3.23 \(\int x^3 \cosh ^4(a+b x) \, dx\)

Optimal. Leaf size=172 \[ -\frac {3 \cosh ^4(a+b x)}{128 b^4}-\frac {45 \cosh ^2(a+b x)}{128 b^4}+\frac {3 x \sinh (a+b x) \cosh ^3(a+b x)}{32 b^3}+\frac {45 x \sinh (a+b x) \cosh (a+b x)}{64 b^3}-\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}-\frac {9 x^2 \cosh ^2(a+b x)}{16 b^2}+\frac {x^3 \sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3 x^3 \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac {45 x^2}{128 b^2}+\frac {3 x^4}{32} \]

[Out]

45/128*x^2/b^2+3/32*x^4-45/128*cosh(b*x+a)^2/b^4-9/16*x^2*cosh(b*x+a)^2/b^2-3/128*cosh(b*x+a)^4/b^4-3/16*x^2*c
osh(b*x+a)^4/b^2+45/64*x*cosh(b*x+a)*sinh(b*x+a)/b^3+3/8*x^3*cosh(b*x+a)*sinh(b*x+a)/b+3/32*x*cosh(b*x+a)^3*si
nh(b*x+a)/b^3+1/4*x^3*cosh(b*x+a)^3*sinh(b*x+a)/b

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Rubi [A]  time = 0.15, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 30, 3310} \[ -\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}-\frac {9 x^2 \cosh ^2(a+b x)}{16 b^2}-\frac {3 \cosh ^4(a+b x)}{128 b^4}-\frac {45 \cosh ^2(a+b x)}{128 b^4}+\frac {3 x \sinh (a+b x) \cosh ^3(a+b x)}{32 b^3}+\frac {45 x \sinh (a+b x) \cosh (a+b x)}{64 b^3}+\frac {x^3 \sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3 x^3 \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac {45 x^2}{128 b^2}+\frac {3 x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cosh[a + b*x]^4,x]

[Out]

(45*x^2)/(128*b^2) + (3*x^4)/32 - (45*Cosh[a + b*x]^2)/(128*b^4) - (9*x^2*Cosh[a + b*x]^2)/(16*b^2) - (3*Cosh[
a + b*x]^4)/(128*b^4) - (3*x^2*Cosh[a + b*x]^4)/(16*b^2) + (45*x*Cosh[a + b*x]*Sinh[a + b*x])/(64*b^3) + (3*x^
3*Cosh[a + b*x]*Sinh[a + b*x])/(8*b) + (3*x*Cosh[a + b*x]^3*Sinh[a + b*x])/(32*b^3) + (x^3*Cosh[a + b*x]^3*Sin
h[a + b*x])/(4*b)

Rule 30

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

Rule 3310

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

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]

Rubi steps

\begin {align*} \int x^3 \cosh ^4(a+b x) \, dx &=-\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}+\frac {x^3 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3}{4} \int x^3 \cosh ^2(a+b x) \, dx+\frac {3 \int x \cosh ^4(a+b x) \, dx}{8 b^2}\\ &=-\frac {9 x^2 \cosh ^2(a+b x)}{16 b^2}-\frac {3 \cosh ^4(a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}+\frac {3 x^3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {3 x \cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac {x^3 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3 \int x^3 \, dx}{8}+\frac {9 \int x \cosh ^2(a+b x) \, dx}{32 b^2}+\frac {9 \int x \cosh ^2(a+b x) \, dx}{8 b^2}\\ &=\frac {3 x^4}{32}-\frac {45 \cosh ^2(a+b x)}{128 b^4}-\frac {9 x^2 \cosh ^2(a+b x)}{16 b^2}-\frac {3 \cosh ^4(a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}+\frac {45 x \cosh (a+b x) \sinh (a+b x)}{64 b^3}+\frac {3 x^3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {3 x \cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac {x^3 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {9 \int x \, dx}{64 b^2}+\frac {9 \int x \, dx}{16 b^2}\\ &=\frac {45 x^2}{128 b^2}+\frac {3 x^4}{32}-\frac {45 \cosh ^2(a+b x)}{128 b^4}-\frac {9 x^2 \cosh ^2(a+b x)}{16 b^2}-\frac {3 \cosh ^4(a+b x)}{128 b^4}-\frac {3 x^2 \cosh ^4(a+b x)}{16 b^2}+\frac {45 x \cosh (a+b x) \sinh (a+b x)}{64 b^3}+\frac {3 x^3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {3 x \cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac {x^3 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 100, normalized size = 0.58 \[ \frac {-192 \left (2 b^2 x^2+1\right ) \cosh (2 (a+b x))-3 \left (8 b^2 x^2+1\right ) \cosh (4 (a+b x))+4 b x \left (32 \left (2 b^2 x^2+3\right ) \sinh (2 (a+b x))+\left (8 b^2 x^2+3\right ) \sinh (4 (a+b x))+24 b^3 x^3\right )}{1024 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cosh[a + b*x]^4,x]

[Out]

(-192*(1 + 2*b^2*x^2)*Cosh[2*(a + b*x)] - 3*(1 + 8*b^2*x^2)*Cosh[4*(a + b*x)] + 4*b*x*(24*b^3*x^3 + 32*(3 + 2*
b^2*x^2)*Sinh[2*(a + b*x)] + (3 + 8*b^2*x^2)*Sinh[4*(a + b*x)]))/(1024*b^4)

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fricas [A]  time = 0.56, size = 195, normalized size = 1.13 \[ \frac {96 \, b^{4} x^{4} - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} + 16 \, {\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4} - 192 \, {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} - 6 \, {\left (64 \, b^{2} x^{2} + 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} + 32\right )} \sinh \left (b x + a\right )^{2} + 16 \, {\left ({\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{3} + 16 \, {\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{1024 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^4,x, algorithm="fricas")

[Out]

1/1024*(96*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)*sinh(b*x + a)^3
- 3*(8*b^2*x^2 + 1)*sinh(b*x + a)^4 - 192*(2*b^2*x^2 + 1)*cosh(b*x + a)^2 - 6*(64*b^2*x^2 + 3*(8*b^2*x^2 + 1)*
cosh(b*x + a)^2 + 32)*sinh(b*x + a)^2 + 16*((8*b^3*x^3 + 3*b*x)*cosh(b*x + a)^3 + 16*(2*b^3*x^3 + 3*b*x)*cosh(
b*x + a))*sinh(b*x + a))/b^4

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giac [A]  time = 0.12, size = 150, normalized size = 0.87 \[ \frac {3}{32} \, x^{4} + \frac {{\left (32 \, b^{3} x^{3} - 24 \, b^{2} x^{2} + 12 \, b x - 3\right )} e^{\left (4 \, b x + 4 \, a\right )}}{2048 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{4}} - \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} - \frac {{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^4,x, algorithm="giac")

[Out]

3/32*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 + 1/32*(4*b^3*x^3 - 6*b^2*x^2 + 6
*b*x - 3)*e^(2*b*x + 2*a)/b^4 - 1/32*(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 = 0.17, size = 400, normalized size = 2.33 \[ \frac {\frac {\left (b x +a \right )^{3} \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \left (b x +a \right )^{3} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}+\frac {3 \left (b x +a \right )^{4}}{32}-\frac {3 \left (b x +a \right )^{2} \left (\cosh ^{4}\left (b x +a \right )\right )}{16}+\frac {3 \left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{32}+\frac {45 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{64}+\frac {45 \left (b x +a \right )^{2}}{128}-\frac {3 \left (\cosh ^{4}\left (b x +a \right )\right )}{128}-\frac {45 \left (\cosh ^{2}\left (b x +a \right )\right )}{128}-\frac {9 \left (b x +a \right )^{2} \left (\cosh ^{2}\left (b x +a \right )\right )}{16}-3 a \left (\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \left (b x +a \right )^{2} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}+\frac {\left (b x +a \right )^{3}}{8}-\frac {\left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{8}+\frac {\sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{32}+\frac {15 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{64}+\frac {15 b x}{64}+\frac {15 a}{64}-\frac {3 \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{8}\right )+3 a^{2} \left (\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}+\frac {3 \left (b x +a \right )^{2}}{16}-\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{16}-\frac {3 \left (\cosh ^{2}\left (b x +a \right )\right )}{16}\right )-a^{3} \left (\left (\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(b*x+a)^4,x)

[Out]

1/b^4*(1/4*(b*x+a)^3*sinh(b*x+a)*cosh(b*x+a)^3+3/8*(b*x+a)^3*cosh(b*x+a)*sinh(b*x+a)+3/32*(b*x+a)^4-3/16*(b*x+
a)^2*cosh(b*x+a)^4+3/32*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3+45/64*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+45/128*(b*x+a)
^2-3/128*cosh(b*x+a)^4-45/128*cosh(b*x+a)^2-9/16*(b*x+a)^2*cosh(b*x+a)^2-3*a*(1/4*(b*x+a)^2*sinh(b*x+a)*cosh(b
*x+a)^3+3/8*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+1/8*(b*x+a)^3-1/8*(b*x+a)*cosh(b*x+a)^4+1/32*sinh(b*x+a)*cosh(b*
x+a)^3+15/64*cosh(b*x+a)*sinh(b*x+a)+15/64*b*x+15/64*a-3/8*(b*x+a)*cosh(b*x+a)^2)+3*a^2*(1/4*(b*x+a)*sinh(b*x+
a)*cosh(b*x+a)^3+3/8*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+3/16*(b*x+a)^2-1/16*cosh(b*x+a)^4-3/16*cosh(b*x+a)^2)-a^3
*((1/4*cosh(b*x+a)^3+3/8*cosh(b*x+a))*sinh(b*x+a)+3/8*b*x+3/8*a))

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maxima [A]  time = 0.36, size = 176, normalized size = 1.02 \[ \frac {3}{32} \, x^{4} + \frac {{\left (32 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 24 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 12 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{2048 \, b^{4}} + \frac {{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{4}} - \frac {{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} - \frac {{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)^4,x, algorithm="maxima")

[Out]

3/32*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 + 1/32*
(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 - 1/32*(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 = 0.37, size = 129, normalized size = 0.75 \[ \frac {3\,x^4}{32}-\frac {\frac {3\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{16}+\frac {3\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{1024}+b^2\,\left (\frac {3\,x^2\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{8}+\frac {3\,x^2\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{128}\right )-b\,\left (\frac {3\,x\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{8}+\frac {3\,x\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{256}\right )-b^3\,\left (\frac {x^3\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4}+\frac {x^3\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{32}\right )}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(a + b*x)^4,x)

[Out]

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

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sympy [A]  time = 5.70, size = 253, normalized size = 1.47 \[ \begin {cases} \frac {3 x^{4} \sinh ^{4}{\left (a + b x \right )}}{32} - \frac {3 x^{4} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16} + \frac {3 x^{4} \cosh ^{4}{\left (a + b x \right )}}{32} - \frac {3 x^{3} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b} + \frac {5 x^{3} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} + \frac {45 x^{2} \sinh ^{4}{\left (a + b x \right )}}{128 b^{2}} - \frac {9 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{64 b^{2}} - \frac {51 x^{2} \cosh ^{4}{\left (a + b x \right )}}{128 b^{2}} - \frac {45 x \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{64 b^{3}} + \frac {51 x \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{64 b^{3}} + \frac {45 \sinh ^{4}{\left (a + b x \right )}}{256 b^{4}} - \frac {51 \cosh ^{4}{\left (a + b x \right )}}{256 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \cosh ^{4}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cosh(b*x+a)**4,x)

[Out]

Piecewise((3*x**4*sinh(a + b*x)**4/32 - 3*x**4*sinh(a + b*x)**2*cosh(a + b*x)**2/16 + 3*x**4*cosh(a + b*x)**4/
32 - 3*x**3*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + 5*x**3*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) + 45*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) - 51*x**2*cosh(a + b*x)**4/(128*b*
*2) - 45*x*sinh(a + b*x)**3*cosh(a + b*x)/(64*b**3) + 51*x*sinh(a + b*x)*cosh(a + b*x)**3/(64*b**3) + 45*sinh(
a + b*x)**4/(256*b**4) - 51*cosh(a + b*x)**4/(256*b**4), Ne(b, 0)), (x**4*cosh(a)**4/4, True))

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