3.11 \(\int \cosh ^5(a+b x) \sinh ^n(a+b x) \, dx\)
Optimal. Leaf size=59 \[ \frac{\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac{2 \sinh ^{n+3}(a+b x)}{b (n+3)}+\frac{\sinh ^{n+5}(a+b x)}{b (n+5)} \]
[Out]
Sinh[a + b*x]^(1 + n)/(b*(1 + n)) + (2*Sinh[a + b*x]^(3 + n))/(b*(3 + n)) + Sinh[a + b*x]^(5 + n)/(b*(5 + n))
________________________________________________________________________________________
Rubi [A] time = 0.0518165, antiderivative size = 59, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used =
{2564, 270} \[ \frac{\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac{2 \sinh ^{n+3}(a+b x)}{b (n+3)}+\frac{\sinh ^{n+5}(a+b x)}{b (n+5)} \]
Antiderivative was successfully verified.
[In]
Int[Cosh[a + b*x]^5*Sinh[a + b*x]^n,x]
[Out]
Sinh[a + b*x]^(1 + n)/(b*(1 + n)) + (2*Sinh[a + b*x]^(3 + n))/(b*(3 + n)) + Sinh[a + b*x]^(5 + n)/(b*(5 + n))
Rule 2564
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])
Rule 270
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Rubi steps
\begin{align*} \int \cosh ^5(a+b x) \sinh ^n(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^n \left (1+x^2\right )^2 \, dx,x,\sinh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^n+2 x^{2+n}+x^{4+n}\right ) \, dx,x,\sinh (a+b x)\right )}{b}\\ &=\frac{\sinh ^{1+n}(a+b x)}{b (1+n)}+\frac{2 \sinh ^{3+n}(a+b x)}{b (3+n)}+\frac{\sinh ^{5+n}(a+b x)}{b (5+n)}\\ \end{align*}
Mathematica [A] time = 0.195388, size = 49, normalized size = 0.83 \[ \frac{\sinh ^{n+1}(a+b x) \left (\frac{\sinh ^4(a+b x)}{n+5}+\frac{2 \sinh ^2(a+b x)}{n+3}+\frac{1}{n+1}\right )}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cosh[a + b*x]^5*Sinh[a + b*x]^n,x]
[Out]
(Sinh[a + b*x]^(1 + n)*((1 + n)^(-1) + (2*Sinh[a + b*x]^2)/(3 + n) + Sinh[a + b*x]^4/(5 + n)))/b
________________________________________________________________________________________
Maple [F] time = 0.478, size = 0, normalized size = 0. \begin{align*} \int \left ( \cosh \left ( bx+a \right ) \right ) ^{5} \left ( \sinh \left ( bx+a \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cosh(b*x+a)^5*sinh(b*x+a)^n,x)
[Out]
int(cosh(b*x+a)^5*sinh(b*x+a)^n,x)
________________________________________________________________________________________
Maxima [B] time = 1.79, size = 926, normalized size = 15.69 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cosh(b*x+a)^5*sinh(b*x+a)^n,x, algorithm="maxima")
[Out]
1/32*n^2*e^((b*x + a)*n + 5*b*x + n*log(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1) + 5*a)/((2^n*n^3 + 9*2^n*
n^2 + 23*2^n*n + 15*2^n)*b) + 1/8*n*e^((b*x + a)*n + 5*b*x + n*log(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1
) + 5*a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2^n)*b) + 1/32*(3*n^2 + 28*n + 25)*e^((b*x + a)*n + 3*b*x + n*l
og(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1) + 3*a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2^n)*b) + 1/16*(n
^2 + 12*n + 75)*e^((b*x + a)*n + b*x + n*log(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1) + a)/((2^n*n^3 + 9*2
^n*n^2 + 23*2^n*n + 15*2^n)*b) - 1/16*(n^2 + 12*n + 75)*e^((b*x + a)*n - b*x + n*log(e^(-b*x - a) + 1) + n*log
(-e^(-b*x - a) + 1) - a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2^n)*b) - 1/32*(3*n^2 + 28*n + 25)*e^((b*x + a)
*n - 3*b*x + n*log(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1) - 3*a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2
^n)*b) - 1/32*(n^2 + 4*n + 3)*e^((b*x + a)*n - 5*b*x + n*log(e^(-b*x - a) + 1) + n*log(-e^(-b*x - a) + 1) - 5*
a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2^n)*b) + 3/32*e^((b*x + a)*n + 5*b*x + n*log(e^(-b*x - a) + 1) + n*l
og(-e^(-b*x - a) + 1) + 5*a)/((2^n*n^3 + 9*2^n*n^2 + 23*2^n*n + 15*2^n)*b)
________________________________________________________________________________________
Fricas [B] time = 2.33701, size = 1015, normalized size = 17.2 \begin{align*} \frac{{\left ({\left (n^{2} + 4 \, n + 3\right )} \sinh \left (b x + a\right )^{5} +{\left (10 \,{\left (n^{2} + 4 \, n + 3\right )} \cosh \left (b x + a\right )^{2} + 3 \, n^{2} + 28 \, n + 25\right )} \sinh \left (b x + a\right )^{3} +{\left (5 \,{\left (n^{2} + 4 \, n + 3\right )} \cosh \left (b x + a\right )^{4} + 3 \,{\left (3 \, n^{2} + 28 \, n + 25\right )} \cosh \left (b x + a\right )^{2} + 2 \, n^{2} + 24 \, n + 150\right )} \sinh \left (b x + a\right )\right )} \cosh \left (n \log \left (\sinh \left (b x + a\right )\right )\right ) +{\left ({\left (n^{2} + 4 \, n + 3\right )} \sinh \left (b x + a\right )^{5} +{\left (10 \,{\left (n^{2} + 4 \, n + 3\right )} \cosh \left (b x + a\right )^{2} + 3 \, n^{2} + 28 \, n + 25\right )} \sinh \left (b x + a\right )^{3} +{\left (5 \,{\left (n^{2} + 4 \, n + 3\right )} \cosh \left (b x + a\right )^{4} + 3 \,{\left (3 \, n^{2} + 28 \, n + 25\right )} \cosh \left (b x + a\right )^{2} + 2 \, n^{2} + 24 \, n + 150\right )} \sinh \left (b x + a\right )\right )} \sinh \left (n \log \left (\sinh \left (b x + a\right )\right )\right )}{16 \,{\left ({\left (b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b\right )} \cosh \left (b x + a\right )^{6} - 3 \,{\left (b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b\right )} \cosh \left (b x + a\right )^{4} \sinh \left (b x + a\right )^{2} + 3 \,{\left (b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} -{\left (b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b\right )} \sinh \left (b x + a\right )^{6}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cosh(b*x+a)^5*sinh(b*x+a)^n,x, algorithm="fricas")
[Out]
1/16*(((n^2 + 4*n + 3)*sinh(b*x + a)^5 + (10*(n^2 + 4*n + 3)*cosh(b*x + a)^2 + 3*n^2 + 28*n + 25)*sinh(b*x + a
)^3 + (5*(n^2 + 4*n + 3)*cosh(b*x + a)^4 + 3*(3*n^2 + 28*n + 25)*cosh(b*x + a)^2 + 2*n^2 + 24*n + 150)*sinh(b*
x + a))*cosh(n*log(sinh(b*x + a))) + ((n^2 + 4*n + 3)*sinh(b*x + a)^5 + (10*(n^2 + 4*n + 3)*cosh(b*x + a)^2 +
3*n^2 + 28*n + 25)*sinh(b*x + a)^3 + (5*(n^2 + 4*n + 3)*cosh(b*x + a)^4 + 3*(3*n^2 + 28*n + 25)*cosh(b*x + a)^
2 + 2*n^2 + 24*n + 150)*sinh(b*x + a))*sinh(n*log(sinh(b*x + a))))/((b*n^3 + 9*b*n^2 + 23*b*n + 15*b)*cosh(b*x
+ a)^6 - 3*(b*n^3 + 9*b*n^2 + 23*b*n + 15*b)*cosh(b*x + a)^4*sinh(b*x + a)^2 + 3*(b*n^3 + 9*b*n^2 + 23*b*n +
15*b)*cosh(b*x + a)^2*sinh(b*x + a)^4 - (b*n^3 + 9*b*n^2 + 23*b*n + 15*b)*sinh(b*x + a)^6)
________________________________________________________________________________________
Sympy [A] time = 80.1033, size = 2574, normalized size = 43.63 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cosh(b*x+a)**5*sinh(b*x+a)**n,x)
[Out]
Piecewise((x*sinh(a)**n*cosh(a)**5, Eq(b, 0)), (log(sinh(a + b*x))/b - cosh(a + b*x)**2/(2*b*sinh(a + b*x)**2)
- cosh(a + b*x)**4/(4*b*sinh(a + b*x)**4), Eq(n, -5)), (16*b*x*tanh(a/2 + b*x/2)**6/(8*b*tanh(a/2 + b*x/2)**6
- 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) - 32*b*x*tanh(a/2 + b*x/2)**4/(8*b*tanh(a/2 + b*x/2)*
*6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) + 16*b*x*tanh(a/2 + b*x/2)**2/(8*b*tanh(a/2 + b*x/2
)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) - 32*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)
**6/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) + 64*log(tanh(a/2 + b*x/
2) + 1)*tanh(a/2 + b*x/2)**4/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2)
- 32*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 +
8*b*tanh(a/2 + b*x/2)**2) + 16*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x/2)**6/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*ta
nh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) - 32*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x/2)**4/(8*b*tanh(a/2
+ b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) + 16*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x
/2)**2/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) - tanh(a/2 + b*x/2)**
8/(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) + 18*tanh(a/2 + b*x/2)**4/
(8*b*tanh(a/2 + b*x/2)**6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2) - 1/(8*b*tanh(a/2 + b*x/2)**
6 - 16*b*tanh(a/2 + b*x/2)**4 + 8*b*tanh(a/2 + b*x/2)**2), Eq(n, -3)), (b*x*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 +
b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*b*x*tanh(
a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 +
b*x/2)**2 + b) + 6*b*x*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2
+ b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*b*x*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a
/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + b*x/(b*tanh(a/2 + b*x/2)**8 - 4*b*
tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tanh(a/2 + b*x/2) + 1)
*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh
(a/2 + b*x/2)**2 + b) + 8*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a
/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 12*log(tanh(a/2 + b*x/2) + 1)*tanh
(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2
+ b*x/2)**2 + b) + 8*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 +
b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 2*log(tanh(a/2 + b*x/2) + 1)/(b*tanh(a/
2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + log(tanh
(a/2 + b*x/2))*tanh(a/2 + b*x/2)**8/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)
**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 -
4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 6*log(tanh(a/2 + b*x/2))
*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh
(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2))*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 +
b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + log(tanh(a/2 + b*x/2))/(b*tanh(a/2 + b
*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 4*tanh(a/2 +
b*x/2)**6/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2
)**2 + b) - 4*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6 + 6*b*tanh(a/2 + b*x/2)*
*4 - 4*b*tanh(a/2 + b*x/2)**2 + b) + 4*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**8 - 4*b*tanh(a/2 + b*x/2)**6
+ 6*b*tanh(a/2 + b*x/2)**4 - 4*b*tanh(a/2 + b*x/2)**2 + b), Eq(n, -1)), (n**2*sinh(a + b*x)*sinh(a + b*x)**n*
cosh(a + b*x)**4/(b*n**3 + 9*b*n**2 + 23*b*n + 15*b) - 4*n*sinh(a + b*x)**3*sinh(a + b*x)**n*cosh(a + b*x)**2/
(b*n**3 + 9*b*n**2 + 23*b*n + 15*b) + 8*n*sinh(a + b*x)*sinh(a + b*x)**n*cosh(a + b*x)**4/(b*n**3 + 9*b*n**2 +
23*b*n + 15*b) + 8*sinh(a + b*x)**5*sinh(a + b*x)**n/(b*n**3 + 9*b*n**2 + 23*b*n + 15*b) - 20*sinh(a + b*x)**
3*sinh(a + b*x)**n*cosh(a + b*x)**2/(b*n**3 + 9*b*n**2 + 23*b*n + 15*b) + 15*sinh(a + b*x)*sinh(a + b*x)**n*co
sh(a + b*x)**4/(b*n**3 + 9*b*n**2 + 23*b*n + 15*b), True))
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (b x + a\right )^{n} \cosh \left (b x + a\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cosh(b*x+a)^5*sinh(b*x+a)^n,x, algorithm="giac")
[Out]
integrate(sinh(b*x + a)^n*cosh(b*x + a)^5, x)