3.10 \(\int \cos ^4(c+d x) (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=89 \[ \frac {(6 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {(6 A+5 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x (6 A+5 C)+\frac {C \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

[Out]

1/16*(6*A+5*C)*x+1/16*(6*A+5*C)*cos(d*x+c)*sin(d*x+c)/d+1/24*(6*A+5*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*C*cos(d*x
+c)^5*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3014, 2635, 8} \[ \frac {(6 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {(6 A+5 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x (6 A+5 C)+\frac {C \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2),x]

[Out]

((6*A + 5*C)*x)/16 + ((6*A + 5*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*A + 5*C)*Cos[c + d*x]^3*Sin[c + d*x]
)/(24*d) + (C*Cos[c + d*x]^5*Sin[c + d*x])/(6*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 ^4(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} (6 A+5 C) \int \cos ^4(c+d x) \, dx\\ &=\frac {(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {C \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{8} (6 A+5 C) \int \cos ^2(c+d x) \, dx\\ &=\frac {(6 A+5 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {C \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{16} (6 A+5 C) \int 1 \, dx\\ &=\frac {1}{16} (6 A+5 C) x+\frac {(6 A+5 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {C \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 68, normalized size = 0.76 \[ \frac {(48 A+45 C) \sin (2 (c+d x))+(6 A+9 C) \sin (4 (c+d x))+72 A c+72 A d x+C \sin (6 (c+d x))+60 c C+60 C d x}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2),x]

[Out]

(72*A*c + 60*c*C + 72*A*d*x + 60*C*d*x + (48*A + 45*C)*Sin[2*(c + d*x)] + (6*A + 9*C)*Sin[4*(c + d*x)] + C*Sin
[6*(c + d*x)])/(192*d)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 68, normalized size = 0.76 \[ \frac {3 \, {\left (6 \, A + 5 \, C\right )} d x + {\left (8 \, C \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(3*(6*A + 5*C)*d*x + (8*C*cos(d*x + c)^5 + 2*(6*A + 5*C)*cos(d*x + c)^3 + 3*(6*A + 5*C)*cos(d*x + c))*sin
(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.25, size = 68, normalized size = 0.76 \[ \frac {1}{16} \, {\left (6 \, A + 5 \, C\right )} x + \frac {C \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (2 \, A + 3 \, C\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, A + 15 \, C\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/16*(6*A + 5*C)*x + 1/192*C*sin(6*d*x + 6*c)/d + 1/64*(2*A + 3*C)*sin(4*d*x + 4*c)/d + 1/64*(16*A + 15*C)*sin
(2*d*x + 2*c)/d

________________________________________________________________________________________

maple [A]  time = 0.05, size = 86, normalized size = 0.97 \[ \frac {C \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+C*cos(d*x+c)^2),x)

[Out]

1/d*(C*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+A*(1/4*(cos(d*x+c)^3+3
/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 103, normalized size = 1.16 \[ \frac {3 \, {\left (d x + c\right )} {\left (6 \, A + 5 \, C\right )} + \frac {3 \, {\left (6 \, A + 5 \, C\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (6 \, A + 5 \, C\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (10 \, A + 11 \, C\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(3*(d*x + c)*(6*A + 5*C) + (3*(6*A + 5*C)*tan(d*x + c)^5 + 8*(6*A + 5*C)*tan(d*x + c)^3 + 3*(10*A + 11*C)
*tan(d*x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

________________________________________________________________________________________

mupad [B]  time = 1.22, size = 91, normalized size = 1.02 \[ x\,\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )+\frac {\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (A+\frac {5\,C}{6}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {5\,A}{8}+\frac {11\,C}{16}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6+3\,{\mathrm {tan}\left (c+d\,x\right )}^4+3\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + C*cos(c + d*x)^2),x)

[Out]

x*((3*A)/8 + (5*C)/16) + (tan(c + d*x)*((5*A)/8 + (11*C)/16) + tan(c + d*x)^3*(A + (5*C)/6) + tan(c + d*x)^5*(
(3*A)/8 + (5*C)/16))/(d*(3*tan(c + d*x)^2 + 3*tan(c + d*x)^4 + tan(c + d*x)^6 + 1))

________________________________________________________________________________________

sympy [A]  time = 3.43, size = 258, normalized size = 2.90 \[ \begin {cases} \frac {3 A x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {5 C x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 C x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 C x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 C x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 C \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 C \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 C \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(A+C*cos(d*x+c)**2),x)

[Out]

Piecewise((3*A*x*sin(c + d*x)**4/8 + 3*A*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*x*cos(c + d*x)**4/8 + 3*A*s
in(c + d*x)**3*cos(c + d*x)/(8*d) + 5*A*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 5*C*x*sin(c + d*x)**6/16 + 15*C*x
*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*C*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*C*x*cos(c + d*x)**6/16 + 5
*C*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*C*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*C*sin(c + d*x)*cos(c +
d*x)**5/(16*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)**4, True))

________________________________________________________________________________________