### 3.320 $$\int \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=31 $\frac{\cosh ^5(a+b x)}{5 b}-\frac{\cosh ^3(a+b x)}{3 b}$

[Out]

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

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Rubi [A]  time = 0.0348252, 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 = {2565, 14} $\frac{\cosh ^5(a+b x)}{5 b}-\frac{\cosh ^3(a+b x)}{3 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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 ^2(a+b x) \sinh ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac{\cosh ^3(a+b x)}{3 b}+\frac{\cosh ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.069528, size = 27, normalized size = 0.87 $\frac{\cosh ^3(a+b x) (3 \cosh (2 (a+b x))-7)}{30 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 48, normalized size = 1.6 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{5}}-{\frac{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{2\,\cosh \left ( bx+a \right ) }{15}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02092, 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)^2*sinh(b*x+a)^3,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 = 1.884, size = 216, normalized size = 6.97 \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} - 5 \, \cosh \left (b x + a\right )^{3} + 15 \,{\left (2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 30 \, \cosh \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)^2*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/240*(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 - 5*cosh(b*x + a)^3 + 15*(2*cosh(b*x + a)^3 - cosh
(b*x + a))*sinh(b*x + a)^2 - 30*cosh(b*x + a))/b

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Sympy [A]  time = 2.12358, size = 44, normalized size = 1.42 \begin{align*} \begin{cases} \frac{\sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac{2 \cosh ^{5}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \sinh ^{3}{\left (a \right )} \cosh ^{2}{\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)**2*sinh(b*x+a)**3,x)

[Out]

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

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Giac [B]  time = 1.15514, 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)^2*sinh(b*x+a)^3,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