3.19 \(\int (c+d x) \cos ^3(a+b x) \, dx\)

Optimal. Leaf size=75 \[ \frac {d \cos ^3(a+b x)}{9 b^2}+\frac {2 d \cos (a+b x)}{3 b^2}+\frac {2 (c+d x) \sin (a+b x)}{3 b}+\frac {(c+d x) \sin (a+b x) \cos ^2(a+b x)}{3 b} \]

[Out]

2/3*d*cos(b*x+a)/b^2+1/9*d*cos(b*x+a)^3/b^2+2/3*(d*x+c)*sin(b*x+a)/b+1/3*(d*x+c)*cos(b*x+a)^2*sin(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3310, 3296, 2638} \[ \frac {d \cos ^3(a+b x)}{9 b^2}+\frac {2 d \cos (a+b x)}{3 b^2}+\frac {2 (c+d x) \sin (a+b x)}{3 b}+\frac {(c+d x) \sin (a+b x) \cos ^2(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cos[a + b*x]^3,x]

[Out]

(2*d*Cos[a + b*x])/(3*b^2) + (d*Cos[a + b*x]^3)/(9*b^2) + (2*(c + d*x)*Sin[a + b*x])/(3*b) + ((c + d*x)*Cos[a
+ b*x]^2*Sin[a + b*x])/(3*b)

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 (c+d x) \cos ^3(a+b x) \, dx &=\frac {d \cos ^3(a+b x)}{9 b^2}+\frac {(c+d x) \cos ^2(a+b x) \sin (a+b x)}{3 b}+\frac {2}{3} \int (c+d x) \cos (a+b x) \, dx\\ &=\frac {d \cos ^3(a+b x)}{9 b^2}+\frac {2 (c+d x) \sin (a+b x)}{3 b}+\frac {(c+d x) \cos ^2(a+b x) \sin (a+b x)}{3 b}-\frac {(2 d) \int \sin (a+b x) \, dx}{3 b}\\ &=\frac {2 d \cos (a+b x)}{3 b^2}+\frac {d \cos ^3(a+b x)}{9 b^2}+\frac {2 (c+d x) \sin (a+b x)}{3 b}+\frac {(c+d x) \cos ^2(a+b x) \sin (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 52, normalized size = 0.69 \[ \frac {3 b (c+d x) (9 \sin (a+b x)+\sin (3 (a+b x)))+27 d \cos (a+b x)+d \cos (3 (a+b x))}{36 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Cos[a + b*x]^3,x]

[Out]

(27*d*Cos[a + b*x] + d*Cos[3*(a + b*x)] + 3*b*(c + d*x)*(9*Sin[a + b*x] + Sin[3*(a + b*x)]))/(36*b^2)

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fricas [A]  time = 0.69, size = 60, normalized size = 0.80 \[ \frac {d \cos \left (b x + a\right )^{3} + 6 \, d \cos \left (b x + a\right ) + 3 \, {\left (2 \, b d x + {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + 2 \, b c\right )} \sin \left (b x + a\right )}{9 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(d*cos(b*x + a)^3 + 6*d*cos(b*x + a) + 3*(2*b*d*x + (b*d*x + b*c)*cos(b*x + a)^2 + 2*b*c)*sin(b*x + a))/b^
2

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giac [A]  time = 0.52, size = 69, normalized size = 0.92 \[ \frac {d \cos \left (3 \, b x + 3 \, a\right )}{36 \, b^{2}} + \frac {3 \, d \cos \left (b x + a\right )}{4 \, b^{2}} + \frac {{\left (b d x + b c\right )} \sin \left (3 \, b x + 3 \, a\right )}{12 \, b^{2}} + \frac {3 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*d*cos(3*b*x + 3*a)/b^2 + 3/4*d*cos(b*x + a)/b^2 + 1/12*(b*d*x + b*c)*sin(3*b*x + 3*a)/b^2 + 3/4*(b*d*x +
b*c)*sin(b*x + a)/b^2

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maple [A]  time = 0.03, size = 95, normalized size = 1.27 \[ \frac {\frac {d \left (\frac {\left (b x +a \right ) \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}+\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \cos \left (b x +a \right )}{3}\right )}{b}-\frac {d a \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3 b}+\frac {c \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b*d*(1/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1/9*cos(b*x+a)^3+2/3*cos(b*x+a))-1/3/b*d*a*(2+cos(b*x+a)^2
)*sin(b*x+a)+1/3*c*(2+cos(b*x+a)^2)*sin(b*x+a))

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maxima [A]  time = 0.35, size = 103, normalized size = 1.37 \[ -\frac {12 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} c - \frac {12 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} a d}{b} - \frac {{\left (3 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 27 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) + 27 \, \cos \left (b x + a\right )\right )} d}{b}}{36 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(12*(sin(b*x + a)^3 - 3*sin(b*x + a))*c - 12*(sin(b*x + a)^3 - 3*sin(b*x + a))*a*d/b - (3*(b*x + a)*sin(
3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*d/b)/b

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mupad [B]  time = 0.26, size = 77, normalized size = 1.03 \[ \frac {\frac {3\,c\,\sin \left (a+b\,x\right )}{4}+\frac {c\,\sin \left (3\,a+3\,b\,x\right )}{12}+\frac {d\,x\,\sin \left (3\,a+3\,b\,x\right )}{12}+\frac {3\,d\,x\,\sin \left (a+b\,x\right )}{4}}{b}+\frac {d\,\cos \left (3\,a+3\,b\,x\right )}{36\,b^2}+\frac {3\,d\,\cos \left (a+b\,x\right )}{4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3*(c + d*x),x)

[Out]

((3*c*sin(a + b*x))/4 + (c*sin(3*a + 3*b*x))/12 + (d*x*sin(3*a + 3*b*x))/12 + (3*d*x*sin(a + b*x))/4)/b + (d*c
os(3*a + 3*b*x))/(36*b^2) + (3*d*cos(a + b*x))/(4*b^2)

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sympy [A]  time = 0.92, size = 126, normalized size = 1.68 \[ \begin {cases} \frac {2 c \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {c \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 d x \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {d x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 d \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{3 b^{2}} + \frac {7 d \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)**3,x)

[Out]

Piecewise((2*c*sin(a + b*x)**3/(3*b) + c*sin(a + b*x)*cos(a + b*x)**2/b + 2*d*x*sin(a + b*x)**3/(3*b) + d*x*si
n(a + b*x)*cos(a + b*x)**2/b + 2*d*sin(a + b*x)**2*cos(a + b*x)/(3*b**2) + 7*d*cos(a + b*x)**3/(9*b**2), Ne(b,
0)), ((c*x + d*x**2/2)*cos(a)**3, True))

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