3.13 \(\int \cos ^6(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, 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 = {4045, 2635, 8} \[ \frac {(5 A+6 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {(5 A+6 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {A \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {1}{16} x (5 A+6 C) \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]

[Out]

((5*A + 6*C)*x)/16 + ((5*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((5*A + 6*C)*Cos[c + d*x]^3*Sin[c + d*x]
)/(24*d) + (A*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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rubi steps

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

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

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(A + C*Sec[c + d*x]^2),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 96, normalized size = 1.08 \[ \frac {3 \, {\left (d x + c\right )} {\left (5 \, A + 6 \, C\right )} + \frac {15 \, A \tan \left (d x + c\right )^{5} + 18 \, C \tan \left (d x + c\right )^{5} + 40 \, A \tan \left (d x + c\right )^{3} + 48 \, C \tan \left (d x + c\right )^{3} + 33 \, A \tan \left (d x + c\right ) + 30 \, C \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/48*(3*(d*x + c)*(5*A + 6*C) + (15*A*tan(d*x + c)^5 + 18*C*tan(d*x + c)^5 + 40*A*tan(d*x + c)^3 + 48*C*tan(d*
x + c)^3 + 33*A*tan(d*x + c) + 30*C*tan(d*x + c))/(tan(d*x + c)^2 + 1)^3)/d

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maple [A]  time = 1.95, size = 86, normalized size = 0.97 \[ \frac {A \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(A*(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)+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))

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maxima [A]  time = 0.43, size = 103, normalized size = 1.16 \[ \frac {3 \, {\left (d x + c\right )} {\left (5 \, A + 6 \, C\right )} + \frac {3 \, {\left (5 \, A + 6 \, C\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (5 \, A + 6 \, C\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (11 \, A + 10 \, C\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.94, size = 91, normalized size = 1.02 \[ x\,\left (\frac {5\,A}{16}+\frac {3\,C}{8}\right )+\frac {\left (\frac {5\,A}{16}+\frac {3\,C}{8}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (\frac {5\,A}{6}+C\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {11\,A}{16}+\frac {5\,C}{8}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6+3\,{\mathrm {tan}\left (c+d\,x\right )}^4+3\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(A + C/cos(c + d*x)^2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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