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]

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

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.43, size = 49, normalized size = 0.80 \[ \frac {{\left (4 \, A + 3 \, C\right )} d x + {\left (2 \, C \cos \left (d x + c\right )^{3} + {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

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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giac [A]  time = 0.33, size = 43, normalized size = 0.70 \[ \frac {1}{8} \, {\left (4 \, A + 3 \, C\right )} x + \frac {C \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (A + C\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

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

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maple [A]  time = 0.05, size = 65, normalized size = 1.07 \[ \frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

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 = 0.43, size = 73, normalized size = 1.20 \[ \frac {{\left (d x + c\right )} {\left (4 \, A + 3 \, C\right )} + \frac {{\left (4 \, A + 3 \, C\right )} \tan \left (d x + c\right )^{3} + {\left (4 \, A + 5 \, C\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

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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mupad [B]  time = 0.78, size = 67, normalized size = 1.10 \[ x\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )+\frac {\left (\frac {A}{2}+\frac {3\,C}{8}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {A}{2}+\frac {5\,C}{8}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.95, size = 158, normalized size = 2.59 \[ \begin {cases} \frac {A x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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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