3.331 \(\int \cos ^7(c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=201 \[ -\frac{a^3 (17 B+19 C) \sin ^3(c+d x)}{15 d}+\frac{a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac{a^3 (21 B+22 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac{a^3 (23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{(4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{1}{16} a^3 x (23 B+26 C)+\frac{a B \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[Out]

(a^3*(23*B + 26*C)*x)/16 + (a^3*(17*B + 19*C)*Sin[c + d*x])/(5*d) + (a^3*(23*B + 26*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(21*B + 22*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*B*Cos[c + d*x]^5*(a + a*Sec[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((4*B + 3*C)*Cos[c + d*x]^4*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d) - (a^3*(17*
B + 19*C)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.479243, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4017, 3996, 3787, 2633, 2635, 8} \[ -\frac{a^3 (17 B+19 C) \sin ^3(c+d x)}{15 d}+\frac{a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac{a^3 (21 B+22 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac{a^3 (23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{(4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac{1}{16} a^3 x (23 B+26 C)+\frac{a B \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(23*B + 26*C)*x)/16 + (a^3*(17*B + 19*C)*Sin[c + d*x])/(5*d) + (a^3*(23*B + 26*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(21*B + 22*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*B*Cos[c + d*x]^5*(a + a*Sec[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((4*B + 3*C)*Cos[c + d*x]^4*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d) - (a^3*(17*
B + 19*C)*Sin[c + d*x]^3)/(15*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 4017

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

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 ^7(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^6(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (2 a (4 B+3 C)+3 a (B+2 C) \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{30} \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (21 B+22 C)+3 a^2 (13 B+16 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{1}{120} \int \cos ^3(c+d x) \left (-24 a^3 (17 B+19 C)-15 a^3 (23 B+26 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{5} \left (a^3 (17 B+19 C)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{8} \left (a^3 (23 B+26 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a^3 (23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{16} \left (a^3 (23 B+26 C)\right ) \int 1 \, dx-\frac{\left (a^3 (17 B+19 C)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{1}{16} a^3 (23 B+26 C) x+\frac{a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac{a^3 (23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{a^3 (17 B+19 C) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.481165, size = 134, normalized size = 0.67 \[ \frac{a^3 (120 (21 B+23 C) \sin (c+d x)+15 (63 B+64 C) \sin (2 (c+d x))+380 B \sin (3 (c+d x))+135 B \sin (4 (c+d x))+36 B \sin (5 (c+d x))+5 B \sin (6 (c+d x))+1380 B c+1380 B d x+340 C \sin (3 (c+d x))+90 C \sin (4 (c+d x))+12 C \sin (5 (c+d x))+1560 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(1380*B*c + 1380*B*d*x + 1560*C*d*x + 120*(21*B + 23*C)*Sin[c + d*x] + 15*(63*B + 64*C)*Sin[2*(c + d*x)]
+ 380*B*Sin[3*(c + d*x)] + 340*C*Sin[3*(c + d*x)] + 135*B*Sin[4*(c + d*x)] + 90*C*Sin[4*(c + d*x)] + 36*B*Sin[
5*(c + d*x)] + 12*C*Sin[5*(c + d*x)] + 5*B*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.105, size = 266, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( B{a}^{3} \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 ) +{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+{\frac{3\,B{a}^{3}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+3\,{a}^{3}C \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +3\,B{a}^{3} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{B{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{3}C \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)^7*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(B*a^3*(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)+1/5*a^3*C*(8/3+cos
(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3/5*B*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3*a^3*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)+3*B*a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3
/8*d*x+3/8*c)+a^3*C*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*B*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+a^3*C*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.953253, size = 354, normalized size = 1.76 \begin{align*} \frac{192 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 90 \,{\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^{3} + 64 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} - 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 90 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/960*(192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^3 - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x -
60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*B*a^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 + 90*(12
*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin
(d*x + c))*C*a^3 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a^3 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3)/d

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Fricas [A]  time = 0.512691, size = 332, normalized size = 1.65 \begin{align*} \frac{15 \,{\left (23 \, B + 26 \, C\right )} a^{3} d x +{\left (40 \, B a^{3} \cos \left (d x + c\right )^{5} + 48 \,{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{4} + 10 \,{\left (23 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \,{\left (17 \, B + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \,{\left (23 \, B + 26 \, C\right )} a^{3} \cos \left (d x + c\right ) + 32 \,{\left (17 \, B + 19 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(23*B + 26*C)*a^3*d*x + (40*B*a^3*cos(d*x + c)^5 + 48*(3*B + C)*a^3*cos(d*x + c)^4 + 10*(23*B + 18*C
)*a^3*cos(d*x + c)^3 + 16*(17*B + 19*C)*a^3*cos(d*x + c)^2 + 15*(23*B + 26*C)*a^3*cos(d*x + c) + 32*(17*B + 19
*C)*a^3)*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)**7*(a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.22676, size = 329, normalized size = 1.64 \begin{align*} \frac{15 \,{\left (23 \, B a^{3} + 26 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (345 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 390 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 1955 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 2210 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 4554 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 5148 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 5814 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 5988 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3165 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4190 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1575 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1530 \, C a^{3} \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 )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(23*B*a^3 + 26*C*a^3)*(d*x + c) + 2*(345*B*a^3*tan(1/2*d*x + 1/2*c)^11 + 390*C*a^3*tan(1/2*d*x + 1/2
*c)^11 + 1955*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 2210*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 4554*B*a^3*tan(1/2*d*x + 1/2*
c)^7 + 5148*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 5814*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 5988*C*a^3*tan(1/2*d*x + 1/2*c)
^5 + 3165*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 4190*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 1575*B*a^3*tan(1/2*d*x + 1/2*c) +
1530*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d