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

Optimal. Leaf size=80 \[ \frac {\cos ^4(a+b x)}{16 b^2}+\frac {3 \cos ^2(a+b x)}{16 b^2}+\frac {x \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x \sin (a+b x) \cos (a+b x)}{8 b}+\frac {3 x^2}{16} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3310, 30} \[ \frac {\cos ^4(a+b x)}{16 b^2}+\frac {3 \cos ^2(a+b x)}{16 b^2}+\frac {x \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x \sin (a+b x) \cos (a+b x)}{8 b}+\frac {3 x^2}{16} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x]^4,x]

[Out]

(3*x^2)/16 + (3*Cos[a + b*x]^2)/(16*b^2) + Cos[a + b*x]^4/(16*b^2) + (3*x*Cos[a + b*x]*Sin[a + b*x])/(8*b) + (
x*Cos[a + b*x]^3*Sin[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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 53, normalized size = 0.66 \[ \frac {4 b x (8 \sin (2 (a+b x))+\sin (4 (a+b x))+6 b x)+16 \cos (2 (a+b x))+\cos (4 (a+b x))}{128 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x]^4,x]

[Out]

(16*Cos[2*(a + b*x)] + Cos[4*(a + b*x)] + 4*b*x*(6*b*x + 8*Sin[2*(a + b*x)] + Sin[4*(a + b*x)]))/(128*b^2)

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fricas [A]  time = 0.86, size = 63, normalized size = 0.79 \[ \frac {3 \, b^{2} x^{2} + \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} + 2 \, {\left (2 \, b x \cos \left (b x + a\right )^{3} + 3 \, b x \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*b^2*x^2 + cos(b*x + a)^4 + 3*cos(b*x + a)^2 + 2*(2*b*x*cos(b*x + a)^3 + 3*b*x*cos(b*x + a))*sin(b*x +
a))/b^2

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giac [A]  time = 0.45, size = 64, normalized size = 0.80 \[ \frac {3}{16} \, x^{2} + \frac {x \sin \left (4 \, b x + 4 \, a\right )}{32 \, b} + \frac {x \sin \left (2 \, b x + 2 \, a\right )}{4 \, b} + \frac {\cos \left (4 \, b x + 4 \, a\right )}{128 \, b^{2}} + \frac {\cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*x^2 + 1/32*x*sin(4*b*x + 4*a)/b + 1/4*x*sin(2*b*x + 2*a)/b + 1/128*cos(4*b*x + 4*a)/b^2 + 1/8*cos(2*b*x +
2*a)/b^2

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maple [A]  time = 0.03, size = 110, normalized size = 1.38 \[ \frac {\left (b x +a \right ) \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {3 \left (b x +a \right )^{2}}{16}+\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{16}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{16}-a \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 98, normalized size = 1.22 \[ \frac {24 \, {\left (b x + a\right )}^{2} - 4 \, {\left (12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a + 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right ) + 16 \, \cos \left (2 \, b x + 2 \, a\right )}{128 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(24*(b*x + a)^2 - 4*(12*b*x + 12*a + sin(4*b*x + 4*a) + 8*sin(2*b*x + 2*a))*a + 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))/b^2

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mupad [B]  time = 0.34, size = 63, normalized size = 0.79 \[ \frac {3\,x^2}{16}-\frac {\frac {{\sin \left (2\,a+2\,b\,x\right )}^2}{64}-b\,\left (\frac {x\,\sin \left (2\,a+2\,b\,x\right )}{4}+\frac {x\,\sin \left (4\,a+4\,b\,x\right )}{32}\right )+\frac {{\sin \left (a+b\,x\right )}^2}{4}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x)^4,x)

[Out]

(3*x^2)/16 - (sin(2*a + 2*b*x)^2/64 - b*((x*sin(2*a + 2*b*x))/4 + (x*sin(4*a + 4*b*x))/32) + sin(a + b*x)^2/4)
/b^2

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sympy [A]  time = 1.81, size = 138, normalized size = 1.72 \[ \begin {cases} \frac {3 x^{2} \sin ^{4}{\left (a + b x \right )}}{16} + \frac {3 x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8} + \frac {3 x^{2} \cos ^{4}{\left (a + b x \right )}}{16} + \frac {3 x \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} + \frac {5 x \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {3 \sin ^{4}{\left (a + b x \right )}}{32 b^{2}} + \frac {5 \cos ^{4}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \cos ^{4}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x**2*sin(a + b*x)**4/16 + 3*x**2*sin(a + b*x)**2*cos(a + b*x)**2/8 + 3*x**2*cos(a + b*x)**4/16 +
3*x*sin(a + b*x)**3*cos(a + b*x)/(8*b) + 5*x*sin(a + b*x)*cos(a + b*x)**3/(8*b) - 3*sin(a + b*x)**4/(32*b**2)
+ 5*cos(a + b*x)**4/(32*b**2), Ne(b, 0)), (x**2*cos(a)**4/2, True))

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