### 3.311 $$\int \cosh (a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=15 $\frac{\sinh ^4(a+b x)}{4 b}$

[Out]

Sinh[a + b*x]^4/(4*b)

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Rubi [A]  time = 0.0207488, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {2564, 30} $\frac{\sinh ^4(a+b x)}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]*Sinh[a + b*x]^3,x]

[Out]

Sinh[a + b*x]^4/(4*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \cosh (a+b x) \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,i \sinh (a+b x)\right )}{b}\\ &=\frac{\sinh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.002444, size = 15, normalized size = 1. $\frac{\sinh ^4(a+b x)}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sinh[a + b*x]^4/(4*b)

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Maple [A]  time = 0.003, size = 14, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{4\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01369, size = 18, normalized size = 1.2 \begin{align*} \frac{\sinh \left (b x + a\right )^{4}}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sinh(b*x + a)^4/b

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Fricas [B]  time = 1.83843, size = 146, normalized size = 9.73 \begin{align*} \frac{\cosh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )^{2} - 4 \, \cosh \left (b x + a\right )^{2}}{32 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(cosh(b*x + a)^4 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2)/b

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Sympy [A]  time = 1.12259, size = 20, normalized size = 1.33 \begin{align*} \begin{cases} \frac{\sinh ^{4}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x \sinh ^{3}{\left (a \right )} \cosh{\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)*sinh(b*x+a)**3,x)

[Out]

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

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Giac [B]  time = 1.20192, size = 66, normalized size = 4.4 \begin{align*} \frac{{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} - 4 \, e^{\left (-2 \, b x - 2 \, a\right )}}{64 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*((e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))^2 - 4*e^(2*b*x + 2*a) - 4*e^(-2*b*x - 2*a))/b