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

Optimal. Leaf size=244 \[ \frac {a^3 (34 A+38 B+45 C) \tan (c+d x)}{15 d}+\frac {a^3 (23 A+26 B+30 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (73 A+86 B+90 C) \tan (c+d x) \sec ^2(c+d x)}{120 d}+\frac {a^3 (23 A+26 B+30 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {(31 A+42 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{120 d}+\frac {(A+2 B) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{10 a d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{6 d} \]

[Out]

1/16*a^3*(23*A+26*B+30*C)*arctanh(sin(d*x+c))/d+1/15*a^3*(34*A+38*B+45*C)*tan(d*x+c)/d+1/16*a^3*(23*A+26*B+30*
C)*sec(d*x+c)*tan(d*x+c)/d+1/120*a^3*(73*A+86*B+90*C)*sec(d*x+c)^2*tan(d*x+c)/d+1/120*(31*A+42*B+30*C)*(a^3+a^
3*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d+1/10*(A+2*B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/a/d+1/6*A*
(a+a*cos(d*x+c))^3*sec(d*x+c)^5*tan(d*x+c)/d

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Rubi [A]  time = 0.70, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {3043, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a^3 (34 A+38 B+45 C) \tan (c+d x)}{15 d}+\frac {a^3 (23 A+26 B+30 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (73 A+86 B+90 C) \tan (c+d x) \sec ^2(c+d x)}{120 d}+\frac {a^3 (23 A+26 B+30 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {(31 A+42 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{120 d}+\frac {(A+2 B) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{10 a d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(23*A + 26*B + 30*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(34*A + 38*B + 45*C)*Tan[c + d*x])/(15*d) + (a^
3*(23*A + 26*B + 30*C)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^3*(73*A + 86*B + 90*C)*Sec[c + d*x]^2*Tan[c + d*
x])/(120*d) + ((31*A + 42*B + 30*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(120*d) + ((A + 2*B)
*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(10*a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^5*T
an[c + d*x])/(6*d)

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

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

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^3 (3 a (A+2 B)+2 a (A+3 C) \cos (c+d x)) \sec ^6(c+d x) \, dx}{6 a}\\ &=\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^2 \left (a^2 (31 A+42 B+30 C)+2 a^2 (8 A+6 B+15 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{30 a}\\ &=\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^3 (73 A+86 B+90 C)+6 a^3 (21 A+22 B+30 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (3 a^4 (73 A+86 B+90 C)+\left (6 a^4 (21 A+22 B+30 C)+3 a^4 (73 A+86 B+90 C)\right ) \cos (c+d x)+6 a^4 (21 A+22 B+30 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac {a^3 (73 A+86 B+90 C) \sec ^2(c+d x) \tan (c+d x)}{120 d}+\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (45 a^4 (23 A+26 B+30 C)+24 a^4 (34 A+38 B+45 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac {a^3 (73 A+86 B+90 C) \sec ^2(c+d x) \tan (c+d x)}{120 d}+\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{8} \left (a^3 (23 A+26 B+30 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{15} \left (a^3 (34 A+38 B+45 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (23 A+26 B+30 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 (73 A+86 B+90 C) \sec ^2(c+d x) \tan (c+d x)}{120 d}+\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{16} \left (a^3 (23 A+26 B+30 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (34 A+38 B+45 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^3 (23 A+26 B+30 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^3 (34 A+38 B+45 C) \tan (c+d x)}{15 d}+\frac {a^3 (23 A+26 B+30 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 (73 A+86 B+90 C) \sec ^2(c+d x) \tan (c+d x)}{120 d}+\frac {(31 A+42 B+30 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{10 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 1.95, size = 265, normalized size = 1.09 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (240 (23 A+26 B+30 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-2 \sin (c+d x) (16 (344 A+328 B+315 C) \cos (c+d x)+20 (115 A+114 B+102 C) \cos (2 (c+d x))+1904 A \cos (3 (c+d x))+345 A \cos (4 (c+d x))+272 A \cos (5 (c+d x))+2275 A+2128 B \cos (3 (c+d x))+390 B \cos (4 (c+d x))+304 B \cos (5 (c+d x))+1890 B+2280 C \cos (3 (c+d x))+450 C \cos (4 (c+d x))+360 C \cos (5 (c+d x))+1590 C)\right )}{30720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30720*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^6*(240*(23*A + 26*B + 30*C)*Cos[c + d*x]^6*
(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 2*(2275*A + 1890*B + 1
590*C + 16*(344*A + 328*B + 315*C)*Cos[c + d*x] + 20*(115*A + 114*B + 102*C)*Cos[2*(c + d*x)] + 1904*A*Cos[3*(
c + d*x)] + 2128*B*Cos[3*(c + d*x)] + 2280*C*Cos[3*(c + d*x)] + 345*A*Cos[4*(c + d*x)] + 390*B*Cos[4*(c + d*x)
] + 450*C*Cos[4*(c + d*x)] + 272*A*Cos[5*(c + d*x)] + 304*B*Cos[5*(c + d*x)] + 360*C*Cos[5*(c + d*x)])*Sin[c +
d*x]))/d

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fricas [A]  time = 0.46, size = 203, normalized size = 0.83 \[ \frac {15 \, {\left (23 \, A + 26 \, B + 30 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (23 \, A + 26 \, B + 30 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (34 \, A + 38 \, B + 45 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \, {\left (23 \, A + 26 \, B + 30 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (17 \, A + 19 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \, {\left (23 \, A + 18 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 48 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 40 \, A a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(23*A + 26*B + 30*C)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(23*A + 26*B + 30*C)*a^3*cos(d*x
+ c)^6*log(-sin(d*x + c) + 1) + 2*(16*(34*A + 38*B + 45*C)*a^3*cos(d*x + c)^5 + 15*(23*A + 26*B + 30*C)*a^3*co
s(d*x + c)^4 + 16*(17*A + 19*B + 15*C)*a^3*cos(d*x + c)^3 + 10*(23*A + 18*B + 6*C)*a^3*cos(d*x + c)^2 + 48*(3*
A + B)*a^3*cos(d*x + c) + 40*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 0.68, size = 392, normalized size = 1.61 \[ \frac {15 \, {\left (23 \, A a^{3} + 26 \, B a^{3} + 30 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (23 \, A a^{3} + 26 \, B a^{3} + 30 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (345 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 390 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 450 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1955 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2210 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2550 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4554 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5148 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5940 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5814 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5988 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 7500 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4190 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5130 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1575 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1530 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1470 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(23*A*a^3 + 26*B*a^3 + 30*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(23*A*a^3 + 26*B*a^3 + 30*C
*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(345*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 390*B*a^3*tan(1/2*d*x + 1/2*
c)^11 + 450*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 1955*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 2210*B*a^3*tan(1/2*d*x + 1/2*c
)^9 - 2550*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 4554*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 5148*B*a^3*tan(1/2*d*x + 1/2*c)^
7 + 5940*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 5814*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 5988*B*a^3*tan(1/2*d*x + 1/2*c)^5
- 7500*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 3165*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 4190*B*a^3*tan(1/2*d*x + 1/2*c)^3 +
5130*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1575*A*a^3*tan(1/2*d*x + 1/2*c) - 1530*B*a^3*tan(1/2*d*x + 1/2*c) - 1470*C
*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

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maple [A]  time = 0.59, size = 385, normalized size = 1.58 \[ \frac {34 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {17 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {13 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 C \,a^{3} \tan \left (d x +c \right )}{d}+\frac {23 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {23 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {23 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {38 a^{3} B \tan \left (d x +c \right )}{15 d}+\frac {19 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {15 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {3 a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

34/15/d*A*a^3*tan(d*x+c)+17/15/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+13/8/d*a^3*B*sec(d*x+c)*tan(d*x+c)+13/8/d*a^3*B
*ln(sec(d*x+c)+tan(d*x+c))+3/d*C*a^3*tan(d*x+c)+23/24/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+23/16/d*A*a^3*sec(d*x+c)
*tan(d*x+c)+23/16/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+38/15/d*a^3*B*tan(d*x+c)+19/15/d*a^3*B*tan(d*x+c)*sec(d*x+
c)^2+15/8/d*C*a^3*sec(d*x+c)*tan(d*x+c)+15/8/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/5/d*A*a^3*tan(d*x+c)*sec(d*x+
c)^4+3/4/d*a^3*B*tan(d*x+c)*sec(d*x+c)^3+1/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+1/6/d*A*a^3*tan(d*x+c)*sec(d*x+c)^5
+1/5/d*a^3*B*tan(d*x+c)*sec(d*x+c)^4+1/4/d*C*a^3*tan(d*x+c)*sec(d*x+c)^3

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maxima [B]  time = 0.37, size = 559, normalized size = 2.29 \[ \frac {96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 32 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 5 \, A a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, C a^{3} \tan \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(96*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^3 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 480*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*B*a^3 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 5*A*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3
+ 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15
*log(sin(d*x + c) - 1)) - 90*A*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 +
1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 90*B*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(si
n(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*C*a^3*(2*(3*sin
(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*
x + c) - 1)) - 120*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1))
- 360*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480*C*a^3
*tan(d*x + c))/d

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mupad [B]  time = 4.71, size = 337, normalized size = 1.38 \[ \frac {a^3\,\mathrm {atanh}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,A+26\,B+30\,C\right )}{4\,\left (\frac {23\,A\,a^3}{4}+\frac {13\,B\,a^3}{2}+\frac {15\,C\,a^3}{2}\right )}\right )\,\left (23\,A+26\,B+30\,C\right )}{8\,d}-\frac {\left (\frac {23\,A\,a^3}{8}+\frac {13\,B\,a^3}{4}+\frac {15\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-\frac {391\,A\,a^3}{24}-\frac {221\,B\,a^3}{12}-\frac {85\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {759\,A\,a^3}{20}+\frac {429\,B\,a^3}{10}+\frac {99\,C\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {969\,A\,a^3}{20}-\frac {499\,B\,a^3}{10}-\frac {125\,C\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {211\,A\,a^3}{8}+\frac {419\,B\,a^3}{12}+\frac {171\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {105\,A\,a^3}{8}-\frac {51\,B\,a^3}{4}-\frac {49\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^7,x)

[Out]

(a^3*atanh((a^3*tan(c/2 + (d*x)/2)*(23*A + 26*B + 30*C))/(4*((23*A*a^3)/4 + (13*B*a^3)/2 + (15*C*a^3)/2)))*(23
*A + 26*B + 30*C))/(8*d) - (tan(c/2 + (d*x)/2)^11*((23*A*a^3)/8 + (13*B*a^3)/4 + (15*C*a^3)/4) - tan(c/2 + (d*
x)/2)^9*((391*A*a^3)/24 + (221*B*a^3)/12 + (85*C*a^3)/4) + tan(c/2 + (d*x)/2)^3*((211*A*a^3)/8 + (419*B*a^3)/1
2 + (171*C*a^3)/4) + tan(c/2 + (d*x)/2)^7*((759*A*a^3)/20 + (429*B*a^3)/10 + (99*C*a^3)/2) - tan(c/2 + (d*x)/2
)^5*((969*A*a^3)/20 + (499*B*a^3)/10 + (125*C*a^3)/2) - tan(c/2 + (d*x)/2)*((105*A*a^3)/8 + (51*B*a^3)/4 + (49
*C*a^3)/4))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x
)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

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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((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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