3.4 \(\int \cos (c+d x) (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=30 \[ \frac{(A+C) \sin (c+d x)}{d}-\frac{C \sin ^3(c+d x)}{3 d} \]

[Out]

((A + C)*Sin[c + d*x])/d - (C*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0232144, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {3013} \[ \frac{(A+C) \sin (c+d x)}{d}-\frac{C \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + C)*Sin[c + d*x])/d - (C*Sin[c + d*x]^3)/(3*d)

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

\begin{align*} \int \cos (c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (A+C-C x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{(A+C) \sin (c+d x)}{d}-\frac{C \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0154197, size = 50, normalized size = 1.67 \[ \frac{A \sin (c) \cos (d x)}{d}+\frac{A \cos (c) \sin (d x)}{d}-\frac{C \sin ^3(c+d x)}{3 d}+\frac{C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*Cos[d*x]*Sin[c])/d + (A*Cos[c]*Sin[d*x])/d + (C*Sin[c + d*x])/d - (C*Sin[c + d*x]^3)/(3*d)

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Maple [A]  time = 0.039, size = 33, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A\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+C*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.01083, size = 46, normalized size = 1.53 \begin{align*} -\frac{{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C - 3 \, A \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((sin(d*x + c)^3 - 3*sin(d*x + c))*C - 3*A*sin(d*x + c))/d

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Fricas [A]  time = 1.65117, size = 69, normalized size = 2.3 \begin{align*} \frac{{\left (C \cos \left (d x + c\right )^{2} + 3 \, A + 2 \, C\right )} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.717456, size = 56, normalized size = 1.87 \begin{align*} \begin{cases} \frac{A \sin{\left (c + d x \right )}}{d} + \frac{2 C \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{C \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \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+C*cos(d*x+c)**2),x)

[Out]

Piecewise((A*sin(c + d*x)/d + 2*C*sin(c + d*x)**3/(3*d) + C*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(A +
C*cos(c)**2)*cos(c), True))

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Giac [A]  time = 1.16038, size = 46, normalized size = 1.53 \begin{align*} -\frac{{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C - 3 \, A \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((sin(d*x + c)^3 - 3*sin(d*x + c))*C - 3*A*sin(d*x + c))/d