3.120 \(\int \cos ^7(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=254 \[ \frac {a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac {a^4 (247 A+308 C) \sin (c+d x) \cos ^2(c+d x)}{210 d}+\frac {a^4 (11 A+14 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac {(109 A+126 C) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{210 d}+\frac {1}{4} a^4 x (11 A+14 C)+\frac {(8 A+7 C) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 d}+\frac {A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac {2 a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{21 d} \]

[Out]

1/4*a^4*(11*A+14*C)*x+1/105*a^4*(454*A+581*C)*sin(d*x+c)/d+1/4*a^4*(11*A+14*C)*cos(d*x+c)*sin(d*x+c)/d+1/210*a
^4*(247*A+308*C)*cos(d*x+c)^2*sin(d*x+c)/d+2/21*a*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/7*A*cos(d*x
+c)^6*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+1/35*(8*A+7*C)*cos(d*x+c)^4*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d+1/210*(1
09*A+126*C)*cos(d*x+c)^3*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.72, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4087, 4017, 3996, 3787, 2635, 8, 2637} \[ \frac {a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac {a^4 (247 A+308 C) \sin (c+d x) \cos ^2(c+d x)}{210 d}+\frac {a^4 (11 A+14 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac {(8 A+7 C) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 d}+\frac {(109 A+126 C) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{210 d}+\frac {1}{4} a^4 x (11 A+14 C)+\frac {A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac {2 a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(11*A + 14*C)*x)/4 + (a^4*(454*A + 581*C)*Sin[c + d*x])/(105*d) + (a^4*(11*A + 14*C)*Cos[c + d*x]*Sin[c +
d*x])/(4*d) + (a^4*(247*A + 308*C)*Cos[c + d*x]^2*Sin[c + d*x])/(210*d) + (2*a*A*Cos[c + d*x]^5*(a + a*Sec[c
+ d*x])^3*Sin[c + d*x])/(21*d) + (A*Cos[c + d*x]^6*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(7*d) + ((8*A + 7*C)*C
os[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*d) + ((109*A + 126*C)*Cos[c + d*x]^3*(a^4 + a^4*Sec
[c + d*x])*Sin[c + d*x])/(210*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 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 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {\int \cos ^6(c+d x) (a+a \sec (c+d x))^4 (4 a A+a (2 A+7 C) \sec (c+d x)) \, dx}{7 a}\\ &=\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (6 a^2 (8 A+7 C)+2 a^2 (10 A+21 C) \sec (c+d x)\right ) \, dx}{42 a}\\ &=\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (4 a^3 (109 A+126 C)+98 a^3 (2 A+3 C) \sec (c+d x)\right ) \, dx}{210 a}\\ &=\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (12 a^4 (247 A+308 C)+24 a^4 (69 A+91 C) \sec (c+d x)\right ) \, dx}{840 a}\\ &=\frac {a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}-\frac {\int \cos ^2(c+d x) \left (-1260 a^5 (11 A+14 C)-24 a^5 (454 A+581 C) \sec (c+d x)\right ) \, dx}{2520 a}\\ &=\frac {a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac {1}{2} \left (a^4 (11 A+14 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{105} \left (a^4 (454 A+581 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac {a^4 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac {1}{4} \left (a^4 (11 A+14 C)\right ) \int 1 \, dx\\ &=\frac {1}{4} a^4 (11 A+14 C) x+\frac {a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac {a^4 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac {2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac {(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac {(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 145, normalized size = 0.57 \[ \frac {a^4 (105 (323 A+392 C) \sin (c+d x)+420 (31 A+32 C) \sin (2 (c+d x))+5495 A \sin (3 (c+d x))+2100 A \sin (4 (c+d x))+651 A \sin (5 (c+d x))+140 A \sin (6 (c+d x))+15 A \sin (7 (c+d x))+11760 A c+18480 A d x+4060 C \sin (3 (c+d x))+840 C \sin (4 (c+d x))+84 C \sin (5 (c+d x))+23520 C d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(11760*A*c + 18480*A*d*x + 23520*C*d*x + 105*(323*A + 392*C)*Sin[c + d*x] + 420*(31*A + 32*C)*Sin[2*(c +
d*x)] + 5495*A*Sin[3*(c + d*x)] + 4060*C*Sin[3*(c + d*x)] + 2100*A*Sin[4*(c + d*x)] + 840*C*Sin[4*(c + d*x)] +
651*A*Sin[5*(c + d*x)] + 84*C*Sin[5*(c + d*x)] + 140*A*Sin[6*(c + d*x)] + 15*A*Sin[7*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.45, size = 146, normalized size = 0.57 \[ \frac {105 \, {\left (11 \, A + 14 \, C\right )} a^{4} d x + {\left (60 \, A a^{4} \cos \left (d x + c\right )^{6} + 280 \, A a^{4} \cos \left (d x + c\right )^{5} + 12 \, {\left (48 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \, {\left (11 \, A + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 4 \, {\left (227 \, A + 238 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 105 \, {\left (11 \, A + 14 \, C\right )} a^{4} \cos \left (d x + c\right ) + 4 \, {\left (454 \, A + 581 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/420*(105*(11*A + 14*C)*a^4*d*x + (60*A*a^4*cos(d*x + c)^6 + 280*A*a^4*cos(d*x + c)^5 + 12*(48*A + 7*C)*a^4*c
os(d*x + c)^4 + 70*(11*A + 6*C)*a^4*cos(d*x + c)^3 + 4*(227*A + 238*C)*a^4*cos(d*x + c)^2 + 105*(11*A + 14*C)*
a^4*cos(d*x + c) + 4*(454*A + 581*C)*a^4)*sin(d*x + c))/d

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giac [A]  time = 0.36, size = 278, normalized size = 1.09 \[ \frac {105 \, {\left (11 \, A a^{4} + 14 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (1155 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 1470 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 7700 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 9800 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 21791 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 27734 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 33792 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 43008 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 31521 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 39914 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 14700 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21560 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5565 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5250 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/420*(105*(11*A*a^4 + 14*C*a^4)*(d*x + c) + 2*(1155*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 1470*C*a^4*tan(1/2*d*x +
1/2*c)^13 + 7700*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 9800*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 21791*A*a^4*tan(1/2*d*x
+ 1/2*c)^9 + 27734*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 33792*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 43008*C*a^4*tan(1/2*d*x
+ 1/2*c)^7 + 31521*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 39914*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 14700*A*a^4*tan(1/2*d*
x + 1/2*c)^3 + 21560*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 5565*A*a^4*tan(1/2*d*x + 1/2*c) + 5250*C*a^4*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d

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maple [A]  time = 2.26, size = 322, normalized size = 1.27 \[ \frac {\frac {A \,a^{4} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}+\frac {a^{4} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 A \,a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+4 a^{4} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {6 A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{4} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/7*A*a^4*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c)+1/5*a^4*C*(8/3+cos(d*x+c)^4+4/
3*cos(d*x+c)^2)*sin(d*x+c)+4*A*a^4*(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)+4*a^4*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)+6/5*A*a^4*(8/3+cos(d*x+c)^4+4/3*cos
(d*x+c)^2)*sin(d*x+c)+2*a^4*C*(2+cos(d*x+c)^2)*sin(d*x+c)+4*A*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c
)+3/8*d*x+3/8*c)+4*a^4*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*A*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*C
*sin(d*x+c))

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maxima [A]  time = 0.44, size = 319, normalized size = 1.26 \[ -\frac {48 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} A a^{4} - 672 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 560 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 112 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{4} + 3360 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 1680 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 1680 \, C a^{4} \sin \left (d x + c\right )}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/1680*(48*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*A*a^4 - 672*(3*sin(d*
x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 + 35*(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))*A*a^4 + 560*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 210*(12*d*x + 12*c + sin(
4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 112*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^4
+ 3360*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C
*a^4 - 1680*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 1680*C*a^4*sin(d*x + c))/d

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mupad [B]  time = 4.78, size = 323, normalized size = 1.27 \[ \frac {\left (\frac {11\,A\,a^4}{2}+7\,C\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (\frac {110\,A\,a^4}{3}+\frac {140\,C\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {3113\,A\,a^4}{30}+\frac {1981\,C\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {5632\,A\,a^4}{35}+\frac {1024\,C\,a^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {1501\,A\,a^4}{10}+\frac {2851\,C\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (70\,A\,a^4+\frac {308\,C\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {53\,A\,a^4}{2}+25\,C\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^4\,\mathrm {atan}\left (\frac {a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,A+14\,C\right )}{2\,\left (\frac {11\,A\,a^4}{2}+7\,C\,a^4\right )}\right )\,\left (11\,A+14\,C\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^4,x)

[Out]

(tan(c/2 + (d*x)/2)*((53*A*a^4)/2 + 25*C*a^4) + tan(c/2 + (d*x)/2)^13*((11*A*a^4)/2 + 7*C*a^4) + tan(c/2 + (d*
x)/2)^11*((110*A*a^4)/3 + (140*C*a^4)/3) + tan(c/2 + (d*x)/2)^3*(70*A*a^4 + (308*C*a^4)/3) + tan(c/2 + (d*x)/2
)^5*((1501*A*a^4)/10 + (2851*C*a^4)/15) + tan(c/2 + (d*x)/2)^9*((3113*A*a^4)/30 + (1981*C*a^4)/15) + tan(c/2 +
(d*x)/2)^7*((5632*A*a^4)/35 + (1024*C*a^4)/5))/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4 + 35*tan(
c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d
*x)/2)^14 + 1)) + (a^4*atan((a^4*tan(c/2 + (d*x)/2)*(11*A + 14*C))/(2*((11*A*a^4)/2 + 7*C*a^4)))*(11*A + 14*C)
)/(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)**7*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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