3.3.71 \(\int x \cosh ^3(a+b x) \sinh (a+b x) \, dx\) [271]

Optimal. Leaf size=65 \[ -\frac {3 x}{32 b}+\frac {x \cosh ^4(a+b x)}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{16 b^2} \]

[Out]

-3/32*x/b+1/4*x*cosh(b*x+a)^4/b-3/32*cosh(b*x+a)*sinh(b*x+a)/b^2-1/16*cosh(b*x+a)^3*sinh(b*x+a)/b^2

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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5481, 2715, 8} \begin {gather*} -\frac {\sinh (a+b x) \cosh ^3(a+b x)}{16 b^2}-\frac {3 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac {x \cosh ^4(a+b x)}{4 b}-\frac {3 x}{32 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(32*b) + (x*Cosh[a + b*x]^4)/(4*b) - (3*Cosh[a + b*x]*Sinh[a + b*x])/(32*b^2) - (Cosh[a + b*x]^3*Sinh[a
+ b*x])/(16*b^2)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 5481

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[x^(m - n
+ 1)*(Cosh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\frac {x \cosh ^4(a+b x)}{4 b}-\frac {\int \cosh ^4(a+b x) \, dx}{4 b}\\ &=\frac {x \cosh ^4(a+b x)}{4 b}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac {3 \int \cosh ^2(a+b x) \, dx}{16 b}\\ &=\frac {x \cosh ^4(a+b x)}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac {3 \int 1 \, dx}{32 b}\\ &=-\frac {3 x}{32 b}+\frac {x \cosh ^4(a+b x)}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 50, normalized size = 0.77 \begin {gather*} -\frac {-16 b x \cosh (2 (a+b x))-4 b x \cosh (4 (a+b x))+8 \sinh (2 (a+b x))+\sinh (4 (a+b x))}{128 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/128*(-16*b*x*Cosh[2*(a + b*x)] - 4*b*x*Cosh[4*(a + b*x)] + 8*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)])/b^2

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Maple [A]
time = 1.02, size = 96, normalized size = 1.48

method	result	size
risch	\(\frac {\left (4 b x -1\right ) {\mathrm e}^{4 b x +4 a}}{256 b^{2}}+\frac {\left (2 b x -1\right ) {\mathrm e}^{2 b x +2 a}}{32 b^{2}}+\frac {\left (2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{32 b^{2}}+\frac {\left (4 b x +1\right ) {\mathrm e}^{-4 b x -4 a}}{256 b^{2}}\)	\(82\)
default	\(\frac {\left (2 b x +2 a \right ) \cosh \left (2 b x +2 a \right )-\sinh \left (2 b x +2 a \right )-2 a \cosh \left (2 b x +2 a \right )}{16 b^{2}}+\frac {\left (4 b x +4 a \right ) \cosh \left (4 b x +4 a \right )-\sinh \left (4 b x +4 a \right )-4 a \cosh \left (4 b x +4 a \right )}{128 b^{2}}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/b^2*((2*b*x+2*a)*cosh(2*b*x+2*a)-sinh(2*b*x+2*a)-2*a*cosh(2*b*x+2*a))+1/128/b^2*((4*b*x+4*a)*cosh(4*b*x+4
*a)-sinh(4*b*x+4*a)-4*a*cosh(4*b*x+4*a))

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Maxima [A]
time = 0.27, size = 91, normalized size = 1.40 \begin {gather*} \frac {{\left (4 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{2}} + \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(4*b*x*e^(4*a) - e^(4*a))*e^(4*b*x)/b^2 + 1/32*(2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 + 1/32*(2*b*x + 1
)*e^(-2*b*x - 2*a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

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Fricas [A]
time = 0.39, size = 108, normalized size = 1.66 \begin {gather*} \frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + 4 \, b x \cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 2 \, {\left (3 \, b x \cosh \left (b x + a\right )^{2} + 2 \, b x\right )} \sinh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{32 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.27, size = 110, normalized size = 1.69 \begin {gather*} \begin {cases} - \frac {3 x \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac {5 x \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{32 b^{2}} - \frac {5 \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh {\left (a \right )} \cosh ^{3}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*x*sinh(a + b*x)**4/(32*b) + 3*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) + 5*x*cosh(a + b*x)**4/
(32*b) + 3*sinh(a + b*x)**3*cosh(a + b*x)/(32*b**2) - 5*sinh(a + b*x)*cosh(a + b*x)**3/(32*b**2), Ne(b, 0)), (
x**2*sinh(a)*cosh(a)**3/2, True))

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Giac [A]
time = 0.41, size = 81, normalized size = 1.25 \begin {gather*} \frac {{\left (4 \, b x - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(4*b*x - 1)*e^(4*b*x + 4*a)/b^2 + 1/32*(2*b*x - 1)*e^(2*b*x + 2*a)/b^2 + 1/32*(2*b*x + 1)*e^(-2*b*x - 2*
a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

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Mupad [B]
time = 0.12, size = 57, normalized size = 0.88 \begin {gather*} -\frac {\frac {3\,x}{32}-\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^4}{4}}{b}-\frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,\mathrm {sinh}\left (a+b\,x\right )}{16\,b^2}-\frac {3\,\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )}{32\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((3*x)/32 - (x*cosh(a + b*x)^4)/4)/b - (cosh(a + b*x)^3*sinh(a + b*x))/(16*b^2) - (3*cosh(a + b*x)*sinh(a +
b*x))/(32*b^2)

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