Integrand size = 17, antiderivative size = 66 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\frac {5 \arctan (\sinh (a+b x))}{2 b}-\frac {5 \sinh (a+b x)}{2 b}+\frac {5 \sinh ^3(a+b x)}{6 b}-\frac {\sinh ^3(a+b x) \tanh ^2(a+b x)}{2 b} \]
5/2*arctan(sinh(b*x+a))/b-5/2*sinh(b*x+a)/b+5/6*sinh(b*x+a)^3/b-1/2*sinh(b *x+a)^3*tanh(b*x+a)^2/b
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\frac {5 \arctan (\sinh (a+b x))}{2 b}-\frac {5 \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {5 \sinh (a+b x) \tanh ^2(a+b x)}{3 b}+\frac {\sinh ^3(a+b x) \tanh ^2(a+b x)}{3 b} \]
(5*ArcTan[Sinh[a + b*x]])/(2*b) - (5*Sech[a + b*x]*Tanh[a + b*x])/(2*b) - (5*Sinh[a + b*x]*Tanh[a + b*x]^2)/(3*b) + (Sinh[a + b*x]^3*Tanh[a + b*x]^2 )/(3*b)
Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 25, 3072, 25, 252, 254, 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.
\(\displaystyle \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin (i a+i b x)^3 \tan (i a+i b x)^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin (i a+i b x)^3 \tan (i a+i b x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle -\frac {\int -\frac {\sinh ^6(a+b x)}{\left (\sinh ^2(a+b x)+1\right )^2}d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sinh ^6(a+b x)}{\left (\sinh ^2(a+b x)+1\right )^2}d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\sinh ^5(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}-\frac {5}{2} \int \frac {\sinh ^4(a+b x)}{\sinh ^2(a+b x)+1}d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\sinh ^5(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}-\frac {5}{2} \int \left (\sinh ^2(a+b x)+\frac {1}{\sinh ^2(a+b x)+1}-1\right )d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\sinh ^5(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}-\frac {5}{2} \left (\arctan (\sinh (a+b x))+\frac {1}{3} \sinh ^3(a+b x)-\sinh (a+b x)\right )}{b}\) |
-((Sinh[a + b*x]^5/(2*(1 + Sinh[a + b*x]^2)) - (5*(ArcTan[Sinh[a + b*x]] - Sinh[a + b*x] + Sinh[a + b*x]^3/3))/2)/b)
3.1.79.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Time = 1.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{5}}{3 \cosh \left (b x +a \right )^{2}}-\frac {5 \sinh \left (b x +a \right )^{3}}{3 \cosh \left (b x +a \right )^{2}}-\frac {5 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}+\frac {5 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}+5 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(81\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{5}}{3 \cosh \left (b x +a \right )^{2}}-\frac {5 \sinh \left (b x +a \right )^{3}}{3 \cosh \left (b x +a \right )^{2}}-\frac {5 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}+\frac {5 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}+5 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(81\) |
risch | \(\frac {{\mathrm e}^{3 b x +3 a}}{24 b}-\frac {9 \,{\mathrm e}^{b x +a}}{8 b}+\frac {9 \,{\mathrm e}^{-b x -a}}{8 b}-\frac {{\mathrm e}^{-3 b x -3 a}}{24 b}-\frac {{\mathrm e}^{b x +a} \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}+\frac {5 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}-\frac {5 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}\) | \(122\) |
1/b*(1/3*sinh(b*x+a)^5/cosh(b*x+a)^2-5/3*sinh(b*x+a)^3/cosh(b*x+a)^2-5/cos h(b*x+a)^2*sinh(b*x+a)+5/2*sech(b*x+a)*tanh(b*x+a)+5*arctan(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 851 vs. \(2 (58) = 116\).
Time = 0.26 (sec) , antiderivative size = 851, normalized size of antiderivative = 12.89 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\text {Too large to display} \]
1/24*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x + a)^ 10 + 5*(9*cosh(b*x + a)^2 - 5)*sinh(b*x + a)^8 - 25*cosh(b*x + a)^8 + 40*( 3*cosh(b*x + a)^3 - 5*cosh(b*x + a))*sinh(b*x + a)^7 + 10*(21*cosh(b*x + a )^4 - 70*cosh(b*x + a)^2 - 5)*sinh(b*x + a)^6 - 50*cosh(b*x + a)^6 + 4*(63 *cosh(b*x + a)^5 - 350*cosh(b*x + a)^3 - 75*cosh(b*x + a))*sinh(b*x + a)^5 + 10*(21*cosh(b*x + a)^6 - 175*cosh(b*x + a)^4 - 75*cosh(b*x + a)^2 + 5)* sinh(b*x + a)^4 + 50*cosh(b*x + a)^4 + 40*(3*cosh(b*x + a)^7 - 35*cosh(b*x + a)^5 - 25*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a)^3 + 5*(9*cos h(b*x + a)^8 - 140*cosh(b*x + a)^6 - 150*cosh(b*x + a)^4 + 60*cosh(b*x + a )^2 + 5)*sinh(b*x + a)^2 + 120*(cosh(b*x + a)^7 + 7*cosh(b*x + a)*sinh(b*x + a)^6 + sinh(b*x + a)^7 + (21*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^5 + 2*c osh(b*x + a)^5 + 5*(7*cosh(b*x + a)^3 + 2*cosh(b*x + a))*sinh(b*x + a)^4 + (35*cosh(b*x + a)^4 + 20*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + cosh(b*x + a)^3 + (21*cosh(b*x + a)^5 + 20*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh( b*x + a)^2 + (7*cosh(b*x + a)^6 + 10*cosh(b*x + a)^4 + 3*cosh(b*x + a)^2)* sinh(b*x + a))*arctan(cosh(b*x + a) + sinh(b*x + a)) + 25*cosh(b*x + a)^2 + 10*(cosh(b*x + a)^9 - 20*cosh(b*x + a)^7 - 30*cosh(b*x + a)^5 + 20*cosh( b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a) - 1)/(b*cosh(b*x + a)^7 + 7*b* cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 + 2*b*cosh(b*x + a)^5 + (21*b*cosh(b*x + a)^2 + 2*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 +...
\[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\int \sinh ^{3}{\left (a + b x \right )} \tanh ^{3}{\left (a + b x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.76 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\frac {27 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac {5 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {25 \, e^{\left (-2 \, b x - 2 \, a\right )} + 77 \, e^{\left (-4 \, b x - 4 \, a\right )} + 3 \, e^{\left (-6 \, b x - 6 \, a\right )} - 1}{24 \, b {\left (e^{\left (-3 \, b x - 3 \, a\right )} + 2 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}\right )}} \]
1/24*(27*e^(-b*x - a) - e^(-3*b*x - 3*a))/b - 5*arctan(e^(-b*x - a))/b - 1 /24*(25*e^(-2*b*x - 2*a) + 77*e^(-4*b*x - 4*a) + 3*e^(-6*b*x - 6*a) - 1)/( b*(e^(-3*b*x - 3*a) + 2*e^(-5*b*x - 5*a) + e^(-7*b*x - 7*a)))
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (58) = 116\).
Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.77 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\frac {30 \, \pi + {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} - \frac {24 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4} + 60 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) - 24 \, e^{\left (b x + a\right )} + 24 \, e^{\left (-b x - a\right )}}{24 \, b} \]
1/24*(30*pi + (e^(b*x + a) - e^(-b*x - a))^3 - 24*(e^(b*x + a) - e^(-b*x - a))/((e^(b*x + a) - e^(-b*x - a))^2 + 4) + 60*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) - 24*e^(b*x + a) + 24*e^(-b*x - a))/b
Time = 2.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.06 \[ \int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx=\frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {9\,{\mathrm {e}}^{a+b\,x}}{8\,b}+\frac {9\,{\mathrm {e}}^{-a-b\,x}}{8\,b}-\frac {{\mathrm {e}}^{-3\,a-3\,b\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{24\,b}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]