3.284 \(\int \cosh (a+b x) \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ \frac{\sinh ^3(a+b x)}{3 b} \]

[Out]

Sinh[a + b*x]^3/(3*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0183197, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2564, 30} \[ \frac{\sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]*Sinh[a + b*x]^2,x]

[Out]

Sinh[a + b*x]^3/(3*b)

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac{i \operatorname{Subst}\left (\int x^2 \, dx,x,i \sinh (a+b x)\right )}{b}\\ &=\frac{\sinh ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0033657, size = 15, normalized size = 1. \[ \frac{\sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]*Sinh[a + b*x]^2,x]

[Out]

Sinh[a + b*x]^3/(3*b)

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 14, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{3\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.12684, size = 18, normalized size = 1.2 \begin{align*} \frac{\sinh \left (b x + a\right )^{3}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*sinh(b*x + a)^3/b

________________________________________________________________________________________

Fricas [B]  time = 1.72191, size = 89, normalized size = 5.93 \begin{align*} \frac{\sinh \left (b x + a\right )^{3} + 3 \,{\left (\cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/12*(sinh(b*x + a)^3 + 3*(cosh(b*x + a)^2 - 1)*sinh(b*x + a))/b

________________________________________________________________________________________

Sympy [A]  time = 0.508895, size = 20, normalized size = 1.33 \begin{align*} \begin{cases} \frac{\sinh ^{3}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x \sinh ^{2}{\left (a \right )} \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)**2,x)

[Out]

Piecewise((sinh(a + b*x)**3/(3*b), Ne(b, 0)), (x*sinh(a)**2*cosh(a), True))

________________________________________________________________________________________

Giac [B]  time = 1.15681, size = 62, normalized size = 4.13 \begin{align*} \frac{{\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

1/24*((3*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) + e^(3*b*x + 3*a) - 3*e^(b*x + a))/b