Integrand size = 23, antiderivative size = 125 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{8 b^{3/2} (a+b)^{5/2} d}-\frac {a \tanh (c+d x)}{4 b (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^2}+\frac {(a+4 b) \tanh (c+d x)}{8 b (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )} \]
1/8*(a+4*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/b^(3/2)/(a+b)^(5/2)/d -1/4*a*tanh(d*x+c)/b/(a+b)/d/(a+b-b*tanh(d*x+c)^2)^2+1/8*(a+4*b)*tanh(d*x+ c)/b/(a+b)^2/d/(a+b-b*tanh(d*x+c)^2)
Time = 5.05 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.00 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^6(c+d x) \left (\frac {(a+4 b) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 (\cosh (2 c)-\sinh (2 c))}{b \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}-\frac {4 (a+b) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a}+\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}(2 c) ((a+4 b) \sinh (2 c)-(a-2 b) \sinh (2 d x))}{b}\right )}{64 (a+b)^2 d \left (a+b \text {sech}^2(c+d x)\right )^3} \]
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^6*(((a + 4*b)*ArcTanh[(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])]*(a + 2*b + a*Cosh[2*(c + d*x )])^2*(Cosh[2*c] - Sinh[2*c]))/(b*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4 ]) - (4*(a + b)*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/a + ((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[2*c]*((a + 4*b)*Sinh[2*c] - (a - 2*b)*Sin h[2*d*x]))/b))/(64*(a + b)^2*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 298, 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.
| \(\displaystyle \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (i c+i d x)^4}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1-\tanh ^2(c+d x)}{\left (-b \tanh ^2(c+d x)+a+b\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {(a+4 b) \int \frac {1}{\left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{4 b (a+b)}-\frac {a \tanh (c+d x)}{4 b (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {(a+4 b) \left (\frac {\int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{2 (a+b)}+\frac {\tanh (c+d x)}{2 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}\right )}{4 b (a+b)}-\frac {a \tanh (c+d x)}{4 b (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {(a+4 b) \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2}}+\frac {\tanh (c+d x)}{2 (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}\right )}{4 b (a+b)}-\frac {a \tanh (c+d x)}{4 b (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\) |
(-1/4*(a*Tanh[c + d*x])/(b*(a + b)*(a + b - b*Tanh[c + d*x]^2)^2) + ((a + 4*b)*(ArcTanh[(Sqrt[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*b*(a + b)) )/d
3.1.98.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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. \(309\) vs. \(2(111)=222\).
Time = 1.30 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.48
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {\left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right ) b}+\frac {\left (3 a^{2}-5 a b +4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2} b}+\frac {\left (3 a^{2}-5 a b +4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2} b}+\frac {\left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right ) b}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {\left (a +4 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{4 b \left (a^{2}+2 a b +b^{2}\right )}}{d}\) | \(310\) |
| default | \(\frac {-\frac {2 \left (\frac {\left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right ) b}+\frac {\left (3 a^{2}-5 a b +4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2} b}+\frac {\left (3 a^{2}-5 a b +4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2} b}+\frac {\left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right ) b}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {\left (a +4 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{4 b \left (a^{2}+2 a b +b^{2}\right )}}{d}\) | \(310\) |
| risch | \(\frac {a^{3} {\mathrm e}^{6 d x +6 c}+4 a^{2} b \,{\mathrm e}^{6 d x +6 c}+3 a^{3} {\mathrm e}^{4 d x +4 c}+2 a^{2} b \,{\mathrm e}^{4 d x +4 c}-8 a \,b^{2} {\mathrm e}^{4 d x +4 c}-16 \,{\mathrm e}^{4 d x +4 c} b^{3}+3 a^{3} {\mathrm e}^{2 d x +2 c}-4 a^{2} b \,{\mathrm e}^{2 d x +2 c}-16 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}+a^{3}-2 a^{2} b}{4 a d b \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) a}{16 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d b}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) a}{16 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d b}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, \left (a +b \right )^{2} d}\) | \(509\) |
1/d*(-2*(1/8*(a-4*b)/(a+b)/b*tanh(1/2*d*x+1/2*c)^7+1/8*(3*a^2-5*a*b+4*b^2) /(a+b)^2/b*tanh(1/2*d*x+1/2*c)^5+1/8*(3*a^2-5*a*b+4*b^2)/(a+b)^2/b*tanh(1/ 2*d*x+1/2*c)^3+1/8*(a-4*b)/(a+b)/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)^2-1/4*(a+4*b)/b/(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. 2603 vs. \(2 (117) = 234\).
Time = 0.31 (sec) , antiderivative size = 5447, normalized size of antiderivative = 43.58 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{4}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (117) = 234\).
Time = 0.36 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.95 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {{\left (a + 4 \, b\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {a^{3} - 2 \, a^{2} b + {\left (3 \, a^{3} - 4 \, a^{2} b - 16 \, a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (3 \, a^{3} + 2 \, a^{2} b - 8 \, a b^{2} - 16 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{3} + 4 \, a^{2} b\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3} + 4 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 5 \, a^{3} b^{3} + 2 \, a^{2} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} b + 14 \, a^{4} b^{2} + 27 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 8 \, a b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} b + 4 \, a^{4} b^{2} + 5 \, a^{3} b^{3} + 2 \, a^{2} b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} \]
-1/16*(a + 4*b)*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*b + 2*a*b^2 + b^3)* sqrt((a + b)*b)*d) - 1/4*(a^3 - 2*a^2*b + (3*a^3 - 4*a^2*b - 16*a*b^2)*e^( -2*d*x - 2*c) + (3*a^3 + 2*a^2*b - 8*a*b^2 - 16*b^3)*e^(-4*d*x - 4*c) + (a ^3 + 4*a^2*b)*e^(-6*d*x - 6*c))/((a^5*b + 2*a^4*b^2 + a^3*b^3 + 4*(a^5*b + 4*a^4*b^2 + 5*a^3*b^3 + 2*a^2*b^4)*e^(-2*d*x - 2*c) + 2*(3*a^5*b + 14*a^4 *b^2 + 27*a^3*b^3 + 24*a^2*b^4 + 8*a*b^5)*e^(-4*d*x - 4*c) + 4*(a^5*b + 4* a^4*b^2 + 5*a^3*b^3 + 2*a^2*b^4)*e^(-6*d*x - 6*c) + (a^5*b + 2*a^4*b^2 + a ^3*b^3)*e^(-8*d*x - 8*c))*d)
\[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{4}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \]