Integrand size = 23, antiderivative size = 104 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {a+b} d}+\frac {\sinh (c+d x)}{4 a d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {3 \sinh (c+d x)}{8 a^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \]
1/4*sinh(d*x+c)/a/d/(a+(a+b)*sinh(d*x+c)^2)^2+3/8*sinh(d*x+c)/a^2/d/(a+(a+ b)*sinh(d*x+c)^2)+3/8*arctan(sinh(d*x+c)*(a+b)^(1/2)/a^(1/2))/a^(5/2)/d/(a +b)^(1/2)
Time = 0.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+\frac {\sqrt {a} \sinh (c+d x) \left (5 a+3 (a+b) \sinh ^2(c+d x)\right )}{\left (a+(a+b) \sinh ^2(c+d x)\right )^2}}{8 a^{5/2} d} \]
((3*ArcTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]])/Sqrt[a + b] + (Sqrt[a]*Si nh[c + d*x]*(5*a + 3*(a + b)*Sinh[c + d*x]^2))/(a + (a + b)*Sinh[c + d*x]^ 2)^2)/(8*a^(5/2)*d)
Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4159, 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.
\(\displaystyle \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^5}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4159 |
\(\displaystyle \frac {\int \frac {1}{\left ((a+b) \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left ((a+b) \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a}+\frac {\sinh (c+d x)}{4 a \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{(a+b) \sinh ^2(c+d x)+a}d\sinh (c+d x)}{2 a}+\frac {\sinh (c+d x)}{2 a \left ((a+b) \sinh ^2(c+d x)+a\right )}\right )}{4 a}+\frac {\sinh (c+d x)}{4 a \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {a+b}}+\frac {\sinh (c+d x)}{2 a \left ((a+b) \sinh ^2(c+d x)+a\right )}\right )}{4 a}+\frac {\sinh (c+d x)}{4 a \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
(Sinh[c + d*x]/(4*a*(a + (a + b)*Sinh[c + d*x]^2)^2) + (3*(ArcTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[a + b]) + Sinh[c + d*x]/(2*a* (a + (a + b)*Sinh[c + d*x]^2))))/(4*a))/d
3.2.31.3.1 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_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + 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] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(275\) vs. \(2(90)=180\).
Time = 0.22 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.65
\[\frac {\frac {-\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a}+\frac {3 \left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a^{2}}-\frac {3 \left (a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a^{2}}+\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a \right )^{2}}+\frac {\frac {3 \left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{8 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {3 \left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{8 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}}{a}}{d}\]
1/d*(2*(-5/8/a*tanh(1/2*d*x+1/2*c)^7+3/8*(a-4*b)/a^2*tanh(1/2*d*x+1/2*c)^5 -3/8*(a-4*b)/a^2*tanh(1/2*d*x+1/2*c)^3+5/8/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*tanh(1/2*d*x+1/2*c)^2*b+a)^2+ 3/4/a*(1/2*(((a+b)*b)^(1/2)+b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)+a+2*b )*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2 ))-1/2*(((a+b)*b)^(1/2)-b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)-a-2*b)*a) ^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2))) )
Leaf count of result is larger than twice the leaf count of optimal. 2712 vs. \(2 (90) = 180\).
Time = 0.32 (sec) , antiderivative size = 5077, normalized size of antiderivative = 48.82 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{5}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
1/4*(3*(a*e^(7*c) + b*e^(7*c))*e^(7*d*x) + (11*a*e^(5*c) - 9*b*e^(5*c))*e^ (5*d*x) - (11*a*e^(3*c) - 9*b*e^(3*c))*e^(3*d*x) - 3*(a*e^c + b*e^c)*e^(d* x))/(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) - a^2*b^2*d*e^(6*c))*e^(6* d*x) + 2*(3*a^4*d*e^(4*c) - 2*a^3*b*d*e^(4*c) + 3*a^2*b^2*d*e^(4*c))*e^(4* d*x) + 4*(a^4*d*e^(2*c) - a^2*b^2*d*e^(2*c))*e^(2*d*x)) + 32*integrate(3/1 28*(e^(3*d*x + 3*c) + e^(d*x + c))/(a^3 + a^2*b + (a^3*e^(4*c) + a^2*b*e^( 4*c))*e^(4*d*x) + 2*(a^3*e^(2*c) - a^2*b*e^(2*c))*e^(2*d*x)), x)
\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^5\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]