Integrand size = 23, antiderivative size = 115 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {(a-3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2} d}+\frac {(a+b) \tanh (c+d x)}{4 a b d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \tanh (c+d x)}{8 a^2 b d \left (a+b \tanh ^2(c+d x)\right )} \]
-1/8*(a-3*b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2)/b^(3/2)/d+1/4*(a+ b)*tanh(d*x+c)/a/b/d/(a+b*tanh(d*x+c)^2)^2-1/8*(a-3*b)*tanh(d*x+c)/a^2/b/d /(a+b*tanh(d*x+c)^2)
Time = 1.96 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {(-a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{b^{3/2}}+\frac {\sqrt {a} \left (a^2+6 a b-3 b^2+\left (a^2+4 a b+3 b^2\right ) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{b (a-b+(a+b) \cosh (2 (c+d x)))^2}}{8 a^{5/2} d} \]
(((-a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/b^(3/2) + (Sqrt[a]*( a^2 + 6*a*b - 3*b^2 + (a^2 + 4*a*b + 3*b^2)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(b*(a - b + (a + b)*Cosh[2*(c + d*x)])^2))/(8*a^(5/2)*d)
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4158, 298, 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.
\(\displaystyle \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^4}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {1-\tanh ^2(c+d x)}{\left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \int \frac {1}{\left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 a b}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \left (\frac {\int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{2 a}+\frac {\tanh (c+d x)}{2 a \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a b}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \left (\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\tanh (c+d x)}{2 a \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a b}}{d}\) |
(((a + b)*Tanh[c + d*x])/(4*a*b*(a + b*Tanh[c + d*x]^2)^2) - ((a - 3*b)*(A rcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]) + Tanh[c + d*x] /(2*a*(a + b*Tanh[c + d*x]^2))))/(4*a*b))/d
3.2.30.3.1 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[((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_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Leaf count of result is larger than twice the leaf count of optimal. \(333\) vs. \(2(101)=202\).
Time = 185.48 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.90
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2} b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2} b}-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a b}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a \right )^{2}}-\frac {\left (a -3 b \right ) \left (-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}+\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{4 a b}}{d}\) | \(334\) |
default | \(\frac {-\frac {2 \left (-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2} b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2} b}-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a b}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a \right )^{2}}-\frac {\left (a -3 b \right ) \left (-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}+\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{4 a b}}{d}\) | \(334\) |
risch | \(-\frac {a^{3} {\mathrm e}^{6 d x +6 c}-a^{2} b \,{\mathrm e}^{6 d x +6 c}-5 a \,b^{2} {\mathrm e}^{6 d x +6 c}-3 \,{\mathrm e}^{6 d x +6 c} b^{3}+3 a^{3} {\mathrm e}^{4 d x +4 c}+7 a^{2} b \,{\mathrm e}^{4 d x +4 c}-3 a \,b^{2} {\mathrm e}^{4 d x +4 c}+9 \,{\mathrm e}^{4 d x +4 c} b^{3}+3 a^{3} {\mathrm e}^{2 d x +2 c}+13 a^{2} b \,{\mathrm e}^{2 d x +2 c}+{\mathrm e}^{2 d x +2 c} a \,b^{2}-9 \,{\mathrm e}^{2 d x +2 c} b^{3}+a^{3}+5 a^{2} b +7 a \,b^{2}+3 b^{3}}{4 \left (a +b \right ) \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} a^{2} b d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, d b a}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, d b a}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, d \,a^{2}}\) | \(504\) |
1/d*(-2*(-1/8*(a+5*b)/a/b*tanh(1/2*d*x+1/2*c)^7-1/8*(3*a^2+11*a*b+12*b^2)/ a^2/b*tanh(1/2*d*x+1/2*c)^5-1/8*(3*a^2+11*a*b+12*b^2)/a^2/b*tanh(1/2*d*x+1 /2*c)^3-1/8*(a+5*b)/a/b*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a+2*ta nh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2-1/4/a*(a-3*b)/b*(-1/2 *(a-((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))+1/2* (-a-((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))))
Leaf count of result is larger than twice the leaf count of optimal. 2677 vs. \(2 (101) = 202\).
Time = 0.34 (sec) , antiderivative size = 5659, normalized size of antiderivative = 49.21 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (101) = 202\).
Time = 0.46 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.13 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {a^{3} + 5 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} + {\left (3 \, a^{3} + 13 \, a^{2} b + a b^{2} - 9 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (3 \, a^{3} + 7 \, a^{2} b - 3 \, a b^{2} + 9 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{3} - a^{2} b - 5 \, a b^{2} - 3 \, b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4} + 4 \, {\left (a^{5} b + a^{4} b^{2} - a^{3} b^{3} - a^{2} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} b + a^{4} b^{2} + a^{3} b^{3} + 3 \, a^{2} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} b + a^{4} b^{2} - a^{3} b^{3} - a^{2} b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {{\left (a - 3 \, b\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b d} \]
1/4*(a^3 + 5*a^2*b + 7*a*b^2 + 3*b^3 + (3*a^3 + 13*a^2*b + a*b^2 - 9*b^3)* e^(-2*d*x - 2*c) + (3*a^3 + 7*a^2*b - 3*a*b^2 + 9*b^3)*e^(-4*d*x - 4*c) + (a^3 - a^2*b - 5*a*b^2 - 3*b^3)*e^(-6*d*x - 6*c))/((a^5*b + 3*a^4*b^2 + 3* a^3*b^3 + a^2*b^4 + 4*(a^5*b + a^4*b^2 - a^3*b^3 - a^2*b^4)*e^(-2*d*x - 2* c) + 2*(3*a^5*b + a^4*b^2 + a^3*b^3 + 3*a^2*b^4)*e^(-4*d*x - 4*c) + 4*(a^5 *b + a^4*b^2 - a^3*b^3 - a^2*b^4)*e^(-6*d*x - 6*c) + (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*e^(-8*d*x - 8*c))*d) + 1/8*(a - 3*b)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^2*b*d)
\[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{4}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]