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

Optimal. Leaf size=85 \[ -\frac{a^3 \sin ^3(c+d x)}{d}+\frac{4 a^3 \sin (c+d x)}{d}+\frac{a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{15 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{15 a^3 x}{8} \]

[Out]

(15*a^3*x)/8 + (4*a^3*Sin[c + d*x])/d + (15*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^3*Cos[c + d*x]^3*Sin[c +
d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/d

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Rubi [A]  time = 0.0780084, antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2751, 2645, 2637, 2635, 8, 2633} \[ -\frac{a^3 \sin ^3(c+d x)}{4 d}+\frac{3 a^3 \sin (c+d x)}{d}+\frac{9 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{15 a^3 x}{8}+\frac{\sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Cos[c + d*x])^3,x]

[Out]

(15*a^3*x)/8 + (3*a^3*Sin[c + d*x])/d + (9*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + ((a + a*Cos[c + d*x])^3*Sin[
c + d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/(4*d)

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \cos (c+d x))^3 \, dx &=\frac{(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{3}{4} \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac{(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{3}{4} \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac{3 a^3 x}{4}+\frac{(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \left (3 a^3\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{4} \left (9 a^3\right ) \int \cos (c+d x) \, dx+\frac{1}{4} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{3 a^3 x}{4}+\frac{9 a^3 \sin (c+d x)}{4 d}+\frac{9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{8} \left (9 a^3\right ) \int 1 \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{15 a^3 x}{8}+\frac{3 a^3 \sin (c+d x)}{d}+\frac{9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{(a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{a^3 \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.118255, size = 51, normalized size = 0.6 \[ \frac{a^3 (104 \sin (c+d x)+32 \sin (2 (c+d x))+8 \sin (3 (c+d x))+\sin (4 (c+d x))+60 d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])^3,x]

[Out]

(a^3*(60*d*x + 104*Sin[c + d*x] + 32*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d)

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Maple [A]  time = 0.04, size = 100, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,{a}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{3}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*a^3*(1
/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^3*sin(d*x+c))

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Maxima [A]  time = 1.14336, size = 127, normalized size = 1.49 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} -{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, a^{3} \sin \left (d x + c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3
- 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 32*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.61081, size = 151, normalized size = 1.78 \begin{align*} \frac{15 \, a^{3} d x +{\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(15*a^3*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^3*cos(d*x + c)^2 + 15*a^3*cos(d*x + c) + 24*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 1.29056, size = 224, normalized size = 2.64 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac{5 a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{3 a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{a^{3} \sin{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\right )^{3} \cos{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**4/8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a**3*x*sin(c + d*x)**2/
2 + 3*a**3*x*cos(c + d*x)**4/8 + 3*a**3*x*cos(c + d*x)**2/2 + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*a*
*3*sin(c + d*x)**3/d + 5*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*a**3*sin(c + d*x)*cos(c + d*x)**2/d + 3*a
**3*sin(c + d*x)*cos(c + d*x)/(2*d) + a**3*sin(c + d*x)/d, Ne(d, 0)), (x*(a*cos(c) + a)**3*cos(c), True))

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Giac [A]  time = 1.3528, size = 96, normalized size = 1.13 \begin{align*} \frac{15}{8} \, a^{3} x + \frac{a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a^{3} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac{a^{3} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac{13 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*a^3*x + 1/32*a^3*sin(4*d*x + 4*c)/d + 1/4*a^3*sin(3*d*x + 3*c)/d + a^3*sin(2*d*x + 2*c)/d + 13/4*a^3*sin(
d*x + c)/d