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

Optimal. Leaf size=136 \[ \frac {\left (a^2 C+2 a b B+b^2 C\right ) \sin (c+d x)}{d}+\frac {\left (3 a^2 B+8 a b C+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 B+8 a b C+4 b^2 B\right )+\frac {a^2 B \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {a (a C+2 b B) \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4072, 4024, 4047, 2635, 8, 4044, 3013} \[ \frac {\left (a^2 C+2 a b B+b^2 C\right ) \sin (c+d x)}{d}+\frac {\left (3 a^2 B+8 a b C+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 B+8 a b C+4 b^2 B\right )+\frac {a^2 B \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {a (a C+2 b B) \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2*B + 4*b^2*B + 8*a*b*C)*x)/8 + ((2*a*b*B + a^2*C + b^2*C)*Sin[c + d*x])/d + ((3*a^2*B + 4*b^2*B + 8*a*b
*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(2*b*B + a*C)*Sin[c + d*
x]^3)/(3*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 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4024

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 4047

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.56, size = 118, normalized size = 0.87 \[ \frac {12 (c+d x) \left (3 a^2 B+8 a b C+4 b^2 B\right )+24 \left (3 a^2 C+6 a b B+4 b^2 C\right ) \sin (c+d x)+24 \left (a^2 B+2 a b C+b^2 B\right ) \sin (2 (c+d x))+3 a^2 B \sin (4 (c+d x))+8 a (a C+2 b B) \sin (3 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^2*B + 4*b^2*B + 8*a*b*C)*(c + d*x) + 24*(6*a*b*B + 3*a^2*C + 4*b^2*C)*Sin[c + d*x] + 24*(a^2*B + b^2*
B + 2*a*b*C)*Sin[2*(c + d*x)] + 8*a*(2*b*B + a*C)*Sin[3*(c + d*x)] + 3*a^2*B*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.49, size = 114, normalized size = 0.84 \[ \frac {3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} d x + {\left (6 \, B a^{2} \cos \left (d x + c\right )^{3} + 16 \, C a^{2} + 32 \, B a b + 24 \, C b^{2} + 8 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*d*x + (6*B*a^2*cos(d*x + c)^3 + 16*C*a^2 + 32*B*a*b + 24*C*b^2 + 8*(C*a^
2 + 2*B*a*b)*cos(d*x + c)^2 + 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.26, size = 437, normalized size = 3.21 \[ \frac {3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C b^{2} \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*(d*x + c) - 2*(15*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*tan(1/2*d*x +
1/2*c)^7 - 48*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b^2*tan(1/2*d*x + 1/2*c)^7
- 24*C*b^2*tan(1/2*d*x + 1/2*c)^7 - 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 40*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 80*B*a
*b*tan(1/2*d*x + 1/2*c)^5 + 24*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*C*b^2*tan(1
/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*B*a*b*tan(1/2*d*x +
1/2*c)^3 - 24*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*C*b^2*tan(1/2*d*x + 1/2*c)^3
- 15*B*a^2*tan(1/2*d*x + 1/2*c) - 24*C*a^2*tan(1/2*d*x + 1/2*c) - 48*B*a*b*tan(1/2*d*x + 1/2*c) - 24*C*a*b*ta
n(1/2*d*x + 1/2*c) - 12*B*b^2*tan(1/2*d*x + 1/2*c) - 24*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 +
1)^4)/d

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maple [A]  time = 1.38, size = 152, normalized size = 1.12 \[ \frac {B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 C a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} C \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 142, normalized size = 1.04 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 96 \, C b^{2} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C
*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b + 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b + 24*(2*d*x + 2*
c + sin(2*d*x + 2*c))*B*b^2 + 96*C*b^2*sin(d*x + c))/d

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mupad [B]  time = 3.78, size = 169, normalized size = 1.24 \[ \frac {3\,B\,a^2\,x}{8}+\frac {B\,b^2\,x}{2}+\frac {3\,C\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{d}+C\,a\,b\,x+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,B\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*B*a^2*x)/8 + (B*b^2*x)/2 + (3*C*a^2*sin(c + d*x))/(4*d) + (C*b^2*sin(c + d*x))/d + C*a*b*x + (B*a^2*sin(2*c
+ 2*d*x))/(4*d) + (B*a^2*sin(4*c + 4*d*x))/(32*d) + (B*b^2*sin(2*c + 2*d*x))/(4*d) + (C*a^2*sin(3*c + 3*d*x))
/(12*d) + (3*B*a*b*sin(c + d*x))/(2*d) + (B*a*b*sin(3*c + 3*d*x))/(6*d) + (C*a*b*sin(2*c + 2*d*x))/(2*d)

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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)**5*(a+b*sec(d*x+c))**2*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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