3.16 \(\int \cosh ^2(a+b x) \sinh ^4(a+b x) \, dx\)

Optimal. Leaf size=69 \[ \frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{8 b}+\frac {\sinh (a+b x) \cosh (a+b x)}{16 b}+\frac {x}{16} \]

[Out]

1/16*x+1/16*cosh(b*x+a)*sinh(b*x+a)/b-1/8*cosh(b*x+a)^3*sinh(b*x+a)/b+1/6*cosh(b*x+a)^3*sinh(b*x+a)^3/b

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2568, 2635, 8} \[ \frac {\sinh ^3(a+b x) \cosh ^3(a+b x)}{6 b}-\frac {\sinh (a+b x) \cosh ^3(a+b x)}{8 b}+\frac {\sinh (a+b x) \cosh (a+b x)}{16 b}+\frac {x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^2*Sinh[a + b*x]^4,x]

[Out]

x/16 + (Cosh[a + b*x]*Sinh[a + b*x])/(16*b) - (Cosh[a + b*x]^3*Sinh[a + b*x])/(8*b) + (Cosh[a + b*x]^3*Sinh[a
+ b*x]^3)/(6*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

\begin {align*} \int \cosh ^2(a+b x) \sinh ^4(a+b x) \, dx &=\frac {\cosh ^3(a+b x) \sinh ^3(a+b x)}{6 b}-\frac {1}{2} \int \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx\\ &=-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh ^3(a+b x)}{6 b}+\frac {1}{8} \int \cosh ^2(a+b x) \, dx\\ &=\frac {\cosh (a+b x) \sinh (a+b x)}{16 b}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh ^3(a+b x)}{6 b}+\frac {\int 1 \, dx}{16}\\ &=\frac {x}{16}+\frac {\cosh (a+b x) \sinh (a+b x)}{16 b}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh ^3(a+b x)}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 40, normalized size = 0.58 \[ \frac {-3 \sinh (2 (a+b x))-3 \sinh (4 (a+b x))+\sinh (6 (a+b x))+12 b x}{192 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^2*Sinh[a + b*x]^4,x]

[Out]

(12*b*x - 3*Sinh[2*(a + b*x)] - 3*Sinh[4*(a + b*x)] + Sinh[6*(a + b*x)])/(192*b)

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fricas [A]  time = 0.39, size = 90, normalized size = 1.30 \[ \frac {3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 6 \, b x + 3 \, {\left (\cosh \left (b x + a\right )^{5} - 2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^4,x, algorithm="fricas")

[Out]

1/96*(3*cosh(b*x + a)*sinh(b*x + a)^5 + 2*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 6*b*x + 3*(c
osh(b*x + a)^5 - 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a))/b

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giac [A]  time = 0.20, size = 88, normalized size = 1.28 \[ \frac {1}{16} \, x + \frac {e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} - \frac {e^{\left (4 \, b x + 4 \, a\right )}}{128 \, b} - \frac {e^{\left (2 \, b x + 2 \, a\right )}}{128 \, b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{128 \, b} + \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{128 \, b} - \frac {e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^4,x, algorithm="giac")

[Out]

1/16*x + 1/384*e^(6*b*x + 6*a)/b - 1/128*e^(4*b*x + 4*a)/b - 1/128*e^(2*b*x + 2*a)/b + 1/128*e^(-2*b*x - 2*a)/
b + 1/128*e^(-4*b*x - 4*a)/b - 1/384*e^(-6*b*x - 6*a)/b

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maple [A]  time = 0.09, size = 61, normalized size = 0.88 \[ \frac {\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{6}-\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{8}+\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{16}+\frac {b x}{16}+\frac {a}{16}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^2*sinh(b*x+a)^4,x)

[Out]

1/b*(1/6*cosh(b*x+a)^3*sinh(b*x+a)^3-1/8*cosh(b*x+a)^3*sinh(b*x+a)+1/16*cosh(b*x+a)*sinh(b*x+a)+1/16*b*x+1/16*
a)

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maxima [A]  time = 0.34, size = 88, normalized size = 1.28 \[ -\frac {{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{384 \, b} + \frac {b x + a}{16 \, b} + \frac {3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (-6 \, b x - 6 \, a\right )}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/384*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4*a) - 1)*e^(6*b*x + 6*a)/b + 1/16*(b*x + a)/b + 1/384*(3*e^(-2*b*x
- 2*a) + 3*e^(-4*b*x - 4*a) - e^(-6*b*x - 6*a))/b

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mupad [B]  time = 1.64, size = 43, normalized size = 0.62 \[ \frac {x}{16}-\frac {\frac {\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{64}+\frac {\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{64}-\frac {\mathrm {sinh}\left (6\,a+6\,b\,x\right )}{192}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^2*sinh(a + b*x)^4,x)

[Out]

x/16 - (sinh(2*a + 2*b*x)/64 + sinh(4*a + 4*b*x)/64 - sinh(6*a + 6*b*x)/192)/b

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sympy [A]  time = 2.91, size = 136, normalized size = 1.97 \[ \begin {cases} - \frac {x \sinh ^{6}{\left (a + b x \right )}}{16} + \frac {3 x \sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16} - \frac {3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{16} + \frac {x \cosh ^{6}{\left (a + b x \right )}}{16} + \frac {\sinh ^{5}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{16 b} + \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{6 b} - \frac {\sinh {\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{16 b} & \text {for}\: b \neq 0 \\x \sinh ^{4}{\relax (a )} \cosh ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2*sinh(b*x+a)**4,x)

[Out]

Piecewise((-x*sinh(a + b*x)**6/16 + 3*x*sinh(a + b*x)**4*cosh(a + b*x)**2/16 - 3*x*sinh(a + b*x)**2*cosh(a + b
*x)**4/16 + x*cosh(a + b*x)**6/16 + sinh(a + b*x)**5*cosh(a + b*x)/(16*b) + sinh(a + b*x)**3*cosh(a + b*x)**3/
(6*b) - sinh(a + b*x)*cosh(a + b*x)**5/(16*b), Ne(b, 0)), (x*sinh(a)**4*cosh(a)**2, True))

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