Integrand size = 23, antiderivative size = 108 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{8 \sqrt {b} (a+b)^{5/2} d}+\frac {\tanh (c+d x)}{4 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^2}+\frac {3 \tanh (c+d x)}{8 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )} \]
3/8*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/(a+b)^(5/2)/d/b^(1/2)+1/4*tan h(d*x+c)/(a+b)/d/(a+b-b*tanh(d*x+c)^2)^2+3/8*tanh(d*x+c)/(a+b)^2/d/(a+b-b* tanh(d*x+c)^2)
Leaf count is larger than twice the leaf count of optimal. \(258\) vs. \(2(108)=216\).
Time = 3.96 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.39 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^6(c+d x) \left (\frac {3 \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 (\cosh (2 c)-\sinh (2 c))}{\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {4 b (a+b) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2}-\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}(2 c) \left (\left (5 a^2+16 a b+8 b^2\right ) \sinh (2 c)-a (5 a+2 b) \sinh (2 d x)\right )}{a^2}\right )}{64 (a+b)^2 d \left (a+b \text {sech}^2(c+d x)\right )^3} \]
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^6*((3*ArcTanh[(Sech[d*x]*(C osh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*(C osh[2*c] - Sinh[2*c]))/(Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + (4*b* (a + b)*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/a^2 - ((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[2*c]*((5*a^2 + 16*a*b + 8*b^2)*Sinh[2*c] - a*(5 *a + 2*b)*Sinh[2*d*x]))/a^2))/(64*(a + b)^2*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 215, 215, 221}
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.
\(\displaystyle \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^2}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {1}{\left (-b \tanh ^2(c+d x)+a+b\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{4 (a+b)}+\frac {\tanh (c+d x)}{4 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{2 (a+b)}+\frac {\tanh (c+d x)}{2 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}\right )}{4 (a+b)}+\frac {\tanh (c+d x)}{4 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2}}+\frac {\tanh (c+d x)}{2 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}\right )}{4 (a+b)}+\frac {\tanh (c+d x)}{4 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
(Tanh[c + d*x]/(4*(a + b)*(a + b - b*Tanh[c + d*x]^2)^2) + (3*(ArcTanh[(Sq rt[b]*Tanh[c + d*x])/Sqrt[a + b]]/(2*Sqrt[b]*(a + b)^(3/2)) + Tanh[c + d*x ]/(2*(a + b)*(a + b - b*Tanh[c + d*x]^2))))/(4*(a + b)))/d
3.1.96.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(259\) vs. \(2(94)=188\).
Time = 1.22 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.41
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {3 \left (5 a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}-\frac {3 \left (5 a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {3 \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{4 \left (a^{2}+2 a b +b^{2}\right )}}{d}\) | \(260\) |
default | \(\frac {-\frac {2 \left (-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {3 \left (5 a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}-\frac {3 \left (5 a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {3 \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{4 \left (a^{2}+2 a b +b^{2}\right )}}{d}\) | \(260\) |
risch | \(-\frac {5 a^{3} {\mathrm e}^{6 d x +6 c}+16 a^{2} b \,{\mathrm e}^{6 d x +6 c}+8 a \,b^{2} {\mathrm e}^{6 d x +6 c}+15 a^{3} {\mathrm e}^{4 d x +4 c}+46 a^{2} b \,{\mathrm e}^{4 d x +4 c}+56 a \,b^{2} {\mathrm e}^{4 d x +4 c}+16 \,{\mathrm e}^{4 d x +4 c} b^{3}+15 a^{3} {\mathrm e}^{2 d x +2 c}+32 a^{2} b \,{\mathrm e}^{2 d x +2 c}+8 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}+5 a^{3}+2 a^{2} b}{4 a^{2} \left (a +b \right )^{2} d \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{16 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{16 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d}\) | \(364\) |
1/d*(-2*(-5/8/(a+b)*tanh(1/2*d*x+1/2*c)^7-3/8*(5*a+b)/(a+b)^2*tanh(1/2*d*x +1/2*c)^5-3/8*(5*a+b)/(a+b)^2*tanh(1/2*d*x+1/2*c)^3-5/8/(a+b)*tanh(1/2*d*x +1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1 /2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)^2-3/4/(a^2+2*a*b+b^2)*(-1/4/b^(1/ 2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)* b^(1/2)+(a+b)^(1/2))+1/4/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1 /2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 2434 vs. \(2 (100) = 200\).
Time = 0.32 (sec) , antiderivative size = 5109, normalized size of antiderivative = 47.31 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (100) = 200\).
Time = 0.34 (sec) , antiderivative size = 353, normalized size of antiderivative = 3.27 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {5 \, a^{3} + 2 \, a^{2} b + {\left (15 \, a^{3} + 32 \, a^{2} b + 8 \, a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (15 \, a^{3} + 46 \, a^{2} b + 56 \, a b^{2} + 16 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (5 \, a^{3} + 16 \, a^{2} b + 8 \, a b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2} + 4 \, {\left (a^{6} + 4 \, a^{5} b + 5 \, a^{4} b^{2} + 2 \, a^{3} b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{6} + 14 \, a^{5} b + 27 \, a^{4} b^{2} + 24 \, a^{3} b^{3} + 8 \, a^{2} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{6} + 4 \, a^{5} b + 5 \, a^{4} b^{2} + 2 \, a^{3} b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} - \frac {3 \, \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b} d} \]
1/4*(5*a^3 + 2*a^2*b + (15*a^3 + 32*a^2*b + 8*a*b^2)*e^(-2*d*x - 2*c) + (1 5*a^3 + 46*a^2*b + 56*a*b^2 + 16*b^3)*e^(-4*d*x - 4*c) + (5*a^3 + 16*a^2*b + 8*a*b^2)*e^(-6*d*x - 6*c))/((a^6 + 2*a^5*b + a^4*b^2 + 4*(a^6 + 4*a^5*b + 5*a^4*b^2 + 2*a^3*b^3)*e^(-2*d*x - 2*c) + 2*(3*a^6 + 14*a^5*b + 27*a^4* b^2 + 24*a^3*b^3 + 8*a^2*b^4)*e^(-4*d*x - 4*c) + 4*(a^6 + 4*a^5*b + 5*a^4* b^2 + 2*a^3*b^3)*e^(-6*d*x - 6*c) + (a^6 + 2*a^5*b + a^4*b^2)*e^(-8*d*x - 8*c))*d) - 3/16*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a* e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + 2*a*b + b^2)*sqrt ((a + b)*b)*d)
\[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{2}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \]