3.272 \(\int \cosh ^3(a+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=15 \[ \frac {\cosh ^4(a+b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 = {2565, 30} \[ \frac {\cosh ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^3*Sinh[a + b*x],x]

[Out]

Cosh[a + b*x]^4/(4*b)

Rule 30

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

Rule 2565

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {\cosh ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cosh[a + b*x]^4/(4*b)

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fricas [B]  time = 1.02, size = 54, normalized size = 3.60 \[ \frac {\cosh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right )^{2}}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a),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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giac [B]  time = 0.14, size = 57, normalized size = 3.80 \[ \frac {e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} + \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*e^(4*b*x + 4*a)/b + 1/16*e^(2*b*x + 2*a)/b + 1/16*e^(-2*b*x - 2*a)/b + 1/64*e^(-4*b*x - 4*a)/b

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maple [A]  time = 0.02, size = 14, normalized size = 0.93 \[ \frac {\cosh ^{4}\left (b x +a \right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 13, normalized size = 0.87 \[ \frac {\cosh \left (b x + a\right )^{4}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*cosh(b*x + a)^4/b

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mupad [B]  time = 1.49, size = 13, normalized size = 0.87 \[ \frac {{\mathrm {cosh}\left (a+b\,x\right )}^4}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*sinh(a + b*x),x)

[Out]

cosh(a + b*x)^4/(4*b)

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sympy [A]  time = 0.74, size = 20, normalized size = 1.33 \[ \begin {cases} \frac {\cosh ^{4}{\left (a + b x \right )}}{4 b} & \text {for}\: b \neq 0 \\x \sinh {\relax (a )} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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