Integrand size = 29, antiderivative size = 113 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh (c+d x)}{b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^3 d}-\frac {a \sinh ^3(c+d x)}{3 b^2 d}+\frac {\sinh ^4(c+d x)}{4 b d} \]
a^2*(a^2+b^2)*ln(a+b*sinh(d*x+c))/b^5/d-a*(a^2+b^2)*sinh(d*x+c)/b^4/d+1/2* (a^2+b^2)*sinh(d*x+c)^2/b^3/d-1/3*a*sinh(d*x+c)^3/b^2/d+1/4*sinh(d*x+c)^4/ b/d
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {12 a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-12 a b \left (a^2+b^2\right ) \sinh (c+d x)+6 b^2 \left (a^2+b^2\right ) \sinh ^2(c+d x)-4 a b^3 \sinh ^3(c+d x)+3 b^4 \sinh ^4(c+d x)}{12 b^5 d} \]
(12*a^2*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]] - 12*a*b*(a^2 + b^2)*Sinh[c + d*x] + 6*b^2*(a^2 + b^2)*Sinh[c + d*x]^2 - 4*a*b^3*Sinh[c + d*x]^3 + 3*b^ 4*Sinh[c + d*x]^4)/(12*b^5*d)
Time = 0.36 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 25, 3316, 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.
\(\displaystyle \int \frac {\sinh ^2(c+d x) \cosh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin (i c+i d x)^2 \cos (i c+i d x)^3}{a-i b \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos (i c+i d x)^3 \sin (i c+i d x)^2}{a-i b \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {\sinh ^2(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^2 \sinh ^2(c+d x) \left (\sinh ^2(c+d x) b^2+b^2\right )}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (b^3 \sinh ^3(c+d x)-a b^2 \sinh ^2(c+d x)+b \left (a^2+b^2\right ) \sinh (c+d x)-a \left (a^2+b^2\right )+\frac {a^2 \left (a^2+b^2\right )}{a+b \sinh (c+d x)}\right )d(b \sinh (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2+b^2\right ) \sinh ^2(c+d x)-a b \left (a^2+b^2\right ) \sinh (c+d x)+a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-\frac {1}{3} a b^3 \sinh ^3(c+d x)+\frac {1}{4} b^4 \sinh ^4(c+d x)}{b^5 d}\) |
(a^2*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]] - a*b*(a^2 + b^2)*Sinh[c + d*x] + (b^2*(a^2 + b^2)*Sinh[c + d*x]^2)/2 - (a*b^3*Sinh[c + d*x]^3)/3 + (b^4*S inh[c + d*x]^4)/4)/(b^5*d)
3.4.75.3.1 Defintions of rubi rules used
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 = 25.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {\sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a \sinh \left (d x +c \right )^{3} b^{2}}{3}-\frac {\left (a^{2}+b^{2}\right ) \sinh \left (d x +c \right )^{2} b}{2}+a \left (a^{2}+b^{2}\right ) \sinh \left (d x +c \right )}{b^{4}}+\frac {a^{2} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{5}}}{d}\) | \(97\) |
default | \(\frac {-\frac {-\frac {\sinh \left (d x +c \right )^{4} b^{3}}{4}+\frac {a \sinh \left (d x +c \right )^{3} b^{2}}{3}-\frac {\left (a^{2}+b^{2}\right ) \sinh \left (d x +c \right )^{2} b}{2}+a \left (a^{2}+b^{2}\right ) \sinh \left (d x +c \right )}{b^{4}}+\frac {a^{2} \left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{5}}}{d}\) | \(97\) |
risch | \(-\frac {a^{4} x}{b^{5}}-\frac {x \,a^{2}}{b^{3}}+\frac {{\mathrm e}^{4 d x +4 c}}{64 b d}-\frac {a \,{\mathrm e}^{3 d x +3 c}}{24 b^{2} d}+\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{2 d x +2 c}}{16 b d}-\frac {a^{3} {\mathrm e}^{d x +c}}{2 b^{4} d}-\frac {3 a \,{\mathrm e}^{d x +c}}{8 b^{2} d}+\frac {a^{3} {\mathrm e}^{-d x -c}}{2 b^{4} d}+\frac {3 a \,{\mathrm e}^{-d x -c}}{8 b^{2} d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{-2 d x -2 c}}{16 b d}+\frac {a \,{\mathrm e}^{-3 d x -3 c}}{24 b^{2} d}+\frac {{\mathrm e}^{-4 d x -4 c}}{64 b d}-\frac {2 a^{4} c}{b^{5} d}-\frac {2 a^{2} c}{b^{3} d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{5} d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{3} d}\) | \(326\) |
1/d*(-1/b^4*(-1/4*sinh(d*x+c)^4*b^3+1/3*a*sinh(d*x+c)^3*b^2-1/2*(a^2+b^2)* sinh(d*x+c)^2*b+a*(a^2+b^2)*sinh(d*x+c))+a^2*(a^2+b^2)/b^5*ln(a+b*sinh(d*x +c)))
Leaf count of result is larger than twice the leaf count of optimal. 1069 vs. \(2 (107) = 214\).
Time = 0.27 (sec) , antiderivative size = 1069, normalized size of antiderivative = 9.46 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]
1/192*(3*b^4*cosh(d*x + c)^8 + 3*b^4*sinh(d*x + c)^8 - 8*a*b^3*cosh(d*x + c)^7 + 8*(3*b^4*cosh(d*x + c) - a*b^3)*sinh(d*x + c)^7 - 192*(a^4 + a^2*b^ 2)*d*x*cosh(d*x + c)^4 + 12*(2*a^2*b^2 + b^4)*cosh(d*x + c)^6 + 4*(21*b^4* cosh(d*x + c)^2 - 14*a*b^3*cosh(d*x + c) + 6*a^2*b^2 + 3*b^4)*sinh(d*x + c )^6 - 24*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^5 + 24*(7*b^4*cosh(d*x + c)^3 - 7*a*b^3*cosh(d*x + c)^2 - 4*a^3*b - 3*a*b^3 + 3*(2*a^2*b^2 + b^4)*cosh(d* x + c))*sinh(d*x + c)^5 + 8*a*b^3*cosh(d*x + c) + 2*(105*b^4*cosh(d*x + c) ^4 - 140*a*b^3*cosh(d*x + c)^3 - 96*(a^4 + a^2*b^2)*d*x + 90*(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 - 60*(4*a^3*b + 3*a*b^3)*cosh(d*x + c))*sinh(d*x + c) ^4 + 3*b^4 + 24*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^3 + 8*(21*b^4*cosh(d*x + c)^5 - 35*a*b^3*cosh(d*x + c)^4 + 12*a^3*b + 9*a*b^3 - 96*(a^4 + a^2*b^2) *d*x*cosh(d*x + c) + 30*(2*a^2*b^2 + b^4)*cosh(d*x + c)^3 - 30*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 12*(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 12*(7*b^4*cosh(d*x + c)^6 - 14*a*b^3*cosh(d*x + c)^5 - 96*(a^4 + a^2*b^2)*d*x*cosh(d*x + c)^2 + 15*(2*a^2*b^2 + b^4)*cosh(d*x + c)^4 + 2*a^ 2*b^2 + b^4 - 20*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^3 + 6*(4*a^3*b + 3*a*b^ 3)*cosh(d*x + c))*sinh(d*x + c)^2 + 192*((a^4 + a^2*b^2)*cosh(d*x + c)^4 + 4*(a^4 + a^2*b^2)*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^4 + a^2*b^2)*cosh( d*x + c)^2*sinh(d*x + c)^2 + 4*(a^4 + a^2*b^2)*cosh(d*x + c)*sinh(d*x + c) ^3 + (a^4 + a^2*b^2)*sinh(d*x + c)^4)*log(2*(b*sinh(d*x + c) + a)/(cosh...
Timed out. \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (107) = 214\).
Time = 0.20 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.07 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (8 \, a b^{2} e^{\left (-d x - c\right )} - 3 \, b^{3} - 12 \, {\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 24 \, {\left (4 \, a^{3} + 3 \, a b^{2}\right )} e^{\left (-3 \, d x - 3 \, c\right )}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{192 \, b^{4} d} + \frac {{\left (a^{4} + a^{2} b^{2}\right )} {\left (d x + c\right )}}{b^{5} d} + \frac {8 \, a b^{2} e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + 24 \, {\left (4 \, a^{3} + 3 \, a b^{2}\right )} e^{\left (-d x - c\right )} + 12 \, {\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{192 \, b^{4} d} + \frac {{\left (a^{4} + a^{2} b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{5} d} \]
-1/192*(8*a*b^2*e^(-d*x - c) - 3*b^3 - 12*(2*a^2*b + b^3)*e^(-2*d*x - 2*c) + 24*(4*a^3 + 3*a*b^2)*e^(-3*d*x - 3*c))*e^(4*d*x + 4*c)/(b^4*d) + (a^4 + a^2*b^2)*(d*x + c)/(b^5*d) + 1/192*(8*a*b^2*e^(-3*d*x - 3*c) + 3*b^3*e^(- 4*d*x - 4*c) + 24*(4*a^3 + 3*a*b^2)*e^(-d*x - c) + 12*(2*a^2*b + b^3)*e^(- 2*d*x - 2*c))/(b^4*d) + (a^4 + a^2*b^2)*log(-2*a*e^(-d*x - c) + b*e^(-2*d* x - 2*c) - b)/(b^5*d)
Time = 0.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.79 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {3 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4} - 8 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 24 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 24 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 96 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 96 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{4}} + \frac {192 \, {\left (a^{4} + a^{2} b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{5}}}{192 \, d} \]
1/192*((3*b^3*(e^(d*x + c) - e^(-d*x - c))^4 - 8*a*b^2*(e^(d*x + c) - e^(- d*x - c))^3 + 24*a^2*b*(e^(d*x + c) - e^(-d*x - c))^2 + 24*b^3*(e^(d*x + c ) - e^(-d*x - c))^2 - 96*a^3*(e^(d*x + c) - e^(-d*x - c)) - 96*a*b^2*(e^(d *x + c) - e^(-d*x - c)))/b^4 + 192*(a^4 + a^2*b^2)*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/b^5)/d
Time = 1.32 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.11 \[ \int \frac {\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,b\,d}-\frac {x\,\left (a^4+a^2\,b^2\right )}{b^5}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,b\,d}+\frac {a\,{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,b^2\,d}-\frac {a\,{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,b^2\,d}+\frac {\ln \left (2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^4+a^2\,b^2\right )}{b^5\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (4\,a^3+3\,a\,b^2\right )}{8\,b^4\,d}+\frac {{\mathrm {e}}^{-c-d\,x}\,\left (4\,a^3+3\,a\,b^2\right )}{8\,b^4\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (2\,a^2+b^2\right )}{16\,b^3\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^2+b^2\right )}{16\,b^3\,d} \]
exp(- 4*c - 4*d*x)/(64*b*d) - (x*(a^4 + a^2*b^2))/b^5 + exp(4*c + 4*d*x)/( 64*b*d) + (a*exp(- 3*c - 3*d*x))/(24*b^2*d) - (a*exp(3*c + 3*d*x))/(24*b^2 *d) + (log(2*a*exp(d*x)*exp(c) - b + b*exp(2*c)*exp(2*d*x))*(a^4 + a^2*b^2 ))/(b^5*d) - (exp(c + d*x)*(3*a*b^2 + 4*a^3))/(8*b^4*d) + (exp(- c - d*x)* (3*a*b^2 + 4*a^3))/(8*b^4*d) + (exp(- 2*c - 2*d*x)*(2*a^2 + b^2))/(16*b^3* d) + (exp(2*c + 2*d*x)*(2*a^2 + b^2))/(16*b^3*d)