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

Optimal. Leaf size=61 \[ \frac{(4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 A+3 C)+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

((4*A + 3*C)*x)/8 + ((4*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.0412907, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3014, 2635, 8} \[ \frac{(4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 A+3 C)+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A + 3*C)*x)/8 + ((4*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} (4 A+3 C) \int \cos ^2(c+d x) \, dx\\ &=\frac{(4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{8} (4 A+3 C) \int 1 \, dx\\ &=\frac{1}{8} (4 A+3 C) x+\frac{(4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0855863, size = 45, normalized size = 0.74 \[ \frac{4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.037, size = 65, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( C \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 \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.68262, size = 99, normalized size = 1.62 \begin{align*} \frac{{\left (d x + c\right )}{\left (4 \, A + 3 \, C\right )} + \frac{{\left (4 \, A + 3 \, C\right )} \tan \left (d x + c\right )^{3} +{\left (4 \, A + 5 \, C\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((d*x + c)*(4*A + 3*C) + ((4*A + 3*C)*tan(d*x + c)^3 + (4*A + 5*C)*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d
*x + c)^2 + 1))/d

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Fricas [A]  time = 1.608, size = 119, normalized size = 1.95 \begin{align*} \frac{{\left (4 \, A + 3 \, C\right )} d x +{\left (2 \, C \cos \left (d x + c\right )^{3} +{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )\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)^2*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.72031, size = 158, normalized size = 2.59 \begin{align*} \begin{cases} \frac{A x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{3 C x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 C x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 C x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 C \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 C \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\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)**2*(A+C*cos(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.12214, size = 58, normalized size = 0.95 \begin{align*} \frac{1}{8} \,{\left (4 \, A + 3 \, C\right )} x + \frac{C \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (A + C\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*A + 3*C)*x + 1/32*C*sin(4*d*x + 4*c)/d + 1/4*(A + C)*sin(2*d*x + 2*c)/d