Integrand size = 23, antiderivative size = 106 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right )}{8 \sqrt {a} (a+b)^{5/2} d}+\frac {\sinh (c+d x)}{4 (a+b) d \left (a+b+a \sinh ^2(c+d x)\right )^2}+\frac {3 \sinh (c+d x)}{8 (a+b)^2 d \left (a+b+a \sinh ^2(c+d x)\right )} \]
1/4*sinh(d*x+c)/(a+b)/d/(a+b+a*sinh(d*x+c)^2)^2+3/8*sinh(d*x+c)/(a+b)^2/d/ (a+b+a*sinh(d*x+c)^2)+3/8*arctan(sinh(d*x+c)*a^(1/2)/(a+b)^(1/2))/(a+b)^(5 /2)/d/a^(1/2)
Time = 1.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x)))^3 \text {sech}^5(c+d x) \left (\frac {3 \arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right ) \text {sech}(c+d x)}{\sqrt {a} (a+b)^{5/2}}+\frac {\left (5 (a+b)+3 a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{(a+b)^2 \left (a+b+a \sinh ^2(c+d x)\right )^2}\right )}{64 d \left (a+b \text {sech}^2(c+d x)\right )^3} \]
((a + 2*b + a*Cosh[2*(c + d*x)])^3*Sech[c + d*x]^5*((3*ArcTan[(Sqrt[a]*Sin h[c + d*x])/Sqrt[a + b]]*Sech[c + d*x])/(Sqrt[a]*(a + b)^(5/2)) + ((5*(a + b) + 3*a*Sinh[c + d*x]^2)*Tanh[c + d*x])/((a + b)^2*(a + b + a*Sinh[c + d *x]^2)^2)))/(64*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.28 (sec) , antiderivative size = 111, 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, 4635, 215, 215, 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.
\(\displaystyle \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^5}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4635 |
\(\displaystyle \frac {\int \frac {1}{\left (a \sinh ^2(c+d x)+a+b\right )^3}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left (a \sinh ^2(c+d x)+a+b\right )^2}d\sinh (c+d x)}{4 (a+b)}+\frac {\sinh (c+d x)}{4 (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{a \sinh ^2(c+d x)+a+b}d\sinh (c+d x)}{2 (a+b)}+\frac {\sinh (c+d x)}{2 (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}\right )}{4 (a+b)}+\frac {\sinh (c+d x)}{4 (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right )}{2 \sqrt {a} (a+b)^{3/2}}+\frac {\sinh (c+d x)}{2 (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}\right )}{4 (a+b)}+\frac {\sinh (c+d x)}{4 (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
(Sinh[c + d*x]/(4*(a + b)*(a + b + a*Sinh[c + d*x]^2)^2) + (3*(ArcTan[(Sqr t[a]*Sinh[c + d*x])/Sqrt[a + b]]/(2*Sqrt[a]*(a + b)^(3/2)) + Sinh[c + d*x] /(2*(a + b)*(a + b + a*Sinh[c + d*x]^2))))/(4*(a + b)))/d
3.1.99.3.1 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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(234\) vs. \(2(92)=184\).
Time = 1.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.22
method | result | size |
risch | \(\frac {{\mathrm e}^{d x +c} \left (3 a \,{\mathrm e}^{6 d x +6 c}+11 a \,{\mathrm e}^{4 d x +4 c}+20 b \,{\mathrm e}^{4 d x +4 c}-11 \,{\mathrm e}^{2 d x +2 c} a -20 b \,{\mathrm e}^{2 d x +2 c}-3 a \right )}{4 d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \left (a +b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (a +b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}\) | \(235\) |
derivativedivides | \(\frac {\frac {-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 \left (a +b \right )}+\frac {3 \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 \left (a +b \right )^{2}}-\frac {3 \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 \left (a +b \right )^{2}}+\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\frac {3 \arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{8 \sqrt {a +b}\, \sqrt {a}}+\frac {3 \arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{8 \sqrt {a +b}\, \sqrt {a}}}{a^{2}+2 a b +b^{2}}}{d}\) | \(240\) |
default | \(\frac {\frac {-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 \left (a +b \right )}+\frac {3 \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 \left (a +b \right )^{2}}-\frac {3 \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 \left (a +b \right )^{2}}+\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\frac {3 \arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{8 \sqrt {a +b}\, \sqrt {a}}+\frac {3 \arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{8 \sqrt {a +b}\, \sqrt {a}}}{a^{2}+2 a b +b^{2}}}{d}\) | \(240\) |
1/4*exp(d*x+c)*(3*a*exp(6*d*x+6*c)+11*a*exp(4*d*x+4*c)+20*b*exp(4*d*x+4*c) -11*exp(2*d*x+2*c)*a-20*b*exp(2*d*x+2*c)-3*a)/d/(a+b)^2/(a*exp(4*d*x+4*c)+ 2*exp(2*d*x+2*c)*a+4*b*exp(2*d*x+2*c)+a)^2-3/16/(-a^2-a*b)^(1/2)/(a+b)^2/d *ln(exp(2*d*x+2*c)-2*(a+b)/(-a^2-a*b)^(1/2)*exp(d*x+c)-1)+3/16/(-a^2-a*b)^ (1/2)/(a+b)^2/d*ln(exp(2*d*x+2*c)+2*(a+b)/(-a^2-a*b)^(1/2)*exp(d*x+c)-1)
Leaf count of result is larger than twice the leaf count of optimal. 2638 vs. \(2 (92) = 184\).
Time = 0.32 (sec) , antiderivative size = 5006, normalized size of antiderivative = 47.23 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{5}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
1/4*((11*a*e^(5*c) + 20*b*e^(5*c))*e^(5*d*x) - (11*a*e^(3*c) + 20*b*e^(3*c ))*e^(3*d*x) + 3*a*e^(7*d*x + 7*c) - 3*a*e^(d*x + c))/(a^4*d + 2*a^3*b*d + a^2*b^2*d + (a^4*d*e^(8*c) + 2*a^3*b*d*e^(8*c) + a^2*b^2*d*e^(8*c))*e^(8* d*x) + 4*(a^4*d*e^(6*c) + 4*a^3*b*d*e^(6*c) + 5*a^2*b^2*d*e^(6*c) + 2*a*b^ 3*d*e^(6*c))*e^(6*d*x) + 2*(3*a^4*d*e^(4*c) + 14*a^3*b*d*e^(4*c) + 27*a^2* b^2*d*e^(4*c) + 24*a*b^3*d*e^(4*c) + 8*b^4*d*e^(4*c))*e^(4*d*x) + 4*(a^4*d *e^(2*c) + 4*a^3*b*d*e^(2*c) + 5*a^2*b^2*d*e^(2*c) + 2*a*b^3*d*e^(2*c))*e^ (2*d*x)) + 32*integrate(3/128*(e^(3*d*x + 3*c) + e^(d*x + c))/(a^3 + 2*a^2 *b + a*b^2 + (a^3*e^(4*c) + 2*a^2*b*e^(4*c) + a*b^2*e^(4*c))*e^(4*d*x) + 2 *(a^3*e^(2*c) + 4*a^2*b*e^(2*c) + 5*a*b^2*e^(2*c) + 2*b^3*e^(2*c))*e^(2*d* x)), x)
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^5\,{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \]