Integrand size = 21, antiderivative size = 48 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}-\frac {(a-b) \tanh ^3(c+d x)}{3 d}-\frac {b \tanh ^5(c+d x)}{5 d} \]
Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.79 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}+\frac {2 b \tanh (c+d x)}{15 d}+\frac {b \text {sech}^2(c+d x) \tanh (c+d x)}{15 d}-\frac {b \text {sech}^4(c+d x) \tanh (c+d x)}{5 d}-\frac {a \tanh ^3(c+d x)}{3 d} \]
(a*Tanh[c + d*x])/d + (2*b*Tanh[c + d*x])/(15*d) + (b*Sech[c + d*x]^2*Tanh [c + d*x])/(15*d) - (b*Sech[c + d*x]^4*Tanh[c + d*x])/(5*d) - (a*Tanh[c + d*x]^3)/(3*d)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4158, 290, 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.
\(\displaystyle \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (i c+i d x)^4 \left (a-b \tan (i c+i d x)^2\right )dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \frac {\int \left (-b \tanh ^4(c+d x)-(a-b) \tanh ^2(c+d x)+a\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} (a-b) \tanh ^3(c+d x)+a \tanh (c+d x)-\frac {1}{5} b \tanh ^5(c+d x)}{d}\) |
3.1.88.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 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])
Time = 8.92 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {\frac {\tanh \left (d x +c \right )^{5} b}{5}+\frac {\left (a -b \right ) \tanh \left (d x +c \right )^{3}}{3}-a \tanh \left (d x +c \right )}{d}\) | \(42\) |
default | \(-\frac {\frac {\tanh \left (d x +c \right )^{5} b}{5}+\frac {\left (a -b \right ) \tanh \left (d x +c \right )^{3}}{3}-a \tanh \left (d x +c \right )}{d}\) | \(42\) |
risch | \(-\frac {4 \left (15 a \,{\mathrm e}^{6 d x +6 c}+15 b \,{\mathrm e}^{6 d x +6 c}+35 a \,{\mathrm e}^{4 d x +4 c}-5 b \,{\mathrm e}^{4 d x +4 c}+25 \,{\mathrm e}^{2 d x +2 c} a +5 b \,{\mathrm e}^{2 d x +2 c}+5 a +b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(96\) |
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (44) = 88\).
Time = 0.26 (sec) , antiderivative size = 345, normalized size of antiderivative = 7.19 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {8 \, {\left (2 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (5 \, a + 7 \, b\right )} \sinh \left (d x + c\right )^{3} + 30 \, a \cosh \left (d x + c\right ) + {\left (3 \, {\left (5 \, a + 7 \, b\right )} \cosh \left (d x + c\right )^{2} + 5 \, a - 5 \, b\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 5 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 11 \, d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + {\left (21 \, d \cosh \left (d x + c\right )^{5} + 50 \, d \cosh \left (d x + c\right )^{3} + 33 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
-8/15*(2*(5*a + 4*b)*cosh(d*x + c)^3 + 6*(5*a + 4*b)*cosh(d*x + c)*sinh(d* x + c)^2 + (5*a + 7*b)*sinh(d*x + c)^3 + 30*a*cosh(d*x + c) + (3*(5*a + 7* b)*cosh(d*x + c)^2 + 5*a - 5*b)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*co sh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 5*d*cosh(d*x + c)^5 + (2 1*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 5*d* cosh(d*x + c))*sinh(d*x + c)^4 + 11*d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c )^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + (21*d*cosh(d*x + c)^5 + 50*d*cosh(d*x + c)^3 + 33*d*cosh(d*x + c))*sinh(d*x + c)^2 + 15*d*cosh(d *x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c) ^2 + 5*d)*sinh(d*x + c))
\[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {sech}^{4}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (44) = 88\).
Time = 0.20 (sec) , antiderivative size = 371, normalized size of antiderivative = 7.73 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {4}{15} \, b {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]
4/15*b*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) - 5*e^ (-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 15*e^(-6*d*x - 6 *c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5 *e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d* x - 10*c) + 1))) + 4/3*a*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^ (-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^( -4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (44) = 88\).
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (15 \, a e^{\left (6 \, d x + 6 \, c\right )} + 15 \, b e^{\left (6 \, d x + 6 \, c\right )} + 35 \, a e^{\left (4 \, d x + 4 \, c\right )} - 5 \, b e^{\left (4 \, d x + 4 \, c\right )} + 25 \, a e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a + b\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]
-4/15*(15*a*e^(6*d*x + 6*c) + 15*b*e^(6*d*x + 6*c) + 35*a*e^(4*d*x + 4*c) - 5*b*e^(4*d*x + 4*c) + 25*a*e^(2*d*x + 2*c) + 5*b*e^(2*d*x + 2*c) + 5*a + b)/(d*(e^(2*d*x + 2*c) + 1)^5)
Time = 0.15 (sec) , antiderivative size = 304, normalized size of antiderivative = 6.33 \[ \int \text {sech}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {\frac {8\,\left (a-b\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {16\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,\left (a+b\right )}{5\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
- ((8*(a - b))/(15*d) + (4*exp(2*c + 2*d*x)*(a + b))/(5*d))/(3*exp(2*c + 2 *d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((8*exp(2*c + 2*d*x)* (a + b))/(5*d) + (8*exp(6*c + 6*d*x)*(a + b))/(5*d) + (16*exp(4*c + 4*d*x) *(a - b))/(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*(a + b))/(5*d) + (6*exp(4*c + 4*d*x)*(a + b))/(5*d) + (8*exp(2*c + 2*d*x)*(a - b))/(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))/(5*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))