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]

1/4*sinh(b*x+a)^4/b+1/6*sinh(b*x+a)^6/b

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 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]]

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 ^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*}

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Mathematica [A]  time = 0.01, 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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fricas [B]  time = 0.67, size = 72, normalized size = 2.32 \[ \frac {\cosh \left (b x + a\right )^{6} + 15 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 3\right )} \sinh \left (b x + a\right )^{2} - 9 \, \cosh \left (b x + a\right )^{2}}{192 \, b} \]

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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giac [B]  time = 0.15, size = 57, normalized size = 1.84 \[ \frac {e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} - \frac {3 \, e^{\left (2 \, b x + 2 \, a\right )}}{128 \, b} - \frac {3 \, e^{\left (-2 \, b x - 2 \, a\right )}}{128 \, b} + \frac {e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} \]

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^(6*b*x + 6*a)/b - 3/128*e^(2*b*x + 2*a)/b - 3/128*e^(-2*b*x - 2*a)/b + 1/384*e^(-6*b*x - 6*a)/b

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maple [A]  time = 0.06, size = 34, normalized size = 1.10 \[ \frac {\frac {\left (\cosh ^{4}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{6}-\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{12}}{b} \]

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)^4)

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maxima [B]  time = 0.34, size = 56, normalized size = 1.81 \[ -\frac {{\left (9 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} - \frac {9 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} \]

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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mupad [B]  time = 1.52, size = 26, normalized size = 0.84 \[ \frac {2\,{\mathrm {sinh}\left (a+b\,x\right )}^6+3\,{\mathrm {sinh}\left (a+b\,x\right )}^4}{12\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*sinh(a + b*x)^4 + 2*sinh(a + b*x)^6)/(12*b)

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sympy [A]  time = 2.59, size = 42, normalized size = 1.35 \[ \begin {cases} \frac {\sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{4 b} - \frac {\cosh ^{6}{\left (a + b x \right )}}{12 b} & \text {for}\: b \neq 0 \\x \sinh ^{3}{\relax (a )} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

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

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