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]

1/3*sinh(b*x+a)^3/b+1/5*sinh(b*x+a)^5/b

________________________________________________________________________________________

Rubi [A]  time = 0.03, 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} \[ \frac {\sinh ^5(a+b x)}{5 b}+\frac {\sinh ^3(a+b x)}{3 b} \]

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 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 ^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 = 0.06, 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)

________________________________________________________________________________________

fricas [B]  time = 0.84, size = 64, normalized size = 2.06 \[ \frac {3 \, \sinh \left (b x + a\right )^{5} + 5 \, {\left (6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 15 \, {\left (\cosh \left (b x + a\right )^{4} + \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )}{240 \, b} \]

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

________________________________________________________________________________________

giac [B]  time = 0.13, size = 82, normalized size = 2.65 \[ \frac {e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac {e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} - \frac {e^{\left (b x + a\right )}}{16 \, b} + \frac {e^{\left (-b x - a\right )}}{16 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac {e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \]

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]

1/160*e^(5*b*x + 5*a)/b + 1/96*e^(3*b*x + 3*a)/b - 1/16*e^(b*x + a)/b + 1/16*e^(-b*x - a)/b - 1/96*e^(-3*b*x -
3*a)/b - 1/160*e^(-5*b*x - 5*a)/b

________________________________________________________________________________________

maple [A]  time = 0.33, size = 42, normalized size = 1.35 \[ \frac {\frac {\sinh \left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{5}-\frac {\left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{3}\right ) \sinh \left (b x +a \right )}{5}}{b} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.62, size = 78, normalized size = 2.52 \[ \frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 30 \, e^{\left (-4 \, b x - 4 \, a\right )} + 3\right )} e^{\left (5 \, b x + 5 \, a\right )}}{480 \, b} + \frac {30 \, e^{\left (-b x - a\right )} - 5 \, e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, e^{\left (-5 \, b x - 5 \, a\right )}}{480 \, b} \]

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

________________________________________________________________________________________

mupad [B]  time = 1.45, size = 26, normalized size = 0.84 \[ \frac {3\,{\mathrm {sinh}\left (a+b\,x\right )}^5+5\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{15\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*sinh(a + b*x)^3 + 3*sinh(a + b*x)^5)/(15*b)

________________________________________________________________________________________

sympy [A]  time = 1.46, size = 44, normalized size = 1.42 \[ \begin {cases} - \frac {2 \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sinh ^{2}{\relax (a )} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**3*sinh(b*x+a)**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))

________________________________________________________________________________________