### 3.329 $$\int \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=31 $\frac{\sinh ^6(a+b x)}{6 b}+\frac{\sinh ^4(a+b x)}{4 b}$

[Out]

Sinh[a + b*x]^4/(4*b) + Sinh[a + b*x]^6/(6*b)

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Rubi [A]  time = 0.036688, 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} $\frac{\sinh ^6(a+b x)}{6 b}+\frac{\sinh ^4(a+b x)}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sinh[a + b*x]^4/(4*b) + Sinh[a + b*x]^6/(6*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 ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,i \sinh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,i \sinh (a+b x)\right )}{b}\\ &=\frac{\sinh ^4(a+b x)}{4 b}+\frac{\sinh ^6(a+b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0121394, size = 35, normalized size = 1.13 $\frac{1}{8} \left (\frac{\cosh (6 (a+b x))}{24 b}-\frac{3 \cosh (2 (a+b x))}{8 b}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-3*Cosh[2*(a + b*x)])/(8*b) + Cosh[6*(a + b*x)]/(24*b))/8

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Maple [A]  time = 0.007, size = 52, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{6}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{12}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{12}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/6*cosh(b*x+a)^4*sinh(b*x+a)^2-1/12*cosh(b*x+a)^2*sinh(b*x+a)^2-1/12*cosh(b*x+a)^2)

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Maxima [B]  time = 1.01174, size = 76, normalized size = 2.45 \begin{align*} -\frac{{\left (9 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} - \frac{9 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(9*e^(-4*b*x - 4*a) - 1)*e^(6*b*x + 6*a)/b - 1/384*(9*e^(-2*b*x - 2*a) - e^(-6*b*x - 6*a))/b

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Fricas [B]  time = 1.75977, size = 197, normalized size = 6.35 \begin{align*} \frac{\cosh \left (b x + a\right )^{6} + 15 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{6} + 3 \,{\left (5 \, \cosh \left (b x + a\right )^{4} - 3\right )} \sinh \left (b x + a\right )^{2} - 9 \, \cosh \left (b x + a\right )^{2}}{192 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(cosh(b*x + a)^6 + 15*cosh(b*x + a)^2*sinh(b*x + a)^4 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^4 - 3)*sinh
(b*x + a)^2 - 9*cosh(b*x + a)^2)/b

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Sympy [A]  time = 3.74648, size = 42, normalized size = 1.35 \begin{align*} \begin{cases} - \frac{\sinh ^{6}{\left (a + b x \right )}}{12 b} + \frac{\sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x \sinh ^{3}{\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)**3,x)

[Out]

Piecewise((-sinh(a + b*x)**6/(12*b) + sinh(a + b*x)**4*cosh(a + b*x)**2/(4*b), Ne(b, 0)), (x*sinh(a)**3*cosh(a
)**3, True))

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Giac [A]  time = 1.16595, size = 66, normalized size = 2.13 \begin{align*} \frac{{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{3} - 12 \, e^{\left (2 \, b x + 2 \, a\right )} - 12 \, e^{\left (-2 \, b x - 2 \, a\right )}}{384 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*((e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))^3 - 12*e^(2*b*x + 2*a) - 12*e^(-2*b*x - 2*a))/b