Integrand size = 17, antiderivative size = 36 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=-\frac {a \left (a+b \sinh ^2(x)\right )^4}{8 b^2}+\frac {\left (a+b \sinh ^2(x)\right )^5}{10 b^2} \] Output:
-1/8*a*(a+b*sinh(x)^2)^4/b^2+1/10*(a+b*sinh(x)^2)^5/b^2
Leaf count is larger than twice the leaf count of optimal. \(114\) vs. \(2(36)=72\).
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.17 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {-20 \left (64 a^3+24 a b^2-7 b^3\right ) \cosh (2 x)+20 \left (16 a^3+18 a b^2-5 b^3\right ) \cosh (4 x)+b \left (-10 (16 a-5 b) b \cosh (6 x)+15 (2 a-b) b \cosh (8 x)+2 b^2 \cosh (10 x)+320 \left ((-4 a+b)^2-b^2 \cosh (2 x)\right ) \sinh ^6(x)\right )}{10240} \] Input:
Integrate[Cosh[x]*Sinh[x]^3*(a + b*Sinh[x]^2)^3,x]
Output:
(-20*(64*a^3 + 24*a*b^2 - 7*b^3)*Cosh[2*x] + 20*(16*a^3 + 18*a*b^2 - 5*b^3 )*Cosh[4*x] + b*(-10*(16*a - 5*b)*b*Cosh[6*x] + 15*(2*a - b)*b*Cosh[8*x] + 2*b^2*Cosh[10*x] + 320*((-4*a + b)^2 - b^2*Cosh[2*x])*Sinh[x]^6))/10240
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 26, 3677, 26, 243, 49, 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.
\(\displaystyle \int \sinh ^3(x) \cosh (x) \left (a+b \sinh ^2(x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int i \sin (i x)^3 \cos (i x) \left (a-b \sin (i x)^2\right )^3dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \cos (i x) \sin (i x)^3 \left (a-b \sin (i x)^2\right )^3dx\) |
\(\Big \downarrow \) 3677 |
\(\displaystyle i \int -i \sinh ^3(x) \left (b \sinh ^2(x)+a\right )^3d\sinh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3d\sinh (x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \sinh ^2(x) \left (b \sinh ^2(x)+a\right )^3d\sinh ^2(x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (b \sinh ^2(x)+a\right )^4}{b}-\frac {a \left (b \sinh ^2(x)+a\right )^3}{b}\right )d\sinh ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a+b \sinh ^2(x)\right )^5}{5 b^2}-\frac {a \left (a+b \sinh ^2(x)\right )^4}{4 b^2}\right )\) |
Input:
Int[Cosh[x]*Sinh[x]^3*(a + b*Sinh[x]^2)^3,x]
Output:
(-1/4*(a*(a + b*Sinh[x]^2)^4)/b^2 + (a + b*Sinh[x]^2)^5/(5*b^2))/2
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_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFa ctors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[(d*ff*x)^n*(1 - ff^2*x^2)^(( m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11
\[\frac {b^{3} \sinh \left (x \right )^{10}}{10}+\frac {3 a \,b^{2} \sinh \left (x \right )^{8}}{8}+\frac {a^{2} b \sinh \left (x \right )^{6}}{2}+\frac {a^{3} \sinh \left (x \right )^{4}}{4}\]
Input:
int(cosh(x)*sinh(x)^3*(a+b*sinh(x)^2)^3,x)
Output:
1/10*b^3*sinh(x)^10+3/8*a*b^2*sinh(x)^8+1/2*a^2*b*sinh(x)^6+1/4*a^3*sinh(x )^4
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (32) = 64\).
Time = 0.08 (sec) , antiderivative size = 386, normalized size of antiderivative = 10.72 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {1}{5120} \, b^{3} \cosh \left (x\right )^{10} + \frac {1}{5120} \, b^{3} \sinh \left (x\right )^{10} + \frac {1}{1024} \, {\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{8} + \frac {1}{1024} \, {\left (9 \, b^{3} \cosh \left (x\right )^{2} + 3 \, a b^{2} - 2 \, b^{3}\right )} \sinh \left (x\right )^{8} + \frac {1}{1024} \, {\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{6} + \frac {1}{1024} \, {\left (42 \, b^{3} \cosh \left (x\right )^{4} + 16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3} + 28 \, {\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{6} + \frac {1}{256} \, {\left (8 \, a^{3} - 24 \, a^{2} b + 21 \, a b^{2} - 6 \, b^{3}\right )} \cosh \left (x\right )^{4} + \frac {1}{1024} \, {\left (42 \, b^{3} \cosh \left (x\right )^{6} + 70 \, {\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{4} + 32 \, a^{3} - 96 \, a^{2} b + 84 \, a b^{2} - 24 \, b^{3} + 15 \, {\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{4} - \frac {1}{512} \, {\left (64 \, a^{3} - 120 \, a^{2} b + 84 \, a b^{2} - 21 \, b^{3}\right )} \cosh \left (x\right )^{2} + \frac {1}{1024} \, {\left (9 \, b^{3} \cosh \left (x\right )^{8} + 28 \, {\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{6} + 15 \, {\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{4} - 128 \, a^{3} + 240 \, a^{2} b - 168 \, a b^{2} + 42 \, b^{3} + 24 \, {\left (8 \, a^{3} - 24 \, a^{2} b + 21 \, a b^{2} - 6 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} \] Input:
integrate(cosh(x)*sinh(x)^3*(a+b*sinh(x)^2)^3,x, algorithm="fricas")
Output:
1/5120*b^3*cosh(x)^10 + 1/5120*b^3*sinh(x)^10 + 1/1024*(3*a*b^2 - 2*b^3)*c osh(x)^8 + 1/1024*(9*b^3*cosh(x)^2 + 3*a*b^2 - 2*b^3)*sinh(x)^8 + 1/1024*( 16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)^6 + 1/1024*(42*b^3*cosh(x)^4 + 16*a^2 *b - 24*a*b^2 + 9*b^3 + 28*(3*a*b^2 - 2*b^3)*cosh(x)^2)*sinh(x)^6 + 1/256* (8*a^3 - 24*a^2*b + 21*a*b^2 - 6*b^3)*cosh(x)^4 + 1/1024*(42*b^3*cosh(x)^6 + 70*(3*a*b^2 - 2*b^3)*cosh(x)^4 + 32*a^3 - 96*a^2*b + 84*a*b^2 - 24*b^3 + 15*(16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)^2)*sinh(x)^4 - 1/512*(64*a^3 - 120*a^2*b + 84*a*b^2 - 21*b^3)*cosh(x)^2 + 1/1024*(9*b^3*cosh(x)^8 + 28*(3 *a*b^2 - 2*b^3)*cosh(x)^6 + 15*(16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)^4 - 1 28*a^3 + 240*a^2*b - 168*a*b^2 + 42*b^3 + 24*(8*a^3 - 24*a^2*b + 21*a*b^2 - 6*b^3)*cosh(x)^2)*sinh(x)^2
Time = 0.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {a^{3} \sinh ^{4}{\left (x \right )}}{4} + \frac {a^{2} b \sinh ^{6}{\left (x \right )}}{2} + \frac {3 a b^{2} \sinh ^{8}{\left (x \right )}}{8} + \frac {b^{3} \sinh ^{10}{\left (x \right )}}{10} \] Input:
integrate(cosh(x)*sinh(x)**3*(a+b*sinh(x)**2)**3,x)
Output:
a**3*sinh(x)**4/4 + a**2*b*sinh(x)**6/2 + 3*a*b**2*sinh(x)**8/8 + b**3*sin h(x)**10/10
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {1}{10} \, b^{3} \sinh \left (x\right )^{10} + \frac {3}{8} \, a b^{2} \sinh \left (x\right )^{8} + \frac {1}{2} \, a^{2} b \sinh \left (x\right )^{6} + \frac {1}{4} \, a^{3} \sinh \left (x\right )^{4} \] Input:
integrate(cosh(x)*sinh(x)^3*(a+b*sinh(x)^2)^3,x, algorithm="maxima")
Output:
1/10*b^3*sinh(x)^10 + 3/8*a*b^2*sinh(x)^8 + 1/2*a^2*b*sinh(x)^6 + 1/4*a^3* sinh(x)^4
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (32) = 64\).
Time = 0.12 (sec) , antiderivative size = 224, normalized size of antiderivative = 6.22 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {1}{10240} \, b^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{5} + \frac {3}{2048} \, a b^{2} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{4} - \frac {1}{1024} \, b^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{4} + \frac {1}{128} \, a^{2} b {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} - \frac {3}{256} \, a b^{2} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} + \frac {1}{256} \, b^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} + \frac {1}{64} \, a^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac {3}{64} \, a^{2} b {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} + \frac {9}{256} \, a b^{2} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac {1}{128} \, b^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac {1}{16} \, a^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} + \frac {3}{32} \, a^{2} b {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} - \frac {3}{64} \, a b^{2} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} + \frac {1}{128} \, b^{3} {\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} \] Input:
integrate(cosh(x)*sinh(x)^3*(a+b*sinh(x)^2)^3,x, algorithm="giac")
Output:
1/10240*b^3*(e^(2*x) + e^(-2*x))^5 + 3/2048*a*b^2*(e^(2*x) + e^(-2*x))^4 - 1/1024*b^3*(e^(2*x) + e^(-2*x))^4 + 1/128*a^2*b*(e^(2*x) + e^(-2*x))^3 - 3/256*a*b^2*(e^(2*x) + e^(-2*x))^3 + 1/256*b^3*(e^(2*x) + e^(-2*x))^3 + 1/ 64*a^3*(e^(2*x) + e^(-2*x))^2 - 3/64*a^2*b*(e^(2*x) + e^(-2*x))^2 + 9/256* a*b^2*(e^(2*x) + e^(-2*x))^2 - 1/128*b^3*(e^(2*x) + e^(-2*x))^2 - 1/16*a^3 *(e^(2*x) + e^(-2*x)) + 3/32*a^2*b*(e^(2*x) + e^(-2*x)) - 3/64*a*b^2*(e^(2 *x) + e^(-2*x)) + 1/128*b^3*(e^(2*x) + e^(-2*x))
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {a^3\,{\mathrm {sinh}\left (x\right )}^4}{4}+\frac {a^2\,b\,{\mathrm {sinh}\left (x\right )}^6}{2}+\frac {3\,a\,b^2\,{\mathrm {sinh}\left (x\right )}^8}{8}+\frac {b^3\,{\mathrm {sinh}\left (x\right )}^{10}}{10} \] Input:
int(cosh(x)*sinh(x)^3*(a + b*sinh(x)^2)^3,x)
Output:
(a^3*sinh(x)^4)/4 + (b^3*sinh(x)^10)/10 + (a^2*b*sinh(x)^6)/2 + (3*a*b^2*s inh(x)^8)/8
Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx=\frac {\sinh \left (x \right )^{4} \left (4 \sinh \left (x \right )^{6} b^{3}+15 \sinh \left (x \right )^{4} a \,b^{2}+20 \sinh \left (x \right )^{2} a^{2} b +10 a^{3}\right )}{40} \] Input:
int(cosh(x)*sinh(x)^3*(a+b*sinh(x)^2)^3,x)
Output:
(sinh(x)**4*(4*sinh(x)**6*b**3 + 15*sinh(x)**4*a*b**2 + 20*sinh(x)**2*a**2 *b + 10*a**3))/40