Integrand size = 21, antiderivative size = 159 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {\arctan (\sinh (c+d x))}{(a-b)^3 d}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a-b)^3 d}-\frac {b \sinh (c+d x)}{4 a (a-b) d \left (a+b \sinh ^2(c+d x)\right )^2}-\frac {(7 a-3 b) b \sinh (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \sinh ^2(c+d x)\right )} \]
arctan(sinh(d*x+c))/(a-b)^3/d-1/4*b*sinh(d*x+c)/a/(a-b)/d/(a+b*sinh(d*x+c) ^2)^2-1/8*(7*a-3*b)*b*sinh(d*x+c)/a^2/(a-b)^2/d/(a+b*sinh(d*x+c)^2)-1/8*(1 5*a^2-10*a*b+3*b^2)*arctan(sinh(d*x+c)*b^(1/2)/a^(1/2))*b^(1/2)/a^(5/2)/(a -b)^3/d
Leaf count is larger than twice the leaf count of optimal. \(321\) vs. \(2(159)=318\).
Time = 0.59 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.02 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {(-2 a+b)^2 \left (\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+16 a^{5/2} \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )+\left (b^{5/2} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+16 a^{5/2} b^2 \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right ) \cosh ^2(2 (c+d x))-2 \sqrt {a} b \left (18 a^3-35 a^2 b+20 a b^2-3 b^3\right ) \sinh (c+d x)-2 b \cosh (2 (c+d x)) \left (-\left ((2 a-b) \left (\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+16 a^{5/2} \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )+\sqrt {a} b \left (7 a^2-10 a b+3 b^2\right ) \sinh (c+d x)\right )}{8 a^{5/2} (a-b)^3 d (2 a-b+b \cosh (2 (c+d x)))^2} \]
((-2*a + b)^2*(Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 16*a^(5/2)*ArcTan[Tanh[(c + d*x)/2]]) + (b^(5/2)*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 16*a^(5/2)*b^2 *ArcTan[Tanh[(c + d*x)/2]])*Cosh[2*(c + d*x)]^2 - 2*Sqrt[a]*b*(18*a^3 - 35 *a^2*b + 20*a*b^2 - 3*b^3)*Sinh[c + d*x] - 2*b*Cosh[2*(c + d*x)]*(-((2*a - b)*(Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt [b]] + 16*a^(5/2)*ArcTan[Tanh[(c + d*x)/2]])) + Sqrt[a]*b*(7*a^2 - 10*a*b + 3*b^2)*Sinh[c + d*x]))/(8*a^(5/2)*(a - b)^3*d*(2*a - b + b*Cosh[2*(c + d *x)])^2)
Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3669, 316, 402, 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.
| \(\displaystyle \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (i c+i d x) \left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {1}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \sinh ^2(c+d x)+4 a-3 b}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a (a-b)}-\frac {b \sinh (c+d x)}{4 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2-7 b a+3 b^2-(7 a-3 b) b \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{2 a (a-b)}-\frac {b (7 a-3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \sinh (c+d x)}{4 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \sinh (c+d x)}{4 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\sinh (c+d x))}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \sinh (c+d x)}{4 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\sinh (c+d x))}{a-b}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b (7 a-3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \sinh (c+d x)}{4 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
(-1/4*(b*Sinh[c + d*x])/(a*(a - b)*(a + b*Sinh[c + d*x]^2)^2) + (((8*a^2*A rcTan[Sinh[c + d*x]])/(a - b) - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[ (Sqrt[b]*Sinh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - ((7*a - 3*b)*b*Sinh[c + d*x])/(2*a*(a - b)*(a + b*Sinh[c + d*x]^2)))/(4*a*(a - b)))/d
3.4.45.3.1 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[((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. \(413\) vs. \(2(145)=290\).
Time = 0.13 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.60
\[\frac {\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{3}}-\frac {2 b \left (\frac {-\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a}+\frac {\left (27 a^{3}-70 a^{2} b +55 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2}}-\frac {\left (27 a^{3}-70 a^{2} b +55 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \left (-\frac {\left (a +\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (-a +\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{8 a}\right )}{\left (a -b \right )^{3}}}{d}\]
1/d*(2/(a-b)^3*arctan(tanh(1/2*d*x+1/2*c))-2*b/(a-b)^3*((-1/8*(9*a^2-14*a* b+5*b^2)/a*tanh(1/2*d*x+1/2*c)^7+1/8*(27*a^3-70*a^2*b+55*a*b^2-12*b^3)/a^2 *tanh(1/2*d*x+1/2*c)^5-1/8*(27*a^3-70*a^2*b+55*a*b^2-12*b^3)/a^2*tanh(1/2* d*x+1/2*c)^3+1/8*(9*a^2-14*a*b+5*b^2)/a*tanh(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)^2+1/8/a *(15*a^2-10*a*b+3*b^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)*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. 4384 vs. \(2 (145) = 290\).
Time = 0.39 (sec) , antiderivative size = 8083, normalized size of antiderivative = 50.84 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
-1/4*((7*a*b^2*e^(7*c) - 3*b^3*e^(7*c))*e^(7*d*x) + (36*a^2*b*e^(5*c) - 41 *a*b^2*e^(5*c) + 9*b^3*e^(5*c))*e^(5*d*x) - (36*a^2*b*e^(3*c) - 41*a*b^2*e ^(3*c) + 9*b^3*e^(3*c))*e^(3*d*x) - (7*a*b^2*e^c - 3*b^3*e^c)*e^(d*x))/(a^ 4*b^2*d - 2*a^3*b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(8*c) - 2*a^3*b^3*d*e^(8* c) + a^2*b^4*d*e^(8*c))*e^(8*d*x) + 4*(2*a^5*b*d*e^(6*c) - 5*a^4*b^2*d*e^( 6*c) + 4*a^3*b^3*d*e^(6*c) - a^2*b^4*d*e^(6*c))*e^(6*d*x) + 2*(8*a^6*d*e^( 4*c) - 24*a^5*b*d*e^(4*c) + 27*a^4*b^2*d*e^(4*c) - 14*a^3*b^3*d*e^(4*c) + 3*a^2*b^4*d*e^(4*c))*e^(4*d*x) + 4*(2*a^5*b*d*e^(2*c) - 5*a^4*b^2*d*e^(2*c ) + 4*a^3*b^3*d*e^(2*c) - a^2*b^4*d*e^(2*c))*e^(2*d*x)) + 2*arctan(e^(d*x + c))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) - 2*integrate(1/8*((15*a^2*b *e^(3*c) - 10*a*b^2*e^(3*c) + 3*b^3*e^(3*c))*e^(3*d*x) + (15*a^2*b*e^c - 1 0*a*b^2*e^c + 3*b^3*e^c)*e^(d*x))/(a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4 + (a^5*b*e^(4*c) - 3*a^4*b^2*e^(4*c) + 3*a^3*b^3*e^(4*c) - a^2*b^4*e^(4*c ))*e^(4*d*x) + 2*(2*a^6*e^(2*c) - 7*a^5*b*e^(2*c) + 9*a^4*b^2*e^(2*c) - 5* a^3*b^3*e^(2*c) + a^2*b^4*e^(2*c))*e^(2*d*x)), x)
\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{\mathrm {cosh}\left (c+d\,x\right )\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]