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

Optimal. Leaf size=67 \[ \frac {3 \sinh (2 a+2 b x)}{128 b^2}-\frac {\sinh (6 a+6 b x)}{1152 b^2}-\frac {3 x \cosh (2 a+2 b x)}{64 b}+\frac {x \cosh (6 a+6 b x)}{192 b} \]

[Out]

-3/64*x*cosh(2*b*x+2*a)/b+1/192*x*cosh(6*b*x+6*a)/b+3/128*sinh(2*b*x+2*a)/b^2-1/1152*sinh(6*b*x+6*a)/b^2

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Rubi [A]  time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5448, 3296, 2637} \[ \frac {3 \sinh (2 a+2 b x)}{128 b^2}-\frac {\sinh (6 a+6 b x)}{1152 b^2}-\frac {3 x \cosh (2 a+2 b x)}{64 b}+\frac {x \cosh (6 a+6 b x)}{192 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Cosh[2*a + 2*b*x])/(64*b) + (x*Cosh[6*a + 6*b*x])/(192*b) + (3*Sinh[2*a + 2*b*x])/(128*b^2) - Sinh[6*a +
6*b*x]/(1152*b^2)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac {3}{32} x \sinh (2 a+2 b x)+\frac {1}{32} x \sinh (6 a+6 b x)\right ) \, dx\\ &=\frac {1}{32} \int x \sinh (6 a+6 b x) \, dx-\frac {3}{32} \int x \sinh (2 a+2 b x) \, dx\\ &=-\frac {3 x \cosh (2 a+2 b x)}{64 b}+\frac {x \cosh (6 a+6 b x)}{192 b}-\frac {\int \cosh (6 a+6 b x) \, dx}{192 b}+\frac {3 \int \cosh (2 a+2 b x) \, dx}{64 b}\\ &=-\frac {3 x \cosh (2 a+2 b x)}{64 b}+\frac {x \cosh (6 a+6 b x)}{192 b}+\frac {3 \sinh (2 a+2 b x)}{128 b^2}-\frac {\sinh (6 a+6 b x)}{1152 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 50, normalized size = 0.75 \[ -\frac {-27 \sinh (2 (a+b x))+\sinh (6 (a+b x))+54 b x \cosh (2 (a+b x))-6 b x \cosh (6 (a+b x))}{1152 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1152*(54*b*x*Cosh[2*(a + b*x)] - 6*b*x*Cosh[6*(a + b*x)] - 27*Sinh[2*(a + b*x)] + Sinh[6*(a + b*x)])/b^2

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fricas [B]  time = 0.65, size = 148, normalized size = 2.21 \[ \frac {3 \, b x \cosh \left (b x + a\right )^{6} + 45 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} + 3 \, b x \sinh \left (b x + a\right )^{6} - 10 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} - 27 \, b x \cosh \left (b x + a\right )^{2} + 9 \, {\left (5 \, b x \cosh \left (b x + a\right )^{4} - 3 \, b x\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{5} - 9 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{576 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/576*(3*b*x*cosh(b*x + a)^6 + 45*b*x*cosh(b*x + a)^2*sinh(b*x + a)^4 + 3*b*x*sinh(b*x + a)^6 - 10*cosh(b*x +
a)^3*sinh(b*x + a)^3 - 3*cosh(b*x + a)*sinh(b*x + a)^5 - 27*b*x*cosh(b*x + a)^2 + 9*(5*b*x*cosh(b*x + a)^4 - 3
*b*x)*sinh(b*x + a)^2 - 3*(cosh(b*x + a)^5 - 9*cosh(b*x + a))*sinh(b*x + a))/b^2

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giac [A]  time = 0.15, size = 81, normalized size = 1.21 \[ \frac {{\left (6 \, b x - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{2304 \, b^{2}} - \frac {3 \, {\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{256 \, b^{2}} - \frac {3 \, {\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{256 \, b^{2}} + \frac {{\left (6 \, b x + 1\right )} e^{\left (-6 \, b x - 6 \, a\right )}}{2304 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2304*(6*b*x - 1)*e^(6*b*x + 6*a)/b^2 - 3/256*(2*b*x - 1)*e^(2*b*x + 2*a)/b^2 - 3/256*(2*b*x + 1)*e^(-2*b*x -
2*a)/b^2 + 1/2304*(6*b*x + 1)*e^(-6*b*x - 6*a)/b^2

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maple [B]  time = 0.33, size = 129, normalized size = 1.93 \[ \frac {\frac {\left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{6}-\frac {\left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{12}-\frac {\sinh \left (b x +a \right ) \left (\cosh ^{5}\left (b x +a \right )\right )}{36}+\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{36}+\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{24}+\frac {b x}{24}+\frac {a}{24}-a \left (\frac {\left (\cosh ^{4}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{6}-\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{12}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/6*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^4-1/12*(b*x+a)*cosh(b*x+a)^4-1/36*sinh(b*x+a)*cosh(b*x+a)^5+1/36*
cosh(b*x+a)^3*sinh(b*x+a)+1/24*cosh(b*x+a)*sinh(b*x+a)+1/24*b*x+1/24*a-a*(1/6*cosh(b*x+a)^4*sinh(b*x+a)^2-1/12
*cosh(b*x+a)^4))

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maxima [A]  time = 0.44, size = 91, normalized size = 1.36 \[ \frac {{\left (6 \, b x e^{\left (6 \, a\right )} - e^{\left (6 \, a\right )}\right )} e^{\left (6 \, b x\right )}}{2304 \, b^{2}} - \frac {3 \, {\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{256 \, b^{2}} - \frac {3 \, {\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{256 \, b^{2}} + \frac {{\left (6 \, b x + 1\right )} e^{\left (-6 \, b x - 6 \, a\right )}}{2304 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2304*(6*b*x*e^(6*a) - e^(6*a))*e^(6*b*x)/b^2 - 3/256*(2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 - 3/256*(2*b*x
+ 1)*e^(-2*b*x - 2*a)/b^2 + 1/2304*(6*b*x + 1)*e^(-6*b*x - 6*a)/b^2

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mupad [B]  time = 0.23, size = 55, normalized size = 0.82 \[ -\frac {\frac {\mathrm {sinh}\left (6\,a+6\,b\,x\right )}{1152}-\frac {3\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{128}+b\,\left (\frac {3\,x\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{64}-\frac {x\,\mathrm {cosh}\left (6\,a+6\,b\,x\right )}{192}\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sinh(6*a + 6*b*x)/1152 - (3*sinh(2*a + 2*b*x))/128 + b*((3*x*cosh(2*a + 2*b*x))/64 - (x*cosh(6*a + 6*b*x))/1
92))/b^2

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sympy [A]  time = 4.94, size = 148, normalized size = 2.21 \[ \begin {cases} - \frac {x \sinh ^{6}{\left (a + b x \right )}}{24 b} + \frac {x \sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8 b} + \frac {x \sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{8 b} - \frac {x \cosh ^{6}{\left (a + b x \right )}}{24 b} + \frac {\sinh ^{5}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{24 b^{2}} - \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {\sinh {\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{24 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh ^{3}{\relax (a )} \cosh ^{3}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x*sinh(a + b*x)**6/(24*b) + x*sinh(a + b*x)**4*cosh(a + b*x)**2/(8*b) + x*sinh(a + b*x)**2*cosh(a
+ b*x)**4/(8*b) - x*cosh(a + b*x)**6/(24*b) + sinh(a + b*x)**5*cosh(a + b*x)/(24*b**2) - sinh(a + b*x)**3*cosh
(a + b*x)**3/(9*b**2) + sinh(a + b*x)*cosh(a + b*x)**5/(24*b**2), Ne(b, 0)), (x**2*sinh(a)**3*cosh(a)**3/2, Tr
ue))

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