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

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

[Out]

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

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Rubi [A]  time = 0.154006, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 30, 3310} \[ \frac{3 x^2 \cos ^4(a+b x)}{16 b^2}+\frac{9 x^2 \cos ^2(a+b x)}{16 b^2}-\frac{3 \cos ^4(a+b x)}{128 b^4}-\frac{45 \cos ^2(a+b x)}{128 b^4}-\frac{3 x \sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac{45 x \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac{x^3 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac{3 x^3 \sin (a+b x) \cos (a+b x)}{8 b}-\frac{45 x^2}{128 b^2}+\frac{3 x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cos[a + b*x]^4,x]

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.414309, size = 100, normalized size = 0.58 \[ \frac{4 b x \left (32 \left (2 b^2 x^2-3\right ) \sin (2 (a+b x))+\left (8 b^2 x^2-3\right ) \sin (4 (a+b x))+24 b^3 x^3\right )+192 \left (2 b^2 x^2-1\right ) \cos (2 (a+b x))+3 \left (8 b^2 x^2-1\right ) \cos (4 (a+b x))}{1024 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cos[a + b*x]^4,x]

[Out]

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

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Maple [B]  time = 0.056, size = 440, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05618, size = 409, normalized size = 2.38 \begin{align*} \frac{96 \,{\left (b x + a\right )}^{4} - 32 \,{\left (12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a^{3} + 24 \,{\left (24 \,{\left (b x + a\right )}^{2} + 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right ) + 16 \, \cos \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 12 \,{\left (32 \,{\left (b x + a\right )}^{3} + 4 \,{\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 64 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (8 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a + 3 \,{\left (8 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + 192 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 4 \,{\left (8 \,{\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 128 \,{\left (2 \,{\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \sin \left (2 \, b x + 2 \, a\right )}{1024 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(96*(b*x + a)^4 - 32*(12*b*x + 12*a + sin(4*b*x + 4*a) + 8*sin(2*b*x + 2*a))*a^3 + 24*(24*(b*x + a)^2 +
4*(b*x + a)*sin(4*b*x + 4*a) + 32*(b*x + a)*sin(2*b*x + 2*a) + cos(4*b*x + 4*a) + 16*cos(2*b*x + 2*a))*a^2 -
12*(32*(b*x + a)^3 + 4*(b*x + a)*cos(4*b*x + 4*a) + 64*(b*x + a)*cos(2*b*x + 2*a) + (8*(b*x + a)^2 - 1)*sin(4*
b*x + 4*a) + 32*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a))*a + 3*(8*(b*x + a)^2 - 1)*cos(4*b*x + 4*a) + 192*(2*(b*x
+ a)^2 - 1)*cos(2*b*x + 2*a) + 4*(8*(b*x + a)^3 - 3*b*x - 3*a)*sin(4*b*x + 4*a) + 128*(2*(b*x + a)^3 - 3*b*x
- 3*a)*sin(2*b*x + 2*a))/b^4

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Fricas [A]  time = 1.41015, size = 271, normalized size = 1.58 \begin{align*} \frac{12 \, b^{4} x^{4} + 3 \,{\left (8 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right )^{4} - 45 \, b^{2} x^{2} + 9 \,{\left (8 \, b^{2} x^{2} - 5\right )} \cos \left (b x + a\right )^{2} + 2 \,{\left (2 \,{\left (8 \, b^{3} x^{3} - 3 \, b x\right )} \cos \left (b x + a\right )^{3} + 3 \,{\left (8 \, b^{3} x^{3} - 15 \, b x\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{128 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(12*b^4*x^4 + 3*(8*b^2*x^2 - 1)*cos(b*x + a)^4 - 45*b^2*x^2 + 9*(8*b^2*x^2 - 5)*cos(b*x + a)^2 + 2*(2*(8
*b^3*x^3 - 3*b*x)*cos(b*x + a)^3 + 3*(8*b^3*x^3 - 15*b*x)*cos(b*x + a))*sin(b*x + a))/b^4

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Sympy [A]  time = 9.4835, size = 253, normalized size = 1.47 \begin{align*} \begin{cases} \frac{3 x^{4} \sin ^{4}{\left (a + b x \right )}}{32} + \frac{3 x^{4} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac{3 x^{4} \cos ^{4}{\left (a + b x \right )}}{32} + \frac{3 x^{3} \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} + \frac{5 x^{3} \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac{45 x^{2} \sin ^{4}{\left (a + b x \right )}}{128 b^{2}} - \frac{9 x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{64 b^{2}} + \frac{51 x^{2} \cos ^{4}{\left (a + b x \right )}}{128 b^{2}} - \frac{45 x \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{64 b^{3}} - \frac{51 x \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} + \frac{45 \sin ^{4}{\left (a + b x \right )}}{256 b^{4}} - \frac{51 \cos ^{4}{\left (a + b x \right )}}{256 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \cos ^{4}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x**4*sin(a + b*x)**4/32 + 3*x**4*sin(a + b*x)**2*cos(a + b*x)**2/16 + 3*x**4*cos(a + b*x)**4/32 +
3*x**3*sin(a + b*x)**3*cos(a + b*x)/(8*b) + 5*x**3*sin(a + b*x)*cos(a + b*x)**3/(8*b) - 45*x**2*sin(a + b*x)*
*4/(128*b**2) - 9*x**2*sin(a + b*x)**2*cos(a + b*x)**2/(64*b**2) + 51*x**2*cos(a + b*x)**4/(128*b**2) - 45*x*s
in(a + b*x)**3*cos(a + b*x)/(64*b**3) - 51*x*sin(a + b*x)*cos(a + b*x)**3/(64*b**3) + 45*sin(a + b*x)**4/(256*
b**4) - 51*cos(a + b*x)**4/(256*b**4), Ne(b, 0)), (x**4*cos(a)**4/4, True))

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Giac [A]  time = 1.11344, size = 146, normalized size = 0.85 \begin{align*} \frac{3}{32} \, x^{4} + \frac{3 \,{\left (8 \, b^{2} x^{2} - 1\right )} \cos \left (4 \, b x + 4 \, a\right )}{1024 \, b^{4}} + \frac{3 \,{\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right )}{16 \, b^{4}} + \frac{{\left (8 \, b^{3} x^{3} - 3 \, b x\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{4}} + \frac{{\left (2 \, b^{3} x^{3} - 3 \, b x\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*x^4 + 3/1024*(8*b^2*x^2 - 1)*cos(4*b*x + 4*a)/b^4 + 3/16*(2*b^2*x^2 - 1)*cos(2*b*x + 2*a)/b^4 + 1/256*(8*
b^3*x^3 - 3*b*x)*sin(4*b*x + 4*a)/b^4 + 1/8*(2*b^3*x^3 - 3*b*x)*sin(2*b*x + 2*a)/b^4