3.16 \(\int \cos (c+d x) (a+a \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ -\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+a^2 x \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2751, 2644} \[ \frac {4 a^2 \sin (c+d x)}{3 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{3 d}+a^2 x+\frac {\sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Cos[c + d*x])^2,x]

[Out]

a^2*x + (4*a^2*Sin[c + d*x])/(3*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(3*d) + ((a + a*Cos[c + d*x])^2*Sin[c + d
*x])/(3*d)

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

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)]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 41, normalized size = 0.72 \[ \frac {a^2 (21 \sin (c+d x)+6 \sin (2 (c+d x))+\sin (3 (c+d x))+12 d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])^2,x]

[Out]

(a^2*(12*d*x + 21*Sin[c + d*x] + 6*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

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fricas [A]  time = 1.56, size = 49, normalized size = 0.86 \[ \frac {3 \, a^{2} d x + {\left (a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) + 5 \, a^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*a^2*d*x + (a^2*cos(d*x + c)^2 + 3*a^2*cos(d*x + c) + 5*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.47, size = 54, normalized size = 0.95 \[ a^{2} x + \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {7 \, a^{2} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

a^2*x + 1/12*a^2*sin(3*d*x + 3*c)/d + 1/2*a^2*sin(2*d*x + 2*c)/d + 7/4*a^2*sin(d*x + c)/d

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maple [A]  time = 0.05, size = 64, normalized size = 1.12 \[ \frac {\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*cos(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.47, size = 61, normalized size = 1.07 \[ -\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 6 \, a^{2} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 6*a^2*sin(d*x + c))/d

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mupad [B]  time = 0.38, size = 61, normalized size = 1.07 \[ a^2\,x+\frac {5\,a^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(a + a*cos(c + d*x))^2,x)

[Out]

a^2*x + (5*a^2*sin(c + d*x))/(3*d) + (a^2*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (a^2*cos(c + d*x)*sin(c + d*x))
/d

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sympy [A]  time = 0.48, size = 107, normalized size = 1.88 \[ \begin {cases} a^{2} x \sin ^{2}{\left (c + d x \right )} + a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{2} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((a**2*x*sin(c + d*x)**2 + a**2*x*cos(c + d*x)**2 + 2*a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)*
cos(c + d*x)**2/d + a**2*sin(c + d*x)*cos(c + d*x)/d + a**2*sin(c + d*x)/d, Ne(d, 0)), (x*(a*cos(c) + a)**2*co
s(c), True))

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