Integrand size = 23, antiderivative size = 49 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]
Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 210, 2009}
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.
\(\displaystyle \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \left (b \tanh ^2(c+d x)+a\right )^2d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \left (b^2 \tanh ^4(c+d x)+2 a b \tanh ^2(c+d x)+a^2\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \tanh (c+d x)+\frac {2}{3} a b \tanh ^3(c+d x)+\frac {1}{5} b^2 \tanh ^5(c+d x)}{d}\) |
3.1.94.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[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. \(125\) vs. \(2(45)=90\).
Time = 6.24 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.57
method | result | size |
derivativedivides | \(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(126\) |
default | \(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(126\) |
risch | \(-\frac {2 \left (15 a^{2} {\mathrm e}^{8 d x +8 c}+30 a b \,{\mathrm e}^{8 d x +8 c}+15 b^{2} {\mathrm e}^{8 d x +8 c}+60 a^{2} {\mathrm e}^{6 d x +6 c}+60 a b \,{\mathrm e}^{6 d x +6 c}+90 a^{2} {\mathrm e}^{4 d x +4 c}+40 a b \,{\mathrm e}^{4 d x +4 c}+30 \,{\mathrm e}^{4 d x +4 c} b^{2}+60 a^{2} {\mathrm e}^{2 d x +2 c}+20 a b \,{\mathrm e}^{2 d x +2 c}+15 a^{2}+10 a b +3 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(170\) |
1/d*(a^2*tanh(d*x+c)+2*a*b*(-1/2*sinh(d*x+c)/cosh(d*x+c)^3+1/2*(2/3+1/3*se ch(d*x+c)^2)*tanh(d*x+c))+b^2*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d *x+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d *x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (45) = 90\).
Time = 0.27 (sec) , antiderivative size = 391, normalized size of antiderivative = 7.98 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 8 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 20 \, a b\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 20 \, a b + 15 \, b^{2} + 8 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \]
-4/15*((15*a^2 + 20*a*b + 9*b^2)*cosh(d*x + c)^4 + 8*(5*a*b + 3*b^2)*cosh( d*x + c)*sinh(d*x + c)^3 + (15*a^2 + 20*a*b + 9*b^2)*sinh(d*x + c)^4 + 20* (3*a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(15*a^2 + 20*a*b + 9*b^2)*cosh(d*x + c)^2 + 30*a^2 + 20*a*b)*sinh(d*x + c)^2 + 45*a^2 + 20*a*b + 15*b^2 + 8*( (5*a*b + 3*b^2)*cosh(d*x + c)^3 + 5*a*b*cosh(d*x + c))*sinh(d*x + c))/(d*c osh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 + 6 *d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^4 + 4*(5* d*cosh(d*x + c)^3 + 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 15*d*cosh(d*x + c )^2 + 3*(5*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 + 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh( d*x + c) + 10*d)
\[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname {sech}^{2}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {b^{2} \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac {2 \, a b \tanh \left (d x + c\right )^{3}}{3 \, d} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (45) = 90\).
Time = 0.33 (sec) , antiderivative size = 169, normalized size of antiderivative = 3.45 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 10 \, a b + 3 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]
-2/15*(15*a^2*e^(8*d*x + 8*c) + 30*a*b*e^(8*d*x + 8*c) + 15*b^2*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) + 60*a*b*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) + 40*a*b*e^(4*d*x + 4*c) + 30*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 20*a*b*e^(2*d*x + 2*c) + 15*a^2 + 10*a*b + 3*b^2)/(d*(e^(2*d*x + 2*c) + 1)^5)
Time = 1.96 (sec) , antiderivative size = 482, normalized size of antiderivative = 9.84 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,{\left (a+b\right )}^2}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,{\left (a+b\right )}^2}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,{\left (a+b\right )}^2}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,{\left (a+b\right )}^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
- ((2*(a^2 - b^2))/(5*d) + (2*exp(2*c + 2*d*x)*(a + b)^2)/(5*d))/(2*exp(2* c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(a^2 - b^2))/(5*d) + (6*exp(4*c + 4*d*x)*(a^2 - b^2))/(5*d) + (2*exp(6*c + 6*d*x)*(a + b)^2)/(5*d) + (2*exp (2*c + 2*d*x)*(3*a^2 - 2*a*b + 3*b^2))/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp( 4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(a + b)^2) /(5*d) + (8*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d) + (8*exp(6*c + 6*d*x)*(a^2 - b^2))/(5*d) + (2*exp(8*c + 8*d*x)*(a + b)^2)/(5*d) + (4*exp(4*c + 4*d*x )*(3*a^2 - 2*a*b + 3*b^2))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x ) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - ( (2*(3*a^2 - 2*a*b + 3*b^2))/(15*d) + (4*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d ) + (2*exp(4*c + 4*d*x)*(a + b)^2)/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*(a + b)^2)/(5*d*(exp(2*c + 2*d*x) + 1))