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

Optimal. Leaf size=155 \[ -\frac {3 \sinh (a+b x) \cosh ^3(a+b x)}{128 b^4}-\frac {45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {9 x \cosh ^2(a+b x)}{32 b^3}-\frac {3 x^2 \sinh (a+b x) \cosh ^3(a+b x)}{16 b^2}-\frac {9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {45 x}{256 b^3}-\frac {3 x^3}{32 b} \]

[Out]

-45/256*x/b^3-3/32*x^3/b+9/32*x*cosh(b*x+a)^2/b^3+3/32*x*cosh(b*x+a)^4/b^3+1/4*x^3*cosh(b*x+a)^4/b-45/256*cosh
(b*x+a)*sinh(b*x+a)/b^4-9/32*x^2*cosh(b*x+a)*sinh(b*x+a)/b^2-3/128*cosh(b*x+a)^3*sinh(b*x+a)/b^4-3/16*x^2*cosh
(b*x+a)^3*sinh(b*x+a)/b^2

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Rubi [A]  time = 0.14, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5373, 3311, 30, 2635, 8} \[ -\frac {3 x^2 \sinh (a+b x) \cosh ^3(a+b x)}{16 b^2}-\frac {9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {9 x \cosh ^2(a+b x)}{32 b^3}-\frac {3 \sinh (a+b x) \cosh ^3(a+b x)}{128 b^4}-\frac {45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {45 x}{256 b^3}-\frac {3 x^3}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*x)/(256*b^3) - (3*x^3)/(32*b) + (9*x*Cosh[a + b*x]^2)/(32*b^3) + (3*x*Cosh[a + b*x]^4)/(32*b^3) + (x^3*Co
sh[a + b*x]^4)/(4*b) - (45*Cosh[a + b*x]*Sinh[a + b*x])/(256*b^4) - (9*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(32*b^
2) - (3*Cosh[a + b*x]^3*Sinh[a + b*x])/(128*b^4) - (3*x^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 30

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

Rule 2635

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] && Integer
Q[2*n]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 5373

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^3 \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {3 \int x^2 \cosh ^4(a+b x) \, dx}{4 b}\\ &=\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac {3 \int \cosh ^4(a+b x) \, dx}{32 b^3}-\frac {9 \int x^2 \cosh ^2(a+b x) \, dx}{16 b}\\ &=\frac {9 x \cosh ^2(a+b x)}{32 b^3}+\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac {9 \int \cosh ^2(a+b x) \, dx}{128 b^3}-\frac {9 \int \cosh ^2(a+b x) \, dx}{32 b^3}-\frac {9 \int x^2 \, dx}{32 b}\\ &=-\frac {3 x^3}{32 b}+\frac {9 x \cosh ^2(a+b x)}{32 b^3}+\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}-\frac {9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}-\frac {9 \int 1 \, dx}{256 b^3}-\frac {9 \int 1 \, dx}{64 b^3}\\ &=-\frac {45 x}{256 b^3}-\frac {3 x^3}{32 b}+\frac {9 x \cosh ^2(a+b x)}{32 b^3}+\frac {3 x \cosh ^4(a+b x)}{32 b^3}+\frac {x^3 \cosh ^4(a+b x)}{4 b}-\frac {45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}-\frac {9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {3 \cosh ^3(a+b x) \sinh (a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^3(a+b x) \sinh (a+b x)}{16 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 91, normalized size = 0.59 \[ \frac {32 b x \left (2 b^2 x^2+3\right ) \cosh (2 (a+b x))+2 b x \left (8 b^2 x^2+3\right ) \cosh (4 (a+b x))-3 \sinh (2 (a+b x)) \left (\left (8 b^2 x^2+1\right ) \cosh (2 (a+b x))+32 b^2 x^2+16\right )}{512 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b*x*(3 + 2*b^2*x^2)*Cosh[2*(a + b*x)] + 2*b*x*(3 + 8*b^2*x^2)*Cosh[4*(a + b*x)] - 3*(16 + 32*b^2*x^2 + (1
+ 8*b^2*x^2)*Cosh[2*(a + b*x)])*Sinh[2*(a + b*x)])/(512*b^4)

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fricas [A]  time = 0.47, size = 191, normalized size = 1.23 \[ \frac {{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{4} - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + {\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \sinh \left (b x + a\right )^{4} + 16 \, {\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (16 \, b^{3} x^{3} + 3 \, {\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} + 24 \, b x\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} + 16 \, {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{256 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*((8*b^3*x^3 + 3*b*x)*cosh(b*x + a)^4 - 3*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (8*b^3*x^3 + 3*
b*x)*sinh(b*x + a)^4 + 16*(2*b^3*x^3 + 3*b*x)*cosh(b*x + a)^2 + 2*(16*b^3*x^3 + 3*(8*b^3*x^3 + 3*b*x)*cosh(b*x
+ a)^2 + 24*b*x)*sinh(b*x + a)^2 - 3*((8*b^2*x^2 + 1)*cosh(b*x + a)^3 + 16*(2*b^2*x^2 + 1)*cosh(b*x + a))*sin
h(b*x + a))/b^4

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giac [A]  time = 0.14, size = 145, normalized size = 0.94 \[ \frac {{\left (32 \, b^{3} x^{3} - 24 \, b^{2} x^{2} + 12 \, b x - 3\right )} e^{\left (4 \, b x + 4 \, a\right )}}{2048 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{64 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{64 \, b^{4}} + \frac {{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(32*b^3*x^3 - 24*b^2*x^2 + 12*b*x - 3)*e^(4*b*x + 4*a)/b^4 + 1/64*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*e
^(2*b*x + 2*a)/b^4 + 1/64*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*b*x - 2*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

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maple [B]  time = 0.33, size = 304, normalized size = 1.96 \[ \frac {\frac {\left (b x +a \right )^{3} \left (\cosh ^{4}\left (b x +a \right )\right )}{4}-\frac {3 \left (b x +a \right )^{2} \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{16}-\frac {9 \left (b x +a \right )^{2} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{32}-\frac {3 \left (b x +a \right )^{3}}{32}+\frac {3 \left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{32}-\frac {3 \left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{128}-\frac {45 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{256}-\frac {45 b x}{256}-\frac {45 a}{256}+\frac {9 \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{32}-3 a \left (\frac {\left (b x +a \right )^{2} \left (\cosh ^{4}\left (b x +a \right )\right )}{4}-\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{8}-\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{16}-\frac {3 \left (b x +a \right )^{2}}{32}+\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{32}+\frac {3 \left (\cosh ^{2}\left (b x +a \right )\right )}{32}\right )+3 a^{2} \left (\frac {\left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{4}-\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{16}-\frac {3 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{32}-\frac {3 b x}{32}-\frac {3 a}{32}\right )-\frac {a^{3} \left (\cosh ^{4}\left (b x +a \right )\right )}{4}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/4*(b*x+a)^3*cosh(b*x+a)^4-3/16*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3-9/32*(b*x+a)^2*cosh(b*x+a)*sinh(b*
x+a)-3/32*(b*x+a)^3+3/32*(b*x+a)*cosh(b*x+a)^4-3/128*cosh(b*x+a)^3*sinh(b*x+a)-45/256*cosh(b*x+a)*sinh(b*x+a)-
45/256*b*x-45/256*a+9/32*(b*x+a)*cosh(b*x+a)^2-3*a*(1/4*(b*x+a)^2*cosh(b*x+a)^4-1/8*(b*x+a)*sinh(b*x+a)*cosh(b
*x+a)^3-3/16*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-3/32*(b*x+a)^2+1/32*cosh(b*x+a)^4+3/32*cosh(b*x+a)^2)+3*a^2*(1/4*
(b*x+a)*cosh(b*x+a)^4-1/16*cosh(b*x+a)^3*sinh(b*x+a)-3/32*cosh(b*x+a)*sinh(b*x+a)-3/32*b*x-3/32*a)-1/4*a^3*cos
h(b*x+a)^4)

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maxima [A]  time = 0.33, size = 171, normalized size = 1.10 \[ \frac {{\left (32 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 24 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 12 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{2048 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{64 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{64 \, b^{4}} + \frac {{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 + 1/64*(4*b^3*x^3*
e^(2*a) - 6*b^2*x^2*e^(2*a) + 6*b*x*e^(2*a) - 3*e^(2*a))*e^(2*b*x)/b^4 + 1/64*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x +
3)*e^(-2*b*x - 2*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

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mupad [B]  time = 1.69, size = 125, normalized size = 0.81 \[ \frac {\frac {x^3\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{8}+\frac {x^3\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{32}}{b}-\frac {\frac {3\,x^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{16}+\frac {3\,x^2\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{128}}{b^2}+\frac {\frac {3\,x\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{16}+\frac {3\,x\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{256}}{b^3}-\frac {3\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{32\,b^4}-\frac {3\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{1024\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^3*cosh(2*a + 2*b*x))/8 + (x^3*cosh(4*a + 4*b*x))/32)/b - ((3*x^2*sinh(2*a + 2*b*x))/16 + (3*x^2*sinh(4*a +
4*b*x))/128)/b^2 + ((3*x*cosh(2*a + 2*b*x))/16 + (3*x*cosh(4*a + 4*b*x))/256)/b^3 - (3*sinh(2*a + 2*b*x))/(32
*b^4) - (3*sinh(4*a + 4*b*x))/(1024*b^4)

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sympy [A]  time = 5.26, size = 226, normalized size = 1.46 \[ \begin {cases} - \frac {3 x^{3} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac {5 x^{3} \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac {9 x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{32 b^{2}} - \frac {15 x^{2} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} - \frac {45 x \sinh ^{4}{\left (a + b x \right )}}{256 b^{3}} + \frac {9 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{128 b^{3}} + \frac {51 x \cosh ^{4}{\left (a + b x \right )}}{256 b^{3}} + \frac {45 \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{256 b^{4}} - \frac {51 \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{256 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh {\relax (a )} \cosh ^{3}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*x**3*sinh(a + b*x)**4/(32*b) + 3*x**3*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) + 5*x**3*cosh(a +
b*x)**4/(32*b) + 9*x**2*sinh(a + b*x)**3*cosh(a + b*x)/(32*b**2) - 15*x**2*sinh(a + b*x)*cosh(a + b*x)**3/(32
*b**2) - 45*x*sinh(a + b*x)**4/(256*b**3) + 9*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(128*b**3) + 51*x*cosh(a + b
*x)**4/(256*b**3) + 45*sinh(a + b*x)**3*cosh(a + b*x)/(256*b**4) - 51*sinh(a + b*x)*cosh(a + b*x)**3/(256*b**4
), Ne(b, 0)), (x**4*sinh(a)*cosh(a)**3/4, True))

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