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]

((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)

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Rubi [A]  time = 0.0586294, antiderivative size = 89, normalized size of antiderivative = 1., 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 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]

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 ^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*}

Mathematica [A]  time = 0.103052, 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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Maple [A]  time = 0.056, size = 86, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( A \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +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 ) \right ) } \end{align*}

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 = 1.40669, size = 139, normalized size = 1.56 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 0.495464, size = 167, normalized size = 1.88 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.20468, size = 130, normalized size = 1.46 \begin{align*} \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} \end{align*}

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