Integrand size = 17, antiderivative size = 92 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=-\frac {5 x}{128}-\frac {5 \cosh (a+b x) \sinh (a+b x)}{128 b}+\frac {5 \cosh ^3(a+b x) \sinh (a+b x)}{64 b}-\frac {5 \cosh ^3(a+b x) \sinh ^3(a+b x)}{48 b}+\frac {\cosh ^3(a+b x) \sinh ^5(a+b x)}{8 b} \]
-5/128*x-5/128*cosh(b*x+a)*sinh(b*x+a)/b+5/64*cosh(b*x+a)^3*sinh(b*x+a)/b- 5/48*cosh(b*x+a)^3*sinh(b*x+a)^3/b+1/8*cosh(b*x+a)^3*sinh(b*x+a)^5/b
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=\frac {-120 b x+48 \sinh (2 (a+b x))+24 \sinh (4 (a+b x))-16 \sinh (6 (a+b x))+3 \sinh (8 (a+b x))}{3072 b} \]
(-120*b*x + 48*Sinh[2*(a + b*x)] + 24*Sinh[4*(a + b*x)] - 16*Sinh[6*(a + b *x)] + 3*Sinh[8*(a + b*x)])/(3072*b)
Time = 0.48 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {3042, 25, 3048, 3042, 3048, 25, 3042, 25, 3048, 3042, 3115, 24}
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.
\(\displaystyle \int \sinh ^6(a+b x) \cosh ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i a+i b x)^6 \left (-\cos (i a+i b x)^2\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cos (i a+i b x)^2 \sin (i a+i b x)^6dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \int \cosh ^2(a+b x) \sinh ^4(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \int \cos (i a+i b x)^2 \sin (i a+i b x)^4dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {1}{2} \int -\cosh ^2(a+b x) \sinh ^2(a+b x)dx+\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}-\frac {1}{2} \int \cosh ^2(a+b x) \sinh ^2(a+b x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}-\frac {1}{2} \int -\cos (i a+i b x)^2 \sin (i a+i b x)^2dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}+\frac {1}{2} \int \cos (i a+i b x)^2 \sin (i a+i b x)^2dx\right )\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \cosh ^2(a+b x)dx-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}\right )+\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}+\frac {1}{2} \left (-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {1}{4} \int \sin \left (i a+i b x+\frac {\pi }{2}\right )^2dx\right )\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int 1dx}{2}+\frac {\sinh (a+b x) \cosh (a+b x)}{2 b}\right )-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}\right )+\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sinh ^5(a+b x) \cosh ^3(a+b x)}{8 b}-\frac {5}{8} \left (\frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}+\frac {1}{2} \left (\frac {1}{4} \left (\frac {\sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x}{2}\right )-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}\right )\right )\) |
(Cosh[a + b*x]^3*Sinh[a + b*x]^5)/(8*b) - (5*((Cosh[a + b*x]^3*Sinh[a + b* x]^3)/(6*b) + (-1/4*(Cosh[a + b*x]^3*Sinh[a + b*x])/b + (x/2 + (Cosh[a + b *x]*Sinh[a + b*x])/(2*b))/4)/2))/8
3.1.17.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.86
\[\frac {\frac {\sinh \left (b x +a \right )^{5} \cosh \left (b x +a \right )^{3}}{8}-\frac {5 \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{3}}{48}+\frac {5 \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )}{64}-\frac {5 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{128}-\frac {5 b x}{128}-\frac {5 a}{128}}{b}\]
1/b*(1/8*sinh(b*x+a)^5*cosh(b*x+a)^3-5/48*cosh(b*x+a)^3*sinh(b*x+a)^3+5/64 *cosh(b*x+a)^3*sinh(b*x+a)-5/128*cosh(b*x+a)*sinh(b*x+a)-5/128*b*x-5/128*a )
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.50 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=\frac {3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + 3 \, {\left (7 \, \cosh \left (b x + a\right )^{3} - 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + {\left (21 \, \cosh \left (b x + a\right )^{5} - 40 \, \cosh \left (b x + a\right )^{3} + 12 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} - 15 \, b x + 3 \, {\left (\cosh \left (b x + a\right )^{7} - 4 \, \cosh \left (b x + a\right )^{5} + 4 \, \cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{384 \, b} \]
1/384*(3*cosh(b*x + a)*sinh(b*x + a)^7 + 3*(7*cosh(b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a)^5 + (21*cosh(b*x + a)^5 - 40*cosh(b*x + a)^3 + 12*cos h(b*x + a))*sinh(b*x + a)^3 - 15*b*x + 3*(cosh(b*x + a)^7 - 4*cosh(b*x + a )^5 + 4*cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a))/b
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (85) = 170\).
Time = 0.68 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.05 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=\begin {cases} - \frac {5 x \sinh ^{8}{\left (a + b x \right )}}{128} + \frac {5 x \sinh ^{6}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{32} - \frac {15 x \sinh ^{4}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{64} + \frac {5 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{6}{\left (a + b x \right )}}{32} - \frac {5 x \cosh ^{8}{\left (a + b x \right )}}{128} + \frac {5 \sinh ^{7}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{128 b} + \frac {73 \sinh ^{5}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{384 b} - \frac {55 \sinh ^{3}{\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{384 b} + \frac {5 \sinh {\left (a + b x \right )} \cosh ^{7}{\left (a + b x \right )}}{128 b} & \text {for}\: b \neq 0 \\x \sinh ^{6}{\left (a \right )} \cosh ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]
Piecewise((-5*x*sinh(a + b*x)**8/128 + 5*x*sinh(a + b*x)**6*cosh(a + b*x)* *2/32 - 15*x*sinh(a + b*x)**4*cosh(a + b*x)**4/64 + 5*x*sinh(a + b*x)**2*c osh(a + b*x)**6/32 - 5*x*cosh(a + b*x)**8/128 + 5*sinh(a + b*x)**7*cosh(a + b*x)/(128*b) + 73*sinh(a + b*x)**5*cosh(a + b*x)**3/(384*b) - 55*sinh(a + b*x)**3*cosh(a + b*x)**5/(384*b) + 5*sinh(a + b*x)*cosh(a + b*x)**7/(128 *b), Ne(b, 0)), (x*sinh(a)**6*cosh(a)**2, True))
Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=-\frac {{\left (16 \, e^{\left (-2 \, b x - 2 \, a\right )} - 24 \, e^{\left (-4 \, b x - 4 \, a\right )} - 48 \, e^{\left (-6 \, b x - 6 \, a\right )} - 3\right )} e^{\left (8 \, b x + 8 \, a\right )}}{6144 \, b} - \frac {5 \, {\left (b x + a\right )}}{128 \, b} - \frac {48 \, e^{\left (-2 \, b x - 2 \, a\right )} + 24 \, e^{\left (-4 \, b x - 4 \, a\right )} - 16 \, e^{\left (-6 \, b x - 6 \, a\right )} + 3 \, e^{\left (-8 \, b x - 8 \, a\right )}}{6144 \, b} \]
-1/6144*(16*e^(-2*b*x - 2*a) - 24*e^(-4*b*x - 4*a) - 48*e^(-6*b*x - 6*a) - 3)*e^(8*b*x + 8*a)/b - 5/128*(b*x + a)/b - 1/6144*(48*e^(-2*b*x - 2*a) + 24*e^(-4*b*x - 4*a) - 16*e^(-6*b*x - 6*a) + 3*e^(-8*b*x - 8*a))/b
Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.26 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=-\frac {5}{128} \, x + \frac {e^{\left (8 \, b x + 8 \, a\right )}}{2048 \, b} - \frac {e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{128 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{128 \, b} - \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b} + \frac {e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} - \frac {e^{\left (-8 \, b x - 8 \, a\right )}}{2048 \, b} \]
-5/128*x + 1/2048*e^(8*b*x + 8*a)/b - 1/384*e^(6*b*x + 6*a)/b + 1/256*e^(4 *b*x + 4*a)/b + 1/128*e^(2*b*x + 2*a)/b - 1/128*e^(-2*b*x - 2*a)/b - 1/256 *e^(-4*b*x - 4*a)/b + 1/384*e^(-6*b*x - 6*a)/b - 1/2048*e^(-8*b*x - 8*a)/b
Time = 2.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \cosh ^2(a+b x) \sinh ^6(a+b x) \, dx=\frac {\frac {\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{64}+\frac {\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{128}-\frac {\mathrm {sinh}\left (6\,a+6\,b\,x\right )}{192}+\frac {\mathrm {sinh}\left (8\,a+8\,b\,x\right )}{1024}}{b}-\frac {5\,x}{128} \]