1/2*a-1/2*b)*tanh(1 /2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2+1)^2+1/2*(a-5*b)*arctan(tanh(1/2*d*x +1/2*c)))+2*b^2/(a-b)^3*((-1/2*(a-b)/a*tanh(1/2*d*x+1/2*c)^3+1/2*(a-b)/a*t anh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*b *tanh(1/2*d*x+1/2*c)^2+a)+1/2*(5*a-b)*(-1/2*(a+(-b*(a-b))^(1/2)-b)/a/(-b*( a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/ 2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/2*(-a+(-b*(a-b))^(1/2)+b)/a/( -b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x +1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)))))
Leaf count of result is larger than twice the leaf count of optimal. 3474 vs. \(2 (141) = 282\).
Time = 0.38 (sec) , antiderivative size = 6548, normalized size of antiderivative = 41.71 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int \frac {\operatorname {sech}^{3}{\left (c + d x \right )}}{\left (a + b \sinh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
(a*e^c - 5*b*e^c)*arctan(e^(d*x + c))*e^(-c)/(a^3*d - 3*a^2*b*d + 3*a*b^2* d - b^3*d) + ((a*b*e^(7*c) + b^2*e^(7*c))*e^(7*d*x) + (4*a^2*e^(5*c) - 3*a *b*e^(5*c) + b^2*e^(5*c))*e^(5*d*x) - (4*a^2*e^(3*c) - 3*a*b*e^(3*c) + b^2 *e^(3*c))*e^(3*d*x) - (a*b*e^c + b^2*e^c)*e^(d*x))/(a^3*b*d - 2*a^2*b^2*d + a*b^3*d + (a^3*b*d*e^(8*c) - 2*a^2*b^2*d*e^(8*c) + a*b^3*d*e^(8*c))*e^(8 *d*x) + 4*(a^4*d*e^(6*c) - 2*a^3*b*d*e^(6*c) + a^2*b^2*d*e^(6*c))*e^(6*d*x ) + 2*(4*a^4*d*e^(4*c) - 9*a^3*b*d*e^(4*c) + 6*a^2*b^2*d*e^(4*c) - a*b^3*d *e^(4*c))*e^(4*d*x) + 4*(a^4*d*e^(2*c) - 2*a^3*b*d*e^(2*c) + a^2*b^2*d*e^( 2*c))*e^(2*d*x)) + 8*integrate(1/8*((5*a*b^2*e^(3*c) - b^3*e^(3*c))*e^(3*d *x) + (5*a*b^2*e^c - b^3*e^c)*e^(d*x))/(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a* b^4 + (a^4*b*e^(4*c) - 3*a^3*b^2*e^(4*c) + 3*a^2*b^3*e^(4*c) - a*b^4*e^(4* c))*e^(4*d*x) + 2*(2*a^5*e^(2*c) - 7*a^4*b*e^(2*c) + 9*a^3*b^2*e^(2*c) - 5 *a^2*b^3*e^(2*c) + a*b^4*e^(2*c))*e^(2*d*x)), x)
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]