3.9.69 \(\int \cos ^5(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [869]

Optimal. Leaf size=156 \[ \frac {1}{8} (3 A b+3 a B+4 b C) x+\frac {(4 a A+5 b B+5 a C) \sin (c+d x)}{5 d}+\frac {(3 A b+3 a B+4 b C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {(4 a A+5 b B+5 a C) \sin ^3(c+d x)}{15 d} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 156, 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 = {4159, 4132, 2713, 4130, 2715, 8} \begin {gather*} -\frac {\sin ^3(c+d x) (4 a A+5 a C+5 b B)}{15 d}+\frac {\sin (c+d x) (4 a A+5 a C+5 b B)}{5 d}+\frac {\sin (c+d x) \cos (c+d x) (3 a B+3 A b+4 b C)}{8 d}+\frac {(a B+A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} x (3 a B+3 A b+4 b C)+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 4130

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 4132

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 4159

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

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Mathematica [A]
time = 0.42, size = 147, normalized size = 0.94 \begin {gather*} \frac {180 A b c+180 a B c+240 b c C+180 A b d x+180 a B d x+240 b C d x+60 (5 a A+8 b B+8 a C) \sin (c+d x)-160 (b B+a C) \sin ^3(c+d x)+120 (A b+a B+b C) \sin (2 (c+d x))+50 a A \sin (3 (c+d x))+15 A b \sin (4 (c+d x))+15 a B \sin (4 (c+d x))+6 a A \sin (5 (c+d x))}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(180*A*b*c + 180*a*B*c + 240*b*c*C + 180*A*b*d*x + 180*a*B*d*x + 240*b*C*d*x + 60*(5*a*A + 8*b*B + 8*a*C)*Sin[
c + d*x] - 160*(b*B + a*C)*Sin[c + d*x]^3 + 120*(A*b + a*B + b*C)*Sin[2*(c + d*x)] + 50*a*A*Sin[3*(c + d*x)] +
15*A*b*Sin[4*(c + d*x)] + 15*a*B*Sin[4*(c + d*x)] + 6*a*A*Sin[5*(c + d*x)])/(480*d)

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Maple [A]
time = 0.12, size = 173, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(173\)
default	\(\frac {\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(173\)
risch	\(\frac {3 A b x}{8}+\frac {3 a B x}{8}+\frac {b x C}{2}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) b B}{4 d}+\frac {3 \sin \left (d x +c \right ) a C}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {A b \sin \left (4 d x +4 c \right )}{32 d}+\frac {B a \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) b B}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a C}{12 d}+\frac {A b \sin \left (2 d x +2 c \right )}{4 d}+\frac {B a \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C b}{4 d}\)	\(200\)
norman	\(\frac {\left (\frac {3}{8} A b +\frac {3}{8} B a +\frac {1}{2} C b \right ) x +\left (-\frac {15}{8} A b -\frac {15}{8} B a -\frac {5}{2} C b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {15}{8} A b -\frac {15}{8} B a -\frac {5}{2} C b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {3}{8} B a +\frac {1}{2} C b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {3}{8} B a +\frac {1}{2} C b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {3}{8} B a +\frac {1}{2} C b \right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{8} A b +\frac {9}{8} B a +\frac {3}{2} C b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{8} A b +\frac {9}{8} B a +\frac {3}{2} C b \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 \left (19 a A +5 b B +5 a C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (2 a A -3 A b -3 B a -2 b B -2 a C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (2 a A +3 A b +3 B a -2 b B -2 a C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (8 a A -5 A b -5 B a +8 b B +8 a C -4 C b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (8 a A +5 A b +5 B a +8 b B +8 a C +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (88 a A -5 A b -5 B a -40 b B -40 a C +60 C b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {\left (88 a A +5 A b +5 B a -40 b B -40 a C -60 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}\)	\(495\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*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)+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*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)+C*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]
time = 0.31, size = 166, normalized size = 1.06 \begin {gather*} \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 )} B a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C 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 b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*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))*B*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*A*b - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*
b)/d

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Fricas [A]
time = 3.11, size = 121, normalized size = 0.78 \begin {gather*} \frac {15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 16 \, {\left (4 \, A + 5 \, C\right )} a + 80 \, B b + 15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*B*a + (3*A + 4*C)*b)*d*x + (24*A*a*cos(d*x + c)^4 + 30*(B*a + A*b)*cos(d*x + c)^3 + 8*((4*A + 5*C
)*a + 5*B*b)*cos(d*x + c)^2 + 16*(4*A + 5*C)*a + 80*B*b + 15*(3*B*a + (3*A + 4*C)*b)*cos(d*x + c))*sin(d*x + c
))/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (144) = 288\).
time = 0.50, size = 437, normalized size = 2.80 \begin {gather*} \frac {15 \, {\left (3 \, B a + 3 \, A b + 4 \, C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, C b \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 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*B*a + 3*A*b + 4*C*b)*(d*x + c) + 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*a*tan(1/2*d*x + 1/2*c)^
9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 + 120*B*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*b*t
an(1/2*d*x + 1/2*c)^9 + 160*A*a*tan(1/2*d*x + 1/2*c)^7 - 30*B*a*tan(1/2*d*x + 1/2*c)^7 + 320*C*a*tan(1/2*d*x +
1/2*c)^7 - 30*A*b*tan(1/2*d*x + 1/2*c)^7 + 320*B*b*tan(1/2*d*x + 1/2*c)^7 - 120*C*b*tan(1/2*d*x + 1/2*c)^7 +
464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*b*tan(1/2*d*x + 1/2*c)^5 + 160*A*a*tan
(1/2*d*x + 1/2*c)^3 + 30*B*a*tan(1/2*d*x + 1/2*c)^3 + 320*C*a*tan(1/2*d*x + 1/2*c)^3 + 30*A*b*tan(1/2*d*x + 1/
2*c)^3 + 320*B*b*tan(1/2*d*x + 1/2*c)^3 + 120*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 75*B
*a*tan(1/2*d*x + 1/2*c) + 120*C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 120*B*b*tan(1/2*d*x + 1
/2*c) + 60*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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Mupad [B]
time = 7.63, size = 258, normalized size = 1.65 \begin {gather*} \frac {x\,\left (\frac {3\,A\,b}{4}+\frac {3\,B\,a}{4}+C\,b\right )}{2}+\frac {\left (2\,A\,a-\frac {5\,A\,b}{4}-\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a}{3}-\frac {A\,b}{2}-\frac {B\,a}{2}+\frac {16\,B\,b}{3}+\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,b}{3}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a}{3}+\frac {A\,b}{2}+\frac {B\,a}{2}+\frac {16\,B\,b}{3}+\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a+C\,b\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*((3*A*b)/4 + (3*B*a)/4 + C*b))/2 + (tan(c/2 + (d*x)/2)^9*(2*A*a - (5*A*b)/4 - (5*B*a)/4 + 2*B*b + 2*C*a - C
*b) + tan(c/2 + (d*x)/2)^3*((8*A*a)/3 + (A*b)/2 + (B*a)/2 + (16*B*b)/3 + (16*C*a)/3 + 2*C*b) + tan(c/2 + (d*x)
/2)^7*((8*A*a)/3 - (A*b)/2 - (B*a)/2 + (16*B*b)/3 + (16*C*a)/3 - 2*C*b) + tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/
4 + (5*B*a)/4 + 2*B*b + 2*C*a + C*b) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*B*b)/3 + (20*C*a)/3))/(d*(5*ta
n(c/2 + (d*x)/2)^2 + 10*tan(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))

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