Integrand size = 21, antiderivative size = 64 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {a^3 \cosh (c+d x)}{d}-\frac {3 a^2 b \text {sech}(c+d x)}{d}-\frac {a b^2 \text {sech}^3(c+d x)}{d}-\frac {b^3 \text {sech}^5(c+d x)}{5 d} \]
Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.45 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {8 \left (b+a \cosh ^2(c+d x)\right )^3 \left (-b^3-5 a b^2 \cosh ^2(c+d x)-15 a^2 b \cosh ^4(c+d x)+5 a^3 \cosh ^6(c+d x)\right ) \text {sech}^5(c+d x)}{5 d (a+2 b+a \cosh (2 (c+d x)))^3} \]
(8*(b + a*Cosh[c + d*x]^2)^3*(-b^3 - 5*a*b^2*Cosh[c + d*x]^2 - 15*a^2*b*Co sh[c + d*x]^4 + 5*a^3*Cosh[c + d*x]^6)*Sech[c + d*x]^5)/(5*d*(a + 2*b + a* Cosh[2*(c + d*x)])^3)
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 26, 4621, 244, 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.
\(\displaystyle \int \sinh (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i \sin (i c+i d x) \left (a+b \sec (i c+i d x)^2\right )^3dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \left (b \sec (i c+i d x)^2+a\right )^3 \sin (i c+i d x)dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle \frac {\int \left (a \cosh ^2(c+d x)+b\right )^3 \text {sech}^6(c+d x)d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (b^3 \text {sech}^6(c+d x)+3 a b^2 \text {sech}^4(c+d x)+3 a^2 b \text {sech}^2(c+d x)+a^3\right )d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \cosh (c+d x)-3 a^2 b \text {sech}(c+d x)-a b^2 \text {sech}^3(c+d x)-\frac {1}{5} b^3 \text {sech}^5(c+d x)}{d}\) |
(a^3*Cosh[c + d*x] - 3*a^2*b*Sech[c + d*x] - a*b^2*Sech[c + d*x]^3 - (b^3* Sech[c + d*x]^5)/5)/d
3.1.20.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 72.64 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(-\frac {\frac {b^{3} \operatorname {sech}\left (d x +c \right )^{5}}{5}+a \,b^{2} \operatorname {sech}\left (d x +c \right )^{3}+3 a^{2} b \,\operatorname {sech}\left (d x +c \right )-\frac {a^{3}}{\operatorname {sech}\left (d x +c \right )}}{d}\) | \(58\) |
default | \(-\frac {\frac {b^{3} \operatorname {sech}\left (d x +c \right )^{5}}{5}+a \,b^{2} \operatorname {sech}\left (d x +c \right )^{3}+3 a^{2} b \,\operatorname {sech}\left (d x +c \right )-\frac {a^{3}}{\operatorname {sech}\left (d x +c \right )}}{d}\) | \(58\) |
parts | \(\frac {a^{3} \cosh \left (d x +c \right )}{d}-\frac {3 a^{2} b \,\operatorname {sech}\left (d x +c \right )}{d}-\frac {a \,b^{2} \operatorname {sech}\left (d x +c \right )^{3}}{d}-\frac {b^{3} \operatorname {sech}\left (d x +c \right )^{5}}{5 d}\) | \(63\) |
risch | \(\frac {a^{3} {\mathrm e}^{d x +c}}{2 d}+\frac {a^{3} {\mathrm e}^{-d x -c}}{2 d}-\frac {2 \,{\mathrm e}^{d x +c} b \left (15 a^{2} {\mathrm e}^{8 d x +8 c}+60 a^{2} {\mathrm e}^{6 d x +6 c}+20 a b \,{\mathrm e}^{6 d x +6 c}+90 a^{2} {\mathrm e}^{4 d x +4 c}+40 a b \,{\mathrm e}^{4 d x +4 c}+16 \,{\mathrm e}^{4 d x +4 c} b^{2}+60 a^{2} {\mathrm e}^{2 d x +2 c}+20 a b \,{\mathrm e}^{2 d x +2 c}+15 a^{2}\right )}{5 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(173\) |
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (62) = 124\).
Time = 0.26 (sec) , antiderivative size = 276, normalized size of antiderivative = 4.31 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {5 \, a^{3} \cosh \left (d x + c\right )^{6} + 5 \, a^{3} \sinh \left (d x + c\right )^{6} + 30 \, {\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{4} + 15 \, {\left (5 \, a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} - 4 \, a^{2} b\right )} \sinh \left (d x + c\right )^{4} + 50 \, a^{3} - 180 \, a^{2} b - 80 \, a b^{2} - 32 \, b^{3} + 5 \, {\left (15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2} + 5 \, {\left (15 \, a^{3} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2} + 36 \, {\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{10 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]
1/10*(5*a^3*cosh(d*x + c)^6 + 5*a^3*sinh(d*x + c)^6 + 30*(a^3 - 2*a^2*b)*c osh(d*x + c)^4 + 15*(5*a^3*cosh(d*x + c)^2 + 2*a^3 - 4*a^2*b)*sinh(d*x + c )^4 + 50*a^3 - 180*a^2*b - 80*a*b^2 - 32*b^3 + 5*(15*a^3 - 48*a^2*b - 16*a *b^2)*cosh(d*x + c)^2 + 5*(15*a^3*cosh(d*x + c)^4 + 15*a^3 - 48*a^2*b - 16 *a*b^2 + 36*(a^3 - 2*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^2)/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d* cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))
\[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3} \sinh {\left (c + d x \right )}\, dx \]
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.47 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {a^{3} \cosh \left (d x + c\right )}{d} - \frac {6 \, a^{2} b}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} - \frac {8 \, a b^{2}}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3}} - \frac {32 \, b^{3}}{5 \, d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}} \]
a^3*cosh(d*x + c)/d - 6*a^2*b/(d*(e^(d*x + c) + e^(-d*x - c))) - 8*a*b^2/( d*(e^(d*x + c) + e^(-d*x - c))^3) - 32/5*b^3/(d*(e^(d*x + c) + e^(-d*x - c ))^5)
Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.58 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {5 \, a^{3} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} - \frac {4 \, {\left (15 \, a^{2} b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4} + 20 \, a b^{2} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 16 \, b^{3}\right )}}{{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}}}{10 \, d} \]
1/10*(5*a^3*(e^(d*x + c) + e^(-d*x - c)) - 4*(15*a^2*b*(e^(d*x + c) + e^(- d*x - c))^4 + 20*a*b^2*(e^(d*x + c) + e^(-d*x - c))^2 + 16*b^3)/(e^(d*x + c) + e^(-d*x - c))^5)/d
Time = 2.14 (sec) , antiderivative size = 288, normalized size of antiderivative = 4.50 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx=\frac {a^3\,{\mathrm {e}}^{c+d\,x}}{2\,d}+\frac {a^3\,{\mathrm {e}}^{-c-d\,x}}{2\,d}+\frac {64\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (5\,a\,b^2-4\,b^3\right )}{5\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {32\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {6\,a^2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {8\,a\,b^2\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
(a^3*exp(c + d*x))/(2*d) + (a^3*exp(- c - d*x))/(2*d) + (64*b^3*exp(c + d* x))/(5*d*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + e xp(8*c + 8*d*x) + 1)) + (8*exp(c + d*x)*(5*a*b^2 - 4*b^3))/(5*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (32*b^3*exp(c + d*x))/(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)) - (6*a^2*b*exp(c + d*x))/ (d*(exp(2*c + 2*d*x) + 1)) - (8*a*b^2*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))