3.24 \(\int x^2 \cos ^4(a+b x) \, dx\)

Optimal. Leaf size=134 \[ -\frac {\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac {15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac {15 x}{64 b^2}+\frac {x^3}{8} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3311, 30, 2635, 8} \[ \frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x \cos ^2(a+b x)}{8 b^2}-\frac {\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac {15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac {x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac {15 x}{64 b^2}+\frac {x^3}{8} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x]^4,x]

[Out]

(-15*x)/(64*b^2) + x^3/8 + (3*x*Cos[a + b*x]^2)/(8*b^2) + (x*Cos[a + b*x]^4)/(8*b^2) - (15*Cos[a + b*x]*Sin[a
+ b*x])/(64*b^3) + (3*x^2*Cos[a + b*x]*Sin[a + b*x])/(8*b) - (Cos[a + b*x]^3*Sin[a + b*x])/(32*b^3) + (x^2*Cos
[a + b*x]^3*Sin[a + b*x])/(4*b)

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]

Rubi steps

\begin {align*} \int x^2 \cos ^4(a+b x) \, dx &=\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {3}{4} \int x^2 \cos ^2(a+b x) \, dx-\frac {\int \cos ^4(a+b x) \, dx}{8 b^2}\\ &=\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {3 \int x^2 \, dx}{8}-\frac {3 \int \cos ^2(a+b x) \, dx}{32 b^2}-\frac {3 \int \cos ^2(a+b x) \, dx}{8 b^2}\\ &=\frac {x^3}{8}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}-\frac {15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}-\frac {3 \int 1 \, dx}{64 b^2}-\frac {3 \int 1 \, dx}{16 b^2}\\ &=-\frac {15 x}{64 b^2}+\frac {x^3}{8}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}-\frac {15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 92, normalized size = 0.69 \[ \frac {64 b^2 x^2 \sin (2 (a+b x))+8 b^2 x^2 \sin (4 (a+b x))-32 \sin (2 (a+b x))-\sin (4 (a+b x))+64 b x \cos (2 (a+b x))+4 b x \cos (4 (a+b x))+32 b^3 x^3}{256 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x]^4,x]

[Out]

(32*b^3*x^3 + 64*b*x*Cos[2*(a + b*x)] + 4*b*x*Cos[4*(a + b*x)] - 32*Sin[2*(a + b*x)] + 64*b^2*x^2*Sin[2*(a + b
*x)] - Sin[4*(a + b*x)] + 8*b^2*x^2*Sin[4*(a + b*x)])/(256*b^3)

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fricas [A]  time = 0.66, size = 88, normalized size = 0.66 \[ \frac {8 \, b^{3} x^{3} + 8 \, b x \cos \left (b x + a\right )^{4} + 24 \, b x \cos \left (b x + a\right )^{2} - 15 \, b x + {\left (2 \, {\left (8 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right )^{3} + 3 \, {\left (8 \, b^{2} x^{2} - 5\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{64 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(b*x+a)^4,x, algorithm="fricas")

[Out]

1/64*(8*b^3*x^3 + 8*b*x*cos(b*x + a)^4 + 24*b*x*cos(b*x + a)^2 - 15*b*x + (2*(8*b^2*x^2 - 1)*cos(b*x + a)^3 +
3*(8*b^2*x^2 - 5)*cos(b*x + a))*sin(b*x + a))/b^3

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giac [A]  time = 0.36, size = 84, normalized size = 0.63 \[ \frac {1}{8} \, x^{3} + \frac {x \cos \left (4 \, b x + 4 \, a\right )}{64 \, b^{2}} + \frac {x \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac {{\left (8 \, b^{2} x^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} + \frac {{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(b*x+a)^4,x, algorithm="giac")

[Out]

1/8*x^3 + 1/64*x*cos(4*b*x + 4*a)/b^2 + 1/4*x*cos(2*b*x + 2*a)/b^2 + 1/256*(8*b^2*x^2 - 1)*sin(4*b*x + 4*a)/b^
3 + 1/8*(2*b^2*x^2 - 1)*sin(2*b*x + 2*a)/b^3

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maple [B]  time = 0.03, size = 241, normalized size = 1.80 \[ \frac {\left (b x +a \right )^{2} \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )+\frac {\left (b x +a \right ) \left (\cos ^{4}\left (b x +a \right )\right )}{8}-\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{32}-\frac {15 b x}{64}-\frac {15 a}{64}+\frac {3 \left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{8}-\frac {3 \cos \left (b x +a \right ) \sin \left (b x +a \right )}{16}-\frac {\left (b x +a \right )^{3}}{4}-2 a \left (\left (b x +a \right ) \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {3 \left (b x +a \right )^{2}}{16}+\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{16}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{16}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(b*x+a)^4,x)

[Out]

1/b^3*((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*(c
os(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)^3-2*a*((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)-3/16*(b*x+a)^2+1/16*cos
(b*x+a)^4+3/16*cos(b*x+a)^2)+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))

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maxima [A]  time = 0.35, size = 188, normalized size = 1.40 \[ \frac {32 \, {\left (b x + a\right )}^{3} + 8 \, {\left (12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 4 \, {\left (24 \, {\left (b x + a\right )}^{2} + 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right ) + 16 \, \cos \left (2 \, b x + 2 \, a\right )\right )} a + 4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 64 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (8 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{256 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(b*x+a)^4,x, algorithm="maxima")

[Out]

1/256*(32*(b*x + a)^3 + 8*(12*b*x + 12*a + sin(4*b*x + 4*a) + 8*sin(2*b*x + 2*a))*a^2 - 4*(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 + 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))/b^3

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mupad [B]  time = 0.54, size = 104, normalized size = 0.78 \[ \frac {x^3}{8}-\frac {\frac {\sin \left (2\,a+2\,b\,x\right )}{8}+\frac {\sin \left (4\,a+4\,b\,x\right )}{256}+b\,\left (\frac {x\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )}{4}+\frac {x\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )}{64}\right )-b^2\,\left (\frac {x^2\,\sin \left (2\,a+2\,b\,x\right )}{4}+\frac {x^2\,\sin \left (4\,a+4\,b\,x\right )}{32}\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*x)^4,x)

[Out]

x^3/8 - (sin(2*a + 2*b*x)/8 + sin(4*a + 4*b*x)/256 + b*((x*(2*sin(a + b*x)^2 - 1))/4 + (x*(2*sin(2*a + 2*b*x)^
2 - 1))/64) - b^2*((x^2*sin(2*a + 2*b*x))/4 + (x^2*sin(4*a + 4*b*x))/32))/b^3

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sympy [A]  time = 3.22, size = 209, normalized size = 1.56 \[ \begin {cases} \frac {x^{3} \sin ^{4}{\left (a + b x \right )}}{8} + \frac {x^{3} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {x^{3} \cos ^{4}{\left (a + b x \right )}}{8} + \frac {3 x^{2} \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} + \frac {5 x^{2} \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {15 x \sin ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {3 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{2}} + \frac {17 x \cos ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {15 \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{64 b^{3}} - \frac {17 \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \cos ^{4}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(b*x+a)**4,x)

[Out]

Piecewise((x**3*sin(a + b*x)**4/8 + x**3*sin(a + b*x)**2*cos(a + b*x)**2/4 + x**3*cos(a + b*x)**4/8 + 3*x**2*s
in(a + b*x)**3*cos(a + b*x)/(8*b) + 5*x**2*sin(a + b*x)*cos(a + b*x)**3/(8*b) - 15*x*sin(a + b*x)**4/(64*b**2)
- 3*x*sin(a + b*x)**2*cos(a + b*x)**2/(32*b**2) + 17*x*cos(a + b*x)**4/(64*b**2) - 15*sin(a + b*x)**3*cos(a +
b*x)/(64*b**3) - 17*sin(a + b*x)*cos(a + b*x)**3/(64*b**3), Ne(b, 0)), (x**3*cos(a)**4/3, True))

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