Integrand size = 27, antiderivative size = 85 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^4 d}+\frac {\left (a^2+b^2\right ) \sinh (c+d x)}{b^3 d}-\frac {a \sinh ^2(c+d x)}{2 b^2 d}+\frac {\sinh ^3(c+d x)}{3 b d} \]
-a*(a^2+b^2)*ln(a+b*sinh(d*x+c))/b^4/d+(a^2+b^2)*sinh(d*x+c)/b^3/d-1/2*a*s inh(d*x+c)^2/b^2/d+1/3*sinh(d*x+c)^3/b/d
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-6 a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))+6 b \left (a^2+b^2\right ) \sinh (c+d x)-3 a b^2 \sinh ^2(c+d x)+2 b^3 \sinh ^3(c+d x)}{6 b^4 d} \]
(-6*a*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]] + 6*b*(a^2 + b^2)*Sinh[c + d*x] - 3*a*b^2*Sinh[c + d*x]^2 + 2*b^3*Sinh[c + d*x]^3)/(6*b^4*d)
Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 26, 3316, 26, 27, 522, 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.
| \(\displaystyle \int \frac {\sinh (c+d x) \cosh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \sin (i c+i d x) \cos (i c+i d x)^3}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\cos (i c+i d x)^3 \sin (i c+i d x)}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {i \int -\frac {i \sinh (c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\sinh (c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b \sinh (c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (\left (\frac {b^2}{a^2}+1\right ) a^2-b \sinh (c+d x) a-\frac {\left (a^2+b^2\right ) a}{a+b \sinh (c+d x)}+b^2 \sinh ^2(c+d x)\right )d(b \sinh (c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (a^2+b^2\right ) \sinh (c+d x)-a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-\frac {1}{2} a b^2 \sinh ^2(c+d x)+\frac {1}{3} b^3 \sinh ^3(c+d x)}{b^4 d}\) |
(-(a*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]]) + b*(a^2 + b^2)*Sinh[c + d*x] - (a*b^2*Sinh[c + d*x]^2)/2 + (b^3*Sinh[c + d*x]^3)/3)/(b^4*d)
3.4.46.3.1 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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 8.98 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\sinh \left (d x +c \right )^{3} b^{2}}{3}-\frac {a \sinh \left (d x +c \right )^{2} b}{2}+a^{2} \sinh \left (d x +c \right )+b^{2} \sinh \left (d x +c \right )}{b^{3}}-\frac {a \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(80\) |
| default | \(\frac {\frac {\frac {\sinh \left (d x +c \right )^{3} b^{2}}{3}-\frac {a \sinh \left (d x +c \right )^{2} b}{2}+a^{2} \sinh \left (d x +c \right )+b^{2} \sinh \left (d x +c \right )}{b^{3}}-\frac {a \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(80\) |
| risch | \(\frac {a^{3} x}{b^{4}}+\frac {a x}{b^{2}}+\frac {{\mathrm e}^{3 d x +3 c}}{24 b d}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 b^{2} d}+\frac {{\mathrm e}^{d x +c} a^{2}}{2 b^{3} d}+\frac {3 \,{\mathrm e}^{d x +c}}{8 b d}-\frac {{\mathrm e}^{-d x -c} a^{2}}{2 b^{3} d}-\frac {3 \,{\mathrm e}^{-d x -c}}{8 b d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 b^{2} d}-\frac {{\mathrm e}^{-3 d x -3 c}}{24 b d}+\frac {2 a^{3} c}{b^{4} d}+\frac {2 a c}{b^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{4} d}-\frac {a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{2} d}\) | \(244\) |
1/d*(1/b^3*(1/3*sinh(d*x+c)^3*b^2-1/2*a*sinh(d*x+c)^2*b+a^2*sinh(d*x+c)+b^ 2*sinh(d*x+c))-a*(a^2+b^2)/b^4*ln(a+b*sinh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 652 vs. \(2 (81) = 162\).
Time = 0.29 (sec) , antiderivative size = 652, normalized size of antiderivative = 7.67 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{3} \cosh \left (d x + c\right )^{6} + b^{3} \sinh \left (d x + c\right )^{6} - 3 \, a b^{2} \cosh \left (d x + c\right )^{5} + 24 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right )^{3} + 3 \, {\left (2 \, b^{3} \cosh \left (d x + c\right ) - a b^{2}\right )} \sinh \left (d x + c\right )^{5} + 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{2} - 5 \, a b^{2} \cosh \left (d x + c\right ) + 4 \, a^{2} b + 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} - 3 \, a b^{2} \cosh \left (d x + c\right ) + 2 \, {\left (10 \, b^{3} \cosh \left (d x + c\right )^{3} - 15 \, a b^{2} \cosh \left (d x + c\right )^{2} + 12 \, {\left (a^{3} + a b^{2}\right )} d x + 6 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - b^{3} - 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} - 10 \, a b^{2} \cosh \left (d x + c\right )^{3} + 24 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right ) - 4 \, a^{2} b - 3 \, b^{3} + 6 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 24 \, {\left ({\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )^{3}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 3 \, {\left (2 \, b^{3} \cosh \left (d x + c\right )^{5} - 5 \, a b^{2} \cosh \left (d x + c\right )^{4} + 24 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right )^{2} + 4 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - a b^{2} - 2 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{24 \, {\left (b^{4} d \cosh \left (d x + c\right )^{3} + 3 \, b^{4} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, b^{4} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + b^{4} d \sinh \left (d x + c\right )^{3}\right )}} \]
1/24*(b^3*cosh(d*x + c)^6 + b^3*sinh(d*x + c)^6 - 3*a*b^2*cosh(d*x + c)^5 + 24*(a^3 + a*b^2)*d*x*cosh(d*x + c)^3 + 3*(2*b^3*cosh(d*x + c) - a*b^2)*s inh(d*x + c)^5 + 3*(4*a^2*b + 3*b^3)*cosh(d*x + c)^4 + 3*(5*b^3*cosh(d*x + c)^2 - 5*a*b^2*cosh(d*x + c) + 4*a^2*b + 3*b^3)*sinh(d*x + c)^4 - 3*a*b^2 *cosh(d*x + c) + 2*(10*b^3*cosh(d*x + c)^3 - 15*a*b^2*cosh(d*x + c)^2 + 12 *(a^3 + a*b^2)*d*x + 6*(4*a^2*b + 3*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - b^3 - 3*(4*a^2*b + 3*b^3)*cosh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 - 10* a*b^2*cosh(d*x + c)^3 + 24*(a^3 + a*b^2)*d*x*cosh(d*x + c) - 4*a^2*b - 3*b ^3 + 6*(4*a^2*b + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 24*((a^3 + a*b ^2)*cosh(d*x + c)^3 + 3*(a^3 + a*b^2)*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a ^3 + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + (a^3 + a*b^2)*sinh(d*x + c)^3) *log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 3*(2*b^3*c osh(d*x + c)^5 - 5*a*b^2*cosh(d*x + c)^4 + 24*(a^3 + a*b^2)*d*x*cosh(d*x + c)^2 + 4*(4*a^2*b + 3*b^3)*cosh(d*x + c)^3 - a*b^2 - 2*(4*a^2*b + 3*b^3)* cosh(d*x + c))*sinh(d*x + c))/(b^4*d*cosh(d*x + c)^3 + 3*b^4*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*b^4*d*cosh(d*x + c)*sinh(d*x + c)^2 + b^4*d*sinh(d* x + c)^3)
Timed out. \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (81) = 162\).
Time = 0.20 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.15 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (3 \, a b e^{\left (-d x - c\right )} - b^{2} - 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{24 \, b^{3} d} - \frac {{\left (a^{3} + a b^{2}\right )} {\left (d x + c\right )}}{b^{4} d} - \frac {3 \, a b e^{\left (-2 \, d x - 2 \, c\right )} + b^{2} e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} e^{\left (-d x - c\right )}}{24 \, b^{3} d} - \frac {{\left (a^{3} + a b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{4} d} \]
-1/24*(3*a*b*e^(-d*x - c) - b^2 - 3*(4*a^2 + 3*b^2)*e^(-2*d*x - 2*c))*e^(3 *d*x + 3*c)/(b^3*d) - (a^3 + a*b^2)*(d*x + c)/(b^4*d) - 1/24*(3*a*b*e^(-2* d*x - 2*c) + b^2*e^(-3*d*x - 3*c) + 3*(4*a^2 + 3*b^2)*e^(-d*x - c))/(b^3*d ) - (a^3 + a*b^2)*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/(b^4*d)
Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.71 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 3 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 12 \, a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 12 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{3}} - \frac {24 \, {\left (a^{3} + a b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{4}}}{24 \, d} \]
1/24*((b^2*(e^(d*x + c) - e^(-d*x - c))^3 - 3*a*b*(e^(d*x + c) - e^(-d*x - c))^2 + 12*a^2*(e^(d*x + c) - e^(-d*x - c)) + 12*b^2*(e^(d*x + c) - e^(-d *x - c)))/b^3 - 24*(a^3 + a*b^2)*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/b^4)/d
Time = 1.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.12 \[ \int \frac {\cosh ^3(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x\,\left (a^3+a\,b^2\right )}{b^4}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,b\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,b\,d}-\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b^2\,d}-\frac {a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b^2\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (4\,a^2+3\,b^2\right )}{8\,b^3\,d}-\frac {\ln \left (2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^3+a\,b^2\right )}{b^4\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (4\,a^2+3\,b^2\right )}{8\,b^3\,d} \]
(x*(a*b^2 + a^3))/b^4 - exp(- 3*c - 3*d*x)/(24*b*d) + exp(3*c + 3*d*x)/(24 *b*d) - (a*exp(- 2*c - 2*d*x))/(8*b^2*d) - (a*exp(2*c + 2*d*x))/(8*b^2*d) - (exp(- c - d*x)*(4*a^2 + 3*b^2))/(8*b^3*d) - (log(2*a*exp(d*x)*exp(c) - b + b*exp(2*c)*exp(2*d*x))*(a*b^2 + a^3))/(b^4*d) + (exp(c + d*x)*(4*a^2 + 3*b^2))/(8*b^3*d)