Integrand size = 17, antiderivative size = 55 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\frac {\arctan (\sinh (a+b x))}{8 b}+\frac {\text {sech}(a+b x) \tanh (a+b x)}{8 b}-\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b} \]
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\frac {\arctan (\sinh (a+b x))}{8 b}+\frac {\text {sech}(a+b x) \tanh (a+b x)}{8 b}-\frac {\text {sech}^3(a+b x) \tanh (a+b x)}{4 b} \]
ArcTan[Sinh[a + b*x]]/(8*b) + (Sech[a + b*x]*Tanh[a + b*x])/(8*b) - (Sech[ a + b*x]^3*Tanh[a + b*x])/(4*b)
Time = 0.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 25, 3091, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \tanh ^2(a+b x) \text {sech}^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (i a+i b x)^2 \left (-\sec (i a+i b x)^3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sec (i a+i b x)^3 \tan (i a+i b x)^2dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {1}{4} \int \text {sech}^3(a+b x)dx-\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}+\frac {1}{4} \int \csc \left (i a+i b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \text {sech}(a+b x)dx+\frac {\tanh (a+b x) \text {sech}(a+b x)}{2 b}\right )-\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}+\frac {1}{4} \left (\frac {\tanh (a+b x) \text {sech}(a+b x)}{2 b}+\frac {1}{2} \int \csc \left (i a+i b x+\frac {\pi }{2}\right )dx\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{4} \left (\frac {\arctan (\sinh (a+b x))}{2 b}+\frac {\tanh (a+b x) \text {sech}(a+b x)}{2 b}\right )-\frac {\tanh (a+b x) \text {sech}^3(a+b x)}{4 b}\) |
-1/4*(Sech[a + b*x]^3*Tanh[a + b*x])/b + (ArcTan[Sinh[a + b*x]]/(2*b) + (S ech[a + b*x]*Tanh[a + b*x])/(2*b))/4
3.1.95.3.1 Defintions of rubi rules used
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.92 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {-\frac {\sinh \left (b x +a \right )}{3 \cosh \left (b x +a \right )^{4}}+\frac {\left (\frac {\operatorname {sech}\left (b x +a \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (b x +a \right )}{8}\right ) \tanh \left (b x +a \right )}{3}+\frac {\arctan \left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) | \(58\) |
default | \(\frac {-\frac {\sinh \left (b x +a \right )}{3 \cosh \left (b x +a \right )^{4}}+\frac {\left (\frac {\operatorname {sech}\left (b x +a \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (b x +a \right )}{8}\right ) \tanh \left (b x +a \right )}{3}+\frac {\arctan \left ({\mathrm e}^{b x +a}\right )}{4}}{b}\) | \(58\) |
risch | \(\frac {{\mathrm e}^{b x +a} \left ({\mathrm e}^{6 b x +6 a}-7 \,{\mathrm e}^{4 b x +4 a}+7 \,{\mathrm e}^{2 b x +2 a}-1\right )}{4 b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{4}}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{8 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{8 b}\) | \(91\) |
1/b*(-1/3/cosh(b*x+a)^4*sinh(b*x+a)+1/3*(1/4*sech(b*x+a)^3+3/8*sech(b*x+a) )*tanh(b*x+a)+1/4*arctan(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 808 vs. \(2 (49) = 98\).
Time = 0.28 (sec) , antiderivative size = 808, normalized size of antiderivative = 14.69 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\text {Too large to display} \]
1/4*(cosh(b*x + a)^7 + 7*cosh(b*x + a)*sinh(b*x + a)^6 + sinh(b*x + a)^7 + 7*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^5 - 7*cosh(b*x + a)^5 + 35*(cosh( b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^4 + 7*(5*cosh(b*x + a)^4 - 10*co sh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 7*cosh(b*x + a)^3 + 7*(3*cosh(b*x + a )^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 + (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 3* cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^ 2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 + 10*cos h(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 + 1 5*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 + 3*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 + cosh(b *x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (7*cos h(b*x + a)^6 - 35*cosh(b*x + a)^4 + 21*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 + 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh (b*x + a)^6 + 8*(7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 + 30*b*cosh(b*x + a)^2 + 3 *b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b* cosh(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*x ...
\[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\int \tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (49) = 98\).
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.00 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=-\frac {\arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac {e^{\left (-b x - a\right )} - 7 \, e^{\left (-3 \, b x - 3 \, a\right )} + 7 \, e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \]
-1/4*arctan(e^(-b*x - a))/b + 1/4*(e^(-b*x - a) - 7*e^(-3*b*x - 3*a) + 7*e ^(-5*b*x - 5*a) - e^(-7*b*x - 7*a))/(b*(4*e^(-2*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.78 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\frac {\pi + \frac {4 \, {\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} - 4 \, e^{\left (b x + a\right )} + 4 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{2}} + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{16 \, b} \]
1/16*(pi + 4*((e^(b*x + a) - e^(-b*x - a))^3 - 4*e^(b*x + a) + 4*e^(-b*x - a))/((e^(b*x + a) - e^(-b*x - a))^2 + 4)^2 + 2*arctan(1/2*(e^(2*b*x + 2*a ) - 1)*e^(-b*x - a)))/b
Time = 2.19 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.91 \[ \int \text {sech}^3(a+b x) \tanh ^2(a+b x) \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{4\,\sqrt {b^2}}-\frac {\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {2\,{\mathrm {e}}^{3\,a+3\,b\,x}}{b}+\frac {{\mathrm {e}}^{5\,a+5\,b\,x}}{b}}{4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{2\,b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{a+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]
atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b)/(4*(b^2)^(1/2)) - (exp(a + b*x)/b - (2*exp(3*a + 3*b*x))/b + exp(5*a + 5*b*x)/b)/(4*exp(2*a + 2*b*x) + 6*exp(4 *a + 4*b*x) + 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1) - (3*exp(a + b*x) )/(2*b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1)) + (2*exp(a + b*x))/(b* (3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1)) + exp(a + b*x)/(4*b*(exp(2*a + 2*b*x) + 1))