3.290 $$\int x^3 \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=79 $-\frac{3 x^2 \cosh (4 a+4 b x)}{128 b^2}+\frac{3 x \sinh (4 a+4 b x)}{256 b^3}-\frac{3 \cosh (4 a+4 b x)}{1024 b^4}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}-\frac{x^4}{32}$

[Out]

-x^4/32 - (3*Cosh[4*a + 4*b*x])/(1024*b^4) - (3*x^2*Cosh[4*a + 4*b*x])/(128*b^2) + (3*x*Sinh[4*a + 4*b*x])/(25
6*b^3) + (x^3*Sinh[4*a + 4*b*x])/(32*b)

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Rubi [A]  time = 0.114185, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {5448, 3296, 2638} $-\frac{3 x^2 \cosh (4 a+4 b x)}{128 b^2}+\frac{3 x \sinh (4 a+4 b x)}{256 b^3}-\frac{3 \cosh (4 a+4 b x)}{1024 b^4}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}-\frac{x^4}{32}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Cosh[a + b*x]^2*Sinh[a + b*x]^2,x]

[Out]

-x^4/32 - (3*Cosh[4*a + 4*b*x])/(1024*b^4) - (3*x^2*Cosh[4*a + 4*b*x])/(128*b^2) + (3*x*Sinh[4*a + 4*b*x])/(25
6*b^3) + (x^3*Sinh[4*a + 4*b*x])/(32*b)

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]

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 2638

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

Rubi steps

\begin{align*} \int x^3 \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{x^3}{8}+\frac{1}{8} x^3 \cosh (4 a+4 b x)\right ) \, dx\\ &=-\frac{x^4}{32}+\frac{1}{8} \int x^3 \cosh (4 a+4 b x) \, dx\\ &=-\frac{x^4}{32}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}-\frac{3 \int x^2 \sinh (4 a+4 b x) \, dx}{32 b}\\ &=-\frac{x^4}{32}-\frac{3 x^2 \cosh (4 a+4 b x)}{128 b^2}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}+\frac{3 \int x \cosh (4 a+4 b x) \, dx}{64 b^2}\\ &=-\frac{x^4}{32}-\frac{3 x^2 \cosh (4 a+4 b x)}{128 b^2}+\frac{3 x \sinh (4 a+4 b x)}{256 b^3}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}-\frac{3 \int \sinh (4 a+4 b x) \, dx}{256 b^3}\\ &=-\frac{x^4}{32}-\frac{3 \cosh (4 a+4 b x)}{1024 b^4}-\frac{3 x^2 \cosh (4 a+4 b x)}{128 b^2}+\frac{3 x \sinh (4 a+4 b x)}{256 b^3}+\frac{x^3 \sinh (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 0.198099, size = 58, normalized size = 0.73 $\frac{4 b x \left (8 b^2 x^2+3\right ) \sinh (4 (a+b x))-3 \left (8 b^2 x^2+1\right ) \cosh (4 (a+b x))-32 b^4 x^4}{1024 b^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Cosh[a + b*x]^2*Sinh[a + b*x]^2,x]

[Out]

(-32*b^4*x^4 - 3*(1 + 8*b^2*x^2)*Cosh[4*(a + b*x)] + 4*b*x*(3 + 8*b^2*x^2)*Sinh[4*(a + b*x)])/(1024*b^4)

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Maple [B]  time = 0.007, size = 384, normalized size = 4.9 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{ \left ( bx+a \right ) ^{3}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}-{\frac{ \left ( bx+a \right ) ^{3}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{ \left ( bx+a \right ) ^{4}}{32}}-{\frac{3\, \left ( bx+a \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{16}}+{\frac{ \left ( 3\,bx+3\,a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{32}}-{\frac{ \left ( 3\,bx+3\,a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{64}}-{\frac{3\, \left ( bx+a \right ) ^{2}}{128}}-{\frac{3\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{128}}-3\,a \left ( 1/4\, \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}-1/8\, \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/24\, \left ( bx+a \right ) ^{3}-1/8\, \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}+1/32\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) -{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{64}}-{\frac{bx}{64}}-{\frac{a}{64}} \right ) +3\,{a}^{2} \left ( 1/4\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}-1/8\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/16\, \left ( bx+a \right ) ^{2}-1/16\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \right ) -{a}^{3} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) }{4}}-{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{bx}{8}}-{\frac{a}{8}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/4*(b*x+a)^3*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)^3*cosh(b*x+a)*sinh(b*x+a)-1/32*(b*x+a)^4-3/16*(b*x+
a)^2*sinh(b*x+a)^2*cosh(b*x+a)^2+3/32*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3-3/64*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-3
/128*(b*x+a)^2-3/128*cosh(b*x+a)^2*sinh(b*x+a)^2-3*a*(1/4*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)^2*co
sh(b*x+a)*sinh(b*x+a)-1/24*(b*x+a)^3-1/8*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^2+1/32*cosh(b*x+a)^3*sinh(b*x+a)-1/
64*cosh(b*x+a)*sinh(b*x+a)-1/64*b*x-1/64*a)+3*a^2*(1/4*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)*cosh(b*x+
a)*sinh(b*x+a)-1/16*(b*x+a)^2-1/16*cosh(b*x+a)^2*sinh(b*x+a)^2)-a^3*(1/4*cosh(b*x+a)^3*sinh(b*x+a)-1/8*cosh(b*
x+a)*sinh(b*x+a)-1/8*b*x-1/8*a))

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Maxima [A]  time = 1.17843, size = 123, normalized size = 1.56 \begin{align*} -\frac{1}{32} \, x^{4} + \frac{{\left (32 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 24 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 12 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{2048 \, b^{4}} - \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/32*x^4 + 1/2048*(32*b^3*x^3*e^(4*a) - 24*b^2*x^2*e^(4*a) + 12*b*x*e^(4*a) - 3*e^(4*a))*e^(4*b*x)/b^4 - 1/20
48*(32*b^3*x^3 + 24*b^2*x^2 + 12*b*x + 3)*e^(-4*b*x - 4*a)/b^4

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Fricas [B]  time = 1.85244, size = 352, normalized size = 4.46 \begin{align*} -\frac{32 \, b^{4} x^{4} + 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} - 16 \,{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 18 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - 16 \,{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4}}{1024 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/1024*(32*b^4*x^4 + 3*(8*b^2*x^2 + 1)*cosh(b*x + a)^4 - 16*(8*b^3*x^3 + 3*b*x)*cosh(b*x + a)^3*sinh(b*x + a)
+ 18*(8*b^2*x^2 + 1)*cosh(b*x + a)^2*sinh(b*x + a)^2 - 16*(8*b^3*x^3 + 3*b*x)*cosh(b*x + a)*sinh(b*x + a)^3 +
3*(8*b^2*x^2 + 1)*sinh(b*x + a)^4)/b^4

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Sympy [A]  time = 8.077, size = 250, normalized size = 3.16 \begin{align*} \begin{cases} - \frac{x^{4} \sinh ^{4}{\left (a + b x \right )}}{32} + \frac{x^{4} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16} - \frac{x^{4} \cosh ^{4}{\left (a + b x \right )}}{32} + \frac{x^{3} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} + \frac{x^{3} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} - \frac{3 x^{2} \sinh ^{4}{\left (a + b x \right )}}{128 b^{2}} - \frac{9 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{64 b^{2}} - \frac{3 x^{2} \cosh ^{4}{\left (a + b x \right )}}{128 b^{2}} + \frac{3 x \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{64 b^{3}} + \frac{3 x \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{64 b^{3}} - \frac{3 \sinh ^{4}{\left (a + b x \right )}}{256 b^{4}} - \frac{3 \cosh ^{4}{\left (a + b x \right )}}{256 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh ^{2}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**4*sinh(a + b*x)**4/32 + x**4*sinh(a + b*x)**2*cosh(a + b*x)**2/16 - x**4*cosh(a + b*x)**4/32 +
x**3*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + x**3*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) - 3*x**2*sinh(a + b*x)**
4/(128*b**2) - 9*x**2*sinh(a + b*x)**2*cosh(a + b*x)**2/(64*b**2) - 3*x**2*cosh(a + b*x)**4/(128*b**2) + 3*x*s
inh(a + b*x)**3*cosh(a + b*x)/(64*b**3) + 3*x*sinh(a + b*x)*cosh(a + b*x)**3/(64*b**3) - 3*sinh(a + b*x)**4/(2
56*b**4) - 3*cosh(a + b*x)**4/(256*b**4), Ne(b, 0)), (x**4*sinh(a)**2*cosh(a)**2/4, True))

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Giac [A]  time = 1.17165, size = 105, normalized size = 1.33 \begin{align*} -\frac{1}{32} \, x^{4} + \frac{{\left (32 \, b^{3} x^{3} - 24 \, b^{2} x^{2} + 12 \, b x - 3\right )} e^{\left (4 \, b x + 4 \, a\right )}}{2048 \, b^{4}} - \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

-1/32*x^4 + 1/2048*(32*b^3*x^3 - 24*b^2*x^2 + 12*b*x - 3)*e^(4*b*x + 4*a)/b^4 - 1/2048*(32*b^3*x^3 + 24*b^2*x^
2 + 12*b*x + 3)*e^(-4*b*x - 4*a)/b^4