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

Optimal. Leaf size=141 \[ -\frac {a (4 A+5 (B+C)) \sin ^3(c+d x)}{15 d}+\frac {a (4 A+5 (B+C)) \sin (c+d x)}{5 d}+\frac {a (3 (A+B)+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} a x (3 (A+B)+4 C)+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

1/8*a*(3*A+3*B+4*C)*x+1/5*a*(4*A+5*B+5*C)*sin(d*x+c)/d+1/8*a*(3*A+3*B+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*(A+B)
*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a*A*cos(d*x+c)^4*sin(d*x+c)/d-1/15*a*(4*A+5*B+5*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.23, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4074, 4047, 2633, 4045, 2635, 8} \[ -\frac {a (4 A+5 (B+C)) \sin ^3(c+d x)}{15 d}+\frac {a (4 A+5 (B+C)) \sin (c+d x)}{5 d}+\frac {a (3 (A+B)+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} a x (3 (A+B)+4 C)+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*(A + B) + 4*C)*x)/8 + (a*(4*A + 5*(B + C))*Sin[c + d*x])/(5*d) + (a*(3*(A + B) + 4*C)*Cos[c + d*x]*Sin[c
+ d*x])/(8*d) + (a*(A + B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*
(4*A + 5*(B + C))*Sin[c + d*x]^3)/(15*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (A+B)-a (4 A+5 (B+C)) \sec (c+d x)-5 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (A+B)-5 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (4 A+5 (B+C))) \int \cos ^3(c+d x) \, dx\\ &=\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{4} (a (3 (A+B)+4 C)) \int \cos ^2(c+d x) \, dx-\frac {(a (4 A+5 (B+C))) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {a (4 A+5 (B+C)) \sin (c+d x)}{5 d}+\frac {a (3 (A+B)+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 (B+C)) \sin ^3(c+d x)}{15 d}+\frac {1}{8} (a (3 (A+B)+4 C)) \int 1 \, dx\\ &=\frac {1}{8} a (3 (A+B)+4 C) x+\frac {a (4 A+5 (B+C)) \sin (c+d x)}{5 d}+\frac {a (3 (A+B)+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 (B+C)) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 94, normalized size = 0.67 \[ \frac {a \left (-160 (2 A+B+C) \sin ^3(c+d x)+480 (A+B+C) \sin (c+d x)+15 (4 (3 A+3 B+4 C) (c+d x)+8 (A+B+C) \sin (2 (c+d x))+(A+B) \sin (4 (c+d x)))+96 A \sin ^5(c+d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(480*(A + B + C)*Sin[c + d*x] - 160*(2*A + B + C)*Sin[c + d*x]^3 + 96*A*Sin[c + d*x]^5 + 15*(4*(3*A + 3*B +
4*C)*(c + d*x) + 8*(A + B + C)*Sin[2*(c + d*x)] + (A + B)*Sin[4*(c + d*x)])))/(480*d)

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fricas [A]  time = 0.44, size = 108, normalized size = 0.77 \[ \frac {15 \, {\left (3 \, A + 3 \, B + 4 \, C\right )} a d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, B + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 3 \, B + 4 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, B + 5 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*A + 3*B + 4*C)*a*d*x + (24*A*a*cos(d*x + c)^4 + 30*(A + B)*a*cos(d*x + c)^3 + 8*(4*A + 5*B + 5*C)
*a*cos(d*x + c)^2 + 15*(3*A + 3*B + 4*C)*a*cos(d*x + c) + 16*(4*A + 5*B + 5*C)*a)*sin(d*x + c))/d

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giac [A]  time = 0.24, size = 263, normalized size = 1.87 \[ \frac {15 \, {\left (3 \, A a + 3 \, B a + 4 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 130 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 290 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 190 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 350 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 440 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*A*a + 3*B*a + 4*C*a)*(d*x + c) + 2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 45*B*a*tan(1/2*d*x + 1/2*c)^9
+ 60*C*a*tan(1/2*d*x + 1/2*c)^9 + 130*A*a*tan(1/2*d*x + 1/2*c)^7 + 290*B*a*tan(1/2*d*x + 1/2*c)^7 + 200*C*a*t
an(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x
+ 1/2*c)^5 + 190*A*a*tan(1/2*d*x + 1/2*c)^3 + 350*B*a*tan(1/2*d*x + 1/2*c)^3 + 440*C*a*tan(1/2*d*x + 1/2*c)^3
+ 195*A*a*tan(1/2*d*x + 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c) + 180*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1
/2*c)^2 + 1)^5)/d

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maple [A]  time = 1.74, size = 173, normalized size = 1.23 \[ \frac {\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*a*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a*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)+a*B*(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*B*(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 = 0.37, size = 166, normalized size = 1.18 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*A*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*B*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*
a)/d

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mupad [B]  time = 5.70, size = 248, normalized size = 1.76 \[ \frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {13\,A\,a}{6}+\frac {29\,B\,a}{6}+\frac {10\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,a}{3}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {19\,A\,a}{6}+\frac {35\,B\,a}{6}+\frac {22\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+3\,B+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}+C\,a\right )}\right )\,\left (3\,A+3\,B+4\,C\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4 + 3*C*a) + tan(c/2 + (d*x)/2)^9*((3*A*a)/4 + (3*B*a)/4 + C*a) + t
an(c/2 + (d*x)/2)^7*((13*A*a)/6 + (29*B*a)/6 + (10*C*a)/3) + tan(c/2 + (d*x)/2)^3*((19*A*a)/6 + (35*B*a)/6 + (
22*C*a)/3) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*B*a)/3 + (20*C*a)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*ta
n(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (a*atan(
(a*tan(c/2 + (d*x)/2)*(3*A + 3*B + 4*C))/(4*((3*A*a)/4 + (3*B*a)/4 + C*a)))*(3*A + 3*B + 4*C))/(4*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+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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