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

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

[Out]

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

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Rubi [A]  time = 0.0169643, antiderivative size = 41, normalized size of antiderivative = 1., 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} \[ \frac{\cosh ^5(a+b x)}{5 b}-\frac{2 \cosh ^3(a+b x)}{3 b}+\frac{\cosh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^5,x]

[Out]

Cosh[a + b*x]/b - (2*Cosh[a + b*x]^3)/(3*b) + Cosh[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 \sinh ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\cosh (a+b x)}{b}-\frac{2 \cosh ^3(a+b x)}{3 b}+\frac{\cosh ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0140689, size = 44, normalized size = 1.07 \[ \frac{5 \cosh (a+b x)}{8 b}-\frac{5 \cosh (3 (a+b x))}{48 b}+\frac{\cosh (5 (a+b x))}{80 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^5,x]

[Out]

(5*Cosh[a + b*x])/(8*b) - (5*Cosh[3*(a + b*x)])/(48*b) + Cosh[5*(a + b*x)]/(80*b)

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Maple [A]  time = 0.039, size = 33, normalized size = 0.8 \begin{align*}{\frac{\cosh \left ( bx+a \right ) }{b} \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.10583, size = 111, normalized size = 2.71 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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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Fricas [B]  time = 2.22009, size = 221, normalized size = 5.39 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{5} + 15 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 25 \, \cosh \left (b x + a\right )^{3} + 15 \,{\left (2 \, \cosh \left (b x + a\right )^{3} - 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 150 \, \cosh \left (b x + a\right )}{240 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 - 25*cosh(b*x + a)^3 + 15*(2*cosh(b*x + a)^3 - 5*c
osh(b*x + a))*sinh(b*x + a)^2 + 150*cosh(b*x + a))/b

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Sympy [A]  time = 2.34914, size = 58, normalized size = 1.41 \begin{align*} \begin{cases} \frac{\sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{b} - \frac{4 \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} + \frac{8 \cosh ^{5}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \sinh ^{5}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.36248, size = 95, normalized size = 2.32 \begin{align*} \frac{{\left (150 \, e^{\left (4 \, b x + 4 \, a\right )} - 25 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )} e^{\left (-5 \, b x - 5 \, a\right )} + 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 25 \, e^{\left (3 \, b x + 3 \, a\right )} + 150 \, e^{\left (b x + a\right )}}{480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*((150*e^(4*b*x + 4*a) - 25*e^(2*b*x + 2*a) + 3)*e^(-5*b*x - 5*a) + 3*e^(5*b*x + 5*a) - 25*e^(3*b*x + 3*a
) + 150*e^(b*x + a))/b