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

Optimal. Leaf size=41 \[ \frac {\sinh ^5(a+b x)}{5 b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2633} \[ \frac {\sinh ^5(a+b x)}{5 b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[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 2633

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*} \int \cosh ^5(a+b x) \, dx &=\frac {i \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \[ \frac {\sinh ^5(a+b x)}{5 b}+\frac {2 \sinh ^3(a+b x)}{3 b}+\frac {\sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.47, size = 66, normalized size = 1.61 \[ \frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.14, size = 82, normalized size = 2.00 \[ \frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.22, size = 33, normalized size = 0.80 \[ \frac {\left (\frac {8}{15}+\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{5}+\frac {4 \left (\cosh ^{2}\left (b x +a \right )\right )}{15}\right ) \sinh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 82, normalized size = 2.00 \[ \frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac {5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac {5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac {5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.92, size = 31, normalized size = 0.76 \[ \frac {\frac {{\mathrm {sinh}\left (a+b\,x\right )}^5}{5}+\frac {2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{3}+\mathrm {sinh}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 1.57, size = 58, normalized size = 1.41 \[ \begin {cases} \frac {8 \sinh ^{5}{\left (a + b x \right )}}{15 b} - \frac {4 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac {\sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \cosh ^{5}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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