3.313 \(\int \cos ^5(c+d x) (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=97 \[ -\frac{a (B+C) \sin ^3(c+d x)}{3 d}+\frac{a (B+C) \sin (c+d x)}{d}+\frac{a (3 B+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} a x (3 B+4 C) \]

[Out]

(a*(3*B + 4*C)*x)/8 + (a*(B + C)*Sin[c + d*x])/d + (a*(3*B + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*B*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(B + C)*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.169343, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4072, 3996, 3787, 2633, 2635, 8} \[ -\frac{a (B+C) \sin ^3(c+d x)}{3 d}+\frac{a (B+C) \sin (c+d x)}{d}+\frac{a (3 B+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} a x (3 B+4 C) \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*(3*B + 4*C)*x)/8 + (a*(B + C)*Sin[c + d*x])/d + (a*(3*B + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*B*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(B + C)*Sin[c + d*x]^3)/(3*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 ^5(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^4(c+d x) (a+a \sec (c+d x)) (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) (-4 a (B+C)-a (3 B+4 C) \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^3(c+d x) \sin (c+d x)}{4 d}+(a (B+C)) \int \cos ^3(c+d x) \, dx+\frac{1}{4} (a (3 B+4 C)) \int \cos ^2(c+d x) \, dx\\ &=\frac{a (3 B+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{8} (a (3 B+4 C)) \int 1 \, dx-\frac{(a (B+C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{1}{8} a (3 B+4 C) x+\frac{a (B+C) \sin (c+d x)}{d}+\frac{a (3 B+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{a (B+C) \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.229937, size = 75, normalized size = 0.77 \[ \frac{a \left (-32 (B+C) \sin ^3(c+d x)+96 (B+C) \sin (c+d x)+24 (B+C) \sin (2 (c+d x))+3 B \sin (4 (c+d x))+36 B c+36 B d x+48 c C+48 C d x\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*(36*B*c + 48*c*C + 36*B*d*x + 48*C*d*x + 96*(B + C)*Sin[c + d*x] - 32*(B + C)*Sin[c + d*x]^3 + 24*(B + C)*S
in[2*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.091, size = 107, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( Ba \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 ) +{\frac{Ba \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{aC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+aC \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(B*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)+1/3*B*a*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
a*C*(2+cos(d*x+c)^2)*sin(d*x+c)+a*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.93707, size = 136, normalized size = 1.4 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*
a + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a)/d

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Fricas [A]  time = 0.492934, size = 193, normalized size = 1.99 \begin{align*} \frac{3 \,{\left (3 \, B + 4 \, C\right )} a d x +{\left (6 \, B a \cos \left (d x + c\right )^{3} + 8 \,{\left (B + C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, B + 4 \, C\right )} a \cos \left (d x + c\right ) + 16 \,{\left (B + C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(3*B + 4*C)*a*d*x + (6*B*a*cos(d*x + c)^3 + 8*(B + C)*a*cos(d*x + c)^2 + 3*(3*B + 4*C)*a*cos(d*x + c)
+ 16*(B + C)*a)*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)**5*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.1221, size = 211, normalized size = 2.18 \begin{align*} \frac{3 \,{\left (3 \, B a + 4 \, C a\right )}{\left (d x + c\right )} + \frac{2 \,{\left (9 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 49 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 28 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 31 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 52 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 39 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*B*a + 4*C*a)*(d*x + c) + 2*(9*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1/2*c)^7 + 49*B*a*t
an(1/2*d*x + 1/2*c)^5 + 28*C*a*tan(1/2*d*x + 1/2*c)^5 + 31*B*a*tan(1/2*d*x + 1/2*c)^3 + 52*C*a*tan(1/2*d*x + 1
/2*c)^3 + 39*B*a*tan(1/2*d*x + 1/2*c) + 36*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d