3.3 \(\int \cos ^3(c+d x) (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=50 \[ -\frac {(A+2 C) \sin ^3(c+d x)}{3 d}+\frac {(A+C) \sin (c+d x)}{d}+\frac {C \sin ^5(c+d x)}{5 d} \]

[Out]

(A+C)*sin(d*x+c)/d-1/3*(A+2*C)*sin(d*x+c)^3/d+1/5*C*sin(d*x+c)^5/d

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Rubi [A]  time = 0.05, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3013, 373} \[ -\frac {(A+2 C) \sin ^3(c+d x)}{3 d}+\frac {(A+C) \sin (c+d x)}{d}+\frac {C \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2),x]

[Out]

((A + C)*Sin[c + d*x])/d - ((A + 2*C)*Sin[c + d*x]^3)/(3*d) + (C*Sin[c + d*x]^5)/(5*d)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

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 ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx &=-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \left (A+C-C x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (A \left (1+\frac {C}{A}\right )-(A+2 C) x^2+C x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {(A+C) \sin (c+d x)}{d}-\frac {(A+2 C) \sin ^3(c+d x)}{3 d}+\frac {C \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 1.42 \[ -\frac {A \sin ^3(c+d x)}{3 d}+\frac {A \sin (c+d x)}{d}+\frac {C \sin ^5(c+d x)}{5 d}-\frac {2 C \sin ^3(c+d x)}{3 d}+\frac {C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2),x]

[Out]

(A*Sin[c + d*x])/d + (C*Sin[c + d*x])/d - (A*Sin[c + d*x]^3)/(3*d) - (2*C*Sin[c + d*x]^3)/(3*d) + (C*Sin[c + d
*x]^5)/(5*d)

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fricas [A]  time = 0.51, size = 45, normalized size = 0.90 \[ \frac {{\left (3 \, C \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 8 \, C\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(3*C*cos(d*x + c)^4 + (5*A + 4*C)*cos(d*x + c)^2 + 10*A + 8*C)*sin(d*x + c)/d

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giac [A]  time = 0.19, size = 57, normalized size = 1.14 \[ \frac {3 \, C \sin \left (d x + c\right )^{5} - 5 \, A \sin \left (d x + c\right )^{3} - 10 \, C \sin \left (d x + c\right )^{3} + 15 \, A \sin \left (d x + c\right ) + 15 \, C \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/15*(3*C*sin(d*x + c)^5 - 5*A*sin(d*x + c)^3 - 10*C*sin(d*x + c)^3 + 15*A*sin(d*x + c) + 15*C*sin(d*x + c))/d

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maple [A]  time = 0.05, size = 54, normalized size = 1.08 \[ \frac {\frac {C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.33, size = 43, normalized size = 0.86 \[ \frac {3 \, C \sin \left (d x + c\right )^{5} - 5 \, {\left (A + 2 \, C\right )} \sin \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/15*(3*C*sin(d*x + c)^5 - 5*(A + 2*C)*sin(d*x + c)^3 + 15*(A + C)*sin(d*x + c))/d

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mupad [B]  time = 0.68, size = 43, normalized size = 0.86 \[ \frac {\frac {C\,{\sin \left (c+d\,x\right )}^5}{5}+\left (-\frac {A}{3}-\frac {2\,C}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (A+C\right )\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + C*cos(c + d*x)^2),x)

[Out]

((C*sin(c + d*x)^5)/5 + sin(c + d*x)*(A + C) - sin(c + d*x)^3*(A/3 + (2*C)/3))/d

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sympy [A]  time = 1.70, size = 105, normalized size = 2.10 \[ \begin {cases} \frac {2 A \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {8 C \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2),x)

[Out]

Piecewise((2*A*sin(c + d*x)**3/(3*d) + A*sin(c + d*x)*cos(c + d*x)**2/d + 8*C*sin(c + d*x)**5/(15*d) + 4*C*sin
(c + d*x)**3*cos(c + d*x)**2/(3*d) + C*sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)*
*3, True))

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