3.302 \(\int \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\sinh ^5(a+b x)}{5 b}+\frac{\sinh ^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0329313, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2564, 14} \[ \frac{\sinh ^5(a+b x)}{5 b}+\frac{\sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

Sinh[a + b*x]^3/(3*b) + Sinh[a + b*x]^5/(5*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0593704, size = 27, normalized size = 0.87 \[ \frac{\sinh ^3(a+b x) (3 \cosh (2 (a+b x))+7)}{30 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

((7 + 3*Cosh[2*(a + b*x)])*Sinh[a + b*x]^3)/(30*b)

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Maple [A]  time = 0.009, size = 42, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{5}}-{\frac{\sinh \left ( bx+a \right ) }{5} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*sinh(b*x+a)^2,x)

[Out]

1/b*(1/5*cosh(b*x+a)^4*sinh(b*x+a)-1/5*(2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a))

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Maxima [B]  time = 1.01285, size = 105, normalized size = 3.39 \begin{align*} \frac{{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 30 \, e^{\left (-4 \, b x - 4 \, a\right )} + 3\right )} e^{\left (5 \, b x + 5 \, a\right )}}{480 \, b} + \frac{30 \, e^{\left (-b x - a\right )} - 5 \, e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, e^{\left (-5 \, b x - 5 \, a\right )}}{480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

1/480*(5*e^(-2*b*x - 2*a) - 30*e^(-4*b*x - 4*a) + 3)*e^(5*b*x + 5*a)/b + 1/480*(30*e^(-b*x - a) - 5*e^(-3*b*x
- 3*a) - 3*e^(-5*b*x - 5*a))/b

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Fricas [B]  time = 1.79896, size = 178, normalized size = 5.74 \begin{align*} \frac{3 \, \sinh \left (b x + a\right )^{5} + 5 \,{\left (6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 15 \,{\left (\cosh \left (b x + a\right )^{4} + \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )}{240 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.11098, size = 44, normalized size = 1.42 \begin{align*} \begin{cases} - \frac{2 \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac{\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x \sinh ^{2}{\left (a \right )} \cosh ^{3}{\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)**3*sinh(b*x+a)**2,x)

[Out]

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

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Giac [B]  time = 1.15691, size = 95, normalized size = 3.06 \begin{align*} \frac{{\left (30 \, e^{\left (4 \, b x + 4 \, a\right )} - 5 \, 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 )} + 5 \, e^{\left (3 \, b x + 3 \, a\right )} - 30 \, e^{\left (b x + a\right )}}{480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

1/480*((30*e^(4*b*x + 4*a) - 5*e^(2*b*x + 2*a) - 3)*e^(-5*b*x - 5*a) + 3*e^(5*b*x + 5*a) + 5*e^(3*b*x + 3*a) -
30*e^(b*x + a))/b