\(\int \cosh ^5(a+b x) \, dx\) [5]

Optimal result

Integrand size = 8, antiderivative size = 41 \[ \int \cosh ^5(a+b x) \, dx=\frac {\sinh (a+b x)}{b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh ^5(a+b x)}{5 b} \]

[Out]

sinh(b*x+a)/b+2/3*sinh(b*x+a)^3/b+1/5*sinh(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2713} \[ \int \cosh ^5(a+b x) \, dx=\frac {\sinh ^5(a+b x)}{5 b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh (a+b x)}{b} \]

[In]

Int[Cosh[a + b*x]^5,x]

[Out]

Sinh[a + b*x]/b + (2*Sinh[a + b*x]^3)/(3*b) + Sinh[a + b*x]^5/(5*b)

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {i \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-i \sinh (a+b x)\right )}{b} \\ & = \frac {\sinh (a+b x)}{b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \cosh ^5(a+b x) \, dx=\frac {\sinh (a+b x)}{b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh ^5(a+b x)}{5 b} \]

[In]

Integrate[Cosh[a + b*x]^5,x]

[Out]

Sinh[a + b*x]/b + (2*Sinh[a + b*x]^3)/(3*b) + Sinh[a + b*x]^5/(5*b)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\left (\frac {8}{15}+\frac {\cosh \left (b x +a \right )^{4}}{5}+\frac {4 \cosh \left (b x +a \right )^{2}}{15}\right ) \sinh \left (b x +a \right )}{b}\)	\(33\)
default	\(\frac {\left (\frac {8}{15}+\frac {\cosh \left (b x +a \right )^{4}}{5}+\frac {4 \cosh \left (b x +a \right )^{2}}{15}\right ) \sinh \left (b x +a \right )}{b}\)	\(33\)
parallelrisch	\(\frac {25 \sinh \left (3 b x +3 a \right )+150 \sinh \left (b x +a \right )+3 \sinh \left (5 b x +5 a \right )}{240 b}\)	\(37\)
risch	\(\frac {{\mathrm e}^{5 b x +5 a}}{160 b}+\frac {5 \,{\mathrm e}^{3 b x +3 a}}{96 b}+\frac {5 \,{\mathrm e}^{b x +a}}{16 b}-\frac {5 \,{\mathrm e}^{-b x -a}}{16 b}-\frac {5 \,{\mathrm e}^{-3 b x -3 a}}{96 b}-\frac {{\mathrm e}^{-5 b x -5 a}}{160 b}\)	\(83\)

[In]

int(cosh(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*(8/15+1/5*cosh(b*x+a)^4+4/15*cosh(b*x+a)^2)*sinh(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.61 \[ \int \cosh ^5(a+b x) \, dx=\frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \]

[In]

integrate(cosh(b*x+a)^5,x, algorithm="fricas")

[Out]

1/240*(3*sinh(b*x + a)^5 + 5*(6*cosh(b*x + a)^2 + 5)*sinh(b*x + a)^3 + 15*(cosh(b*x + a)^4 + 5*cosh(b*x + a)^2
+ 10)*sinh(b*x + a))/b

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41 \[ \int \cosh ^5(a+b x) \, dx=\begin {cases} \frac {8 \sinh ^{5}{\left (a + b x \right )}}{15 b} - \frac {4 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac {\sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \cosh ^{5}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cosh(b*x+a)**5,x)

[Out]

Piecewise((8*sinh(a + b*x)**5/(15*b) - 4*sinh(a + b*x)**3*cosh(a + b*x)**2/(3*b) + sinh(a + b*x)*cosh(a + b*x)
**4/b, Ne(b, 0)), (x*cosh(a)**5, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (37) = 74\).

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.00 \[ \int \cosh ^5(a+b x) \, dx=\frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

[In]

integrate(cosh(b*x+a)^5,x, algorithm="maxima")

[Out]

1/160*e^(5*b*x + 5*a)/b + 5/96*e^(3*b*x + 3*a)/b + 5/16*e^(b*x + a)/b - 5/16*e^(-b*x - a)/b - 5/96*e^(-3*b*x -
3*a)/b - 1/160*e^(-5*b*x - 5*a)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (37) = 74\).

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.00 \[ \int \cosh ^5(a+b x) \, dx=\frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

[In]

integrate(cosh(b*x+a)^5,x, algorithm="giac")

[Out]

1/160*e^(5*b*x + 5*a)/b + 5/96*e^(3*b*x + 3*a)/b + 5/16*e^(b*x + a)/b - 5/16*e^(-b*x - a)/b - 5/96*e^(-3*b*x -
3*a)/b - 1/160*e^(-5*b*x - 5*a)/b

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \cosh ^5(a+b x) \, dx=\frac {\frac {{\mathrm {sinh}\left (a+b\,x\right )}^5}{5}+\frac {2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{3}+\mathrm {sinh}\left (a+b\,x\right )}{b} \]

[In]

int(cosh(a + b*x)^5,x)

[Out]

(sinh(a + b*x) + (2*sinh(a + b*x)^3)/3 + sinh(a + b*x)^5/5)/b