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]

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

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 30

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

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])

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*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {\sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sinh[a + b*x]^3/(3*b)

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fricas [B]  time = 0.58, size = 32, normalized size = 2.13 \[ \frac {\sinh \left (b x + a\right )^{3} + 3 \, {\left (\cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )}{12 \, b} \]

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

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giac [B]  time = 0.14, size = 54, normalized size = 3.60 \[ \frac {e^{\left (3 \, b x + 3 \, a\right )}}{24 \, b} - \frac {e^{\left (b x + a\right )}}{8 \, b} + \frac {e^{\left (-b x - a\right )}}{8 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} \]

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*e^(3*b*x + 3*a)/b - 1/8*e^(b*x + a)/b + 1/8*e^(-b*x - a)/b - 1/24*e^(-3*b*x - 3*a)/b

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maple [A]  time = 0.02, size = 14, normalized size = 0.93 \[ \frac {\sinh ^{3}\left (b x +a \right )}{3 b} \]

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

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maxima [A]  time = 0.33, size = 13, normalized size = 0.87 \[ \frac {\sinh \left (b x + a\right )^{3}}{3 \, b} \]

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

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mupad [B]  time = 1.64, size = 13, normalized size = 0.87 \[ \frac {{\mathrm {sinh}\left (a+b\,x\right )}^3}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*sinh(a + b*x)^2,x)

[Out]

sinh(a + b*x)^3/(3*b)

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sympy [A]  time = 0.37, size = 20, normalized size = 1.33 \[ \begin {cases} \frac {\sinh ^{3}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sinh ^{2}{\relax (a )} \cosh {\relax (a )} & \text {otherwise} \end {cases} \]

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))

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