3.326 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=212 \[ \frac {a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+135 B+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac {(3 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

[Out]

1/8*a^3*(13*A+15*B+20*C)*arctanh(sin(d*x+c))/d+1/15*a^3*(38*A+45*B+55*C)*tan(d*x+c)/d+1/120*a^3*(109*A+135*B+1
40*C)*sec(d*x+c)*tan(d*x+c)/d+1/30*(11*A+15*B+10*C)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/20*(3*A+5
*B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/a/d+1/5*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d

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Rubi [A]  time = 0.64, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3043, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+135 B+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac {(3 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

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]^6,x]

[Out]

(a^3*(13*A + 15*B + 20*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(38*A + 45*B + 55*C)*Tan[c + d*x])/(15*d) + (a^3
*(109*A + 135*B + 140*C)*Sec[c + d*x]*Tan[c + d*x])/(120*d) + ((11*A + 15*B + 10*C)*(a^3 + a^3*Cos[c + d*x])*S
ec[c + d*x]^2*Tan[c + d*x])/(30*d) + ((3*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*
a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*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 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 ^6(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^3 (a (3 A+5 B)+a (A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^2 \left (2 a^2 (11 A+15 B+10 C)+a^2 (7 A+5 B+20 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x)) \left (a^3 (109 A+135 B+140 C)+a^3 (43 A+45 B+80 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (a^4 (109 A+135 B+140 C)+\left (a^4 (43 A+45 B+80 C)+a^4 (109 A+135 B+140 C)\right ) \cos (c+d x)+a^4 (43 A+45 B+80 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (8 a^4 (38 A+45 B+55 C)+15 a^4 (13 A+15 B+20 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac {a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^3 (13 A+15 B+20 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (a^3 (38 A+45 B+55 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (a^3 (38 A+45 B+55 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [B]  time = 6.21, size = 931, normalized size = 4.39 \[ \frac {(-13 A-15 B-20 C) (\cos (c+d x) a+a)^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{64 d}+\frac {(13 A+15 B+20 C) (\cos (c+d x) a+a)^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{64 d}+\frac {A (\cos (c+d x) a+a)^3 \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{160 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}+\frac {(\cos (c+d x) a+a)^3 \left (79 A \sin \left (\frac {1}{2} (c+d x)\right )+60 B \sin \left (\frac {1}{2} (c+d x)\right )+20 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{960 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {(\cos (c+d x) a+a)^3 \left (79 A \sin \left (\frac {1}{2} (c+d x)\right )+60 B \sin \left (\frac {1}{2} (c+d x)\right )+20 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{960 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {(\cos (c+d x) a+a)^3 \left (38 A \sin \left (\frac {1}{2} (c+d x)\right )+45 B \sin \left (\frac {1}{2} (c+d x)\right )+55 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{120 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(\cos (c+d x) a+a)^3 \left (38 A \sin \left (\frac {1}{2} (c+d x)\right )+45 B \sin \left (\frac {1}{2} (c+d x)\right )+55 C \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{120 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(274 A+285 B+200 C) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{1920 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(-274 A-285 B-200 C) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{1920 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(17 A+5 B) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{640 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {(-17 A-5 B) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{640 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {A (\cos (c+d x) a+a)^3 \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{160 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

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]^6,x]

[Out]

((-13*A - 15*B - 20*C)*(a + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(
64*d) + ((13*A + 15*B + 20*C)*(a + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/
2]^6)/(64*d) + ((17*A + 5*B)*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/(640*d*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^4) + ((274*A + 285*B + 200*C)*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/(1920*d*(Cos[(c + d*x)/2]
- Sin[(c + d*x)/2])^2) + (A*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*Sin[(c + d*x)/2])/(160*d*(Cos[(c + d*x
)/2] - Sin[(c + d*x)/2])^5) + (A*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*Sin[(c + d*x)/2])/(160*d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^5) + ((-17*A - 5*B)*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/(640*d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^4) + ((-274*A - 285*B - 200*C)*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/(192
0*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(79*A*Sin[(c + d*x
)/2] + 60*B*Sin[(c + d*x)/2] + 20*C*Sin[(c + d*x)/2]))/(960*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + ((a +
a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(79*A*Sin[(c + d*x)/2] + 60*B*Sin[(c + d*x)/2] + 20*C*Sin[(c + d*x)/2]
))/(960*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(38*A*Sin[(c
+ d*x)/2] + 45*B*Sin[(c + d*x)/2] + 55*C*Sin[(c + d*x)/2]))/(120*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (
(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(38*A*Sin[(c + d*x)/2] + 45*B*Sin[(c + d*x)/2] + 55*C*Sin[(c + d*x
)/2]))/(120*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.44, size = 180, normalized size = 0.85 \[ \frac {15 \, {\left (13 \, A + 15 \, B + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (13 \, A + 15 \, B + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (38 \, A + 45 \, B + 55 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (13 \, A + 15 \, B + 12 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 15 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

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)^6,x, algorithm="fricas")

[Out]

1/240*(15*(13*A + 15*B + 20*C)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(13*A + 15*B + 20*C)*a^3*cos(d*x
+ c)^5*log(-sin(d*x + c) + 1) + 2*(8*(38*A + 45*B + 55*C)*a^3*cos(d*x + c)^4 + 15*(13*A + 15*B + 12*C)*a^3*cos
(d*x + c)^3 + 8*(19*A + 15*B + 5*C)*a^3*cos(d*x + c)^2 + 30*(3*A + B)*a^3*cos(d*x + c) + 24*A*a^3)*sin(d*x + c
))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.67, size = 341, normalized size = 1.61 \[ \frac {15 \, {\left (13 \, A a^{3} + 15 \, B a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (13 \, A a^{3} + 15 \, B a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 300 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2560 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 660 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

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)^6,x, algorithm="giac")

[Out]

1/120*(15*(13*A*a^3 + 15*B*a^3 + 20*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(13*A*a^3 + 15*B*a^3 + 20*C
*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(195*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 225*B*a^3*tan(1/2*d*x + 1/2*c
)^9 + 300*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 910*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 1050*B*a^3*tan(1/2*d*x + 1/2*c)^7
- 1400*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1920*B*a^3*tan(1/2*d*x + 1/2*c)^5 +
2560*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 1330*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1830*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 21
20*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*A*a^3*tan(1/2*d*x + 1/2*c) + 735*B*a^3*tan(1/2*d*x + 1/2*c) + 660*C*a^3*
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 = 0.54, size = 316, normalized size = 1.49 \[ \frac {13 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {5 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {38 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {19 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {15 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {11 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

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)^6,x)

[Out]

13/8/d*A*a^3*sec(d*x+c)*tan(d*x+c)+13/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3*B*tan(d*x+c)+5/2/d*C*a^3*ln(
sec(d*x+c)+tan(d*x+c))+38/15/d*A*a^3*tan(d*x+c)+19/15/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+15/8/d*a^3*B*sec(d*x+c)*
tan(d*x+c)+15/8/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+11/3/d*C*a^3*tan(d*x+c)+3/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+
1/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+3/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)+1/5/d*A*a^3*tan(d*x+c)*sec(d*x+c)^4+1/4/d*
a^3*B*tan(d*x+c)*sec(d*x+c)^3+1/3/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2

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maxima [B]  time = 0.37, size = 446, normalized size = 2.10 \[ \frac {16 \, {\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} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, 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 )} - 15 \, 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 )} - 60 \, A 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 )} - 180 \, 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 )} - 180 \, 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 )} + 120 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right ) + 720 \, C a^{3} \tan \left (d x + c\right )}{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)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 45*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)) - 15*B*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)) - 60*A*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)) - 180*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)) - 180*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)) + 120*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 240*B*a^3*tan(d*x + c) + 720*
C*a^3*tan(d*x + c))/d

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mupad [B]  time = 4.73, size = 292, normalized size = 1.38 \[ \frac {a^3\,\mathrm {atanh}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (13\,A+15\,B+20\,C\right )}{2\,\left (\frac {13\,A\,a^3}{2}+\frac {15\,B\,a^3}{2}+10\,C\,a^3\right )}\right )\,\left (13\,A+15\,B+20\,C\right )}{4\,d}-\frac {\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}+5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {91\,A\,a^3}{6}-\frac {35\,B\,a^3}{2}-\frac {70\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+32\,B\,a^3+\frac {128\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {133\,A\,a^3}{6}-\frac {61\,B\,a^3}{2}-\frac {106\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {49\,B\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

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)^6,x)

[Out]

(a^3*atanh((a^3*tan(c/2 + (d*x)/2)*(13*A + 15*B + 20*C))/(2*((13*A*a^3)/2 + (15*B*a^3)/2 + 10*C*a^3)))*(13*A +
15*B + 20*C))/(4*d) - (tan(c/2 + (d*x)/2)^9*((13*A*a^3)/4 + (15*B*a^3)/4 + 5*C*a^3) - tan(c/2 + (d*x)/2)^7*((
91*A*a^3)/6 + (35*B*a^3)/2 + (70*C*a^3)/3) - tan(c/2 + (d*x)/2)^3*((133*A*a^3)/6 + (61*B*a^3)/2 + (106*C*a^3)/
3) + tan(c/2 + (d*x)/2)^5*((416*A*a^3)/15 + 32*B*a^3 + (128*C*a^3)/3) + tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (49
*B*a^3)/4 + 11*C*a^3))/(d*(5*tan(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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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)**6,x)

[Out]

Timed out

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