Integrand size = 18, antiderivative size = 109 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\frac {1}{8} a \left (8 a^2-3 b^2\right ) x+\frac {b \left (16 a^2-b^2\right ) \cosh (2 c+2 d x)}{24 d}+\frac {5 a b^2 \cosh (2 c+2 d x) \sinh (2 c+2 d x)}{48 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d} \]
1/8*a*(8*a^2-3*b^2)*x+1/24*b*(16*a^2-b^2)*cosh(2*d*x+2*c)/d+5/48*a*b^2*cos h(2*d*x+2*c)*sinh(2*d*x+2*c)/d+1/48*b*cosh(2*d*x+2*c)*(2*a+b*sinh(2*d*x+2* c))^2/d
Time = 0.65 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\frac {9 \left (16 a^2 b-b^3\right ) \cosh (2 (c+d x))+b^3 \cosh (6 (c+d x))+6 a \left (4 \left (8 a^2-3 b^2\right ) (c+d x)+3 b^2 \sinh (4 (c+d x))\right )}{192 d} \]
(9*(16*a^2*b - b^3)*Cosh[2*(c + d*x)] + b^3*Cosh[6*(c + d*x)] + 6*a*(4*(8* a^2 - 3*b^2)*(c + d*x) + 3*b^2*Sinh[4*(c + d*x)]))/(192*d)
Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 3145, 3042, 3135, 27, 3042, 3213}
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.
\(\displaystyle \int (a+b \sinh (c+d x) \cosh (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a-i b \sin (i c+i d x) \cos (i c+i d x))^3dx\) |
\(\Big \downarrow \) 3145 |
\(\displaystyle \int \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-\frac {1}{2} i b \sin (2 i c+2 i d x)\right )^3dx\) |
\(\Big \downarrow \) 3135 |
\(\displaystyle \frac {1}{3} \int \frac {1}{4} (2 a+b \sinh (2 c+2 d x)) \left (6 a^2+5 b \sinh (2 c+2 d x) a-b^2\right )dx+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \int (2 a+b \sinh (2 c+2 d x)) \left (6 a^2+5 b \sinh (2 c+2 d x) a-b^2\right )dx+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}+\frac {1}{12} \int (2 a-i b \sin (2 i c+2 i d x)) \left (6 a^2-5 i b \sin (2 i c+2 i d x) a-b^2\right )dx\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{12} \left (\frac {b \left (16 a^2-b^2\right ) \cosh (2 c+2 d x)}{2 d}+\frac {3}{2} a x \left (8 a^2-3 b^2\right )+\frac {5 a b^2 \sinh (2 c+2 d x) \cosh (2 c+2 d x)}{4 d}\right )+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}\) |
(b*Cosh[2*c + 2*d*x]*(2*a + b*Sinh[2*c + 2*d*x])^2)/(48*d) + ((3*a*(8*a^2 - 3*b^2)*x)/2 + (b*(16*a^2 - b^2)*Cosh[2*c + 2*d*x])/(2*d) + (5*a*b^2*Cosh [2*c + 2*d*x]*Sinh[2*c + 2*d*x])/(4*d))/12
3.9.55.3.1 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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[1/n Int[(a + b* Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*x] , x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 88.70 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91
method | result | size |
parts | \(a^{3} x +\frac {b^{3} \left (\frac {\cosh \left (d x +c \right )^{6}}{6}-\frac {\cosh \left (d x +c \right )^{4}}{4}\right )}{d}+\frac {3 a^{2} b \cosh \left (d x +c \right )^{2}}{2 d}+\frac {3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(99\) |
derivativedivides | \(\frac {a^{3} \left (d x +c \right )+\frac {3 a^{2} b \cosh \left (d x +c \right )^{2}}{2}+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{2} \cosh \left (d x +c \right )^{4}}{6}-\frac {\cosh \left (d x +c \right )^{4}}{12}\right )}{d}\) | \(106\) |
default | \(\frac {a^{3} \left (d x +c \right )+\frac {3 a^{2} b \cosh \left (d x +c \right )^{2}}{2}+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{2} \cosh \left (d x +c \right )^{4}}{6}-\frac {\cosh \left (d x +c \right )^{4}}{12}\right )}{d}\) | \(106\) |
risch | \(a^{3} x -\frac {3 a \,b^{2} x}{8}+\frac {b^{3} {\mathrm e}^{6 d x +6 c}}{384 d}+\frac {3 a \,b^{2} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {3 b \,{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}-\frac {3 b^{3} {\mathrm e}^{2 d x +2 c}}{128 d}+\frac {3 b \,{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}-\frac {3 b^{3} {\mathrm e}^{-2 d x -2 c}}{128 d}-\frac {3 a \,b^{2} {\mathrm e}^{-4 d x -4 c}}{64 d}+\frac {b^{3} {\mathrm e}^{-6 d x -6 c}}{384 d}\) | \(154\) |
a^3*x+b^3/d*(1/6*cosh(d*x+c)^6-1/4*cosh(d*x+c)^4)+3/2*a^2*b/d*cosh(d*x+c)^ 2+3*a*b^2/d*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8 *d*x-1/8*c)
Time = 0.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.50 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\frac {b^{3} \cosh \left (d x + c\right )^{6} + 15 \, b^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{4} + b^{3} \sinh \left (d x + c\right )^{6} + 72 \, a b^{2} \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 72 \, a b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 24 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} d x + 9 \, {\left (16 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 48 \, a^{2} b - 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{2}}{192 \, d} \]
1/192*(b^3*cosh(d*x + c)^6 + 15*b^3*cosh(d*x + c)^2*sinh(d*x + c)^4 + b^3* sinh(d*x + c)^6 + 72*a*b^2*cosh(d*x + c)^3*sinh(d*x + c) + 72*a*b^2*cosh(d *x + c)*sinh(d*x + c)^3 + 24*(8*a^3 - 3*a*b^2)*d*x + 9*(16*a^2*b - b^3)*co sh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 + 48*a^2*b - 3*b^3)*sinh(d*x + c) ^2)/d
Time = 0.36 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.74 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\begin {cases} a^{3} x + \frac {3 a^{2} b \sinh ^{2}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} x \sinh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {3 a b^{2} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {3 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {b^{3} \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \cosh ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh {\left (c \right )} \cosh {\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \]
Piecewise((a**3*x + 3*a**2*b*sinh(c + d*x)**2/(2*d) - 3*a*b**2*x*sinh(c + d*x)**4/8 + 3*a*b**2*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 - 3*a*b**2*x*co sh(c + d*x)**4/8 + 3*a*b**2*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) + 3*a*b** 2*sinh(c + d*x)*cosh(c + d*x)**3/(8*d) + b**3*sinh(c + d*x)**2*cosh(c + d* x)**4/(4*d) - b**3*cosh(c + d*x)**6/(12*d), Ne(d, 0)), (x*(a + b*sinh(c)*c osh(c))**3, True))
Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=a^{3} x - \frac {1}{384} \, b^{3} {\left (\frac {{\left (9 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {9 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac {3}{64} \, a b^{2} {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3 \, a^{2} b \cosh \left (d x + c\right )^{2}}{2 \, d} \]
a^3*x - 1/384*b^3*((9*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + (9*e^(-2*d *x - 2*c) - e^(-6*d*x - 6*c))/d) - 3/64*a*b^2*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d) + 3/2*a^2*b*cosh(d*x + c)^2/d
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.27 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\frac {b^{3} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} + \frac {3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} - \frac {3 \, a b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac {b^{3} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} + \frac {1}{8} \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} x + \frac {3 \, {\left (16 \, a^{2} b - b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} + \frac {3 \, {\left (16 \, a^{2} b - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} \]
1/384*b^3*e^(6*d*x + 6*c)/d + 3/64*a*b^2*e^(4*d*x + 4*c)/d - 3/64*a*b^2*e^ (-4*d*x - 4*c)/d + 1/384*b^3*e^(-6*d*x - 6*c)/d + 1/8*(8*a^3 - 3*a*b^2)*x + 3/128*(16*a^2*b - b^3)*e^(2*d*x + 2*c)/d + 3/128*(16*a^2*b - b^3)*e^(-2* d*x - 2*c)/d
Time = 2.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx=\frac {\frac {b^3\,\mathrm {cosh}\left (6\,c+6\,d\,x\right )}{8}-\frac {9\,b^3\,\mathrm {cosh}\left (2\,c+2\,d\,x\right )}{8}+18\,a^2\,b\,\mathrm {cosh}\left (2\,c+2\,d\,x\right )+\frac {9\,a\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+24\,a^3\,d\,x-9\,a\,b^2\,d\,x}{24\,d} \]