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

Optimal. Leaf size=185 \[ \frac{a^3 (38 B+45 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (43 B+45 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^3 (13 B+15 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(7 B+5 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

[Out]

(a^3*(13*B + 15*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(38*B + 45*C)*Tan[c + d*x])/(15*d) + (a^3*(13*B + 15*C)
*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^3*(43*B + 45*C)*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + ((7*B + 5*C)*(a^3
+ a^3*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c +
d*x])/(5*d)

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Rubi [A]  time = 0.533425, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.225, Rules used = {3029, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{a^3 (38 B+45 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (43 B+45 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^3 (13 B+15 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(7 B+5 C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(13*B + 15*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(38*B + 45*C)*Tan[c + d*x])/(15*d) + (a^3*(13*B + 15*C)
*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^3*(43*B + 45*C)*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + ((7*B + 5*C)*(a^3
+ a^3*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c +
d*x])/(5*d)

Rule 3029

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

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 8

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

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\int (a+a \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^6(c+d x) \, dx\\ &=\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+a \cos (c+d x))^2 (a (7 B+5 C)+a (2 B+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+a \cos (c+d x)) \left (a^2 (43 B+45 C)+2 a^2 (11 B+15 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int \left (a^3 (43 B+45 C)+\left (2 a^3 (11 B+15 C)+a^3 (43 B+45 C)\right ) \cos (c+d x)+2 a^3 (11 B+15 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^3 (43 B+45 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{60} \int \left (15 a^3 (13 B+15 C)+4 a^3 (38 B+45 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (43 B+45 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} \left (a^3 (13 B+15 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{15} \left (a^3 (38 B+45 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (13 B+15 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 (43 B+45 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^3 (13 B+15 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (38 B+45 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^3 (13 B+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (38 B+45 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 B+15 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 (43 B+45 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 B+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.42365, size = 294, normalized size = 1.59 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (13 B+15 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 B+5 C) \sin (2 c+d x)+750 B \sin (c+2 d x)+750 B \sin (3 c+2 d x)+1520 B \sin (2 c+3 d x)+195 B \sin (3 c+4 d x)+195 B \sin (5 c+4 d x)+304 B \sin (4 c+5 d x)+80 (29 B+30 C) \sin (d x)+570 C \sin (c+2 d x)+570 C \sin (3 c+2 d x)+1680 C \sin (2 c+3 d x)-120 C \sin (4 c+3 d x)+225 C \sin (3 c+4 d x)+225 C \sin (5 c+4 d x)+360 C \sin (4 c+5 d x))\right )}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^5*(240*(13*B + 15*C)*Cos[c + d*x]^5*(Log[Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(80*(29*B + 30*C)*Sin[d*x] - 2
40*(3*B + 5*C)*Sin[2*c + d*x] + 750*B*Sin[c + 2*d*x] + 570*C*Sin[c + 2*d*x] + 750*B*Sin[3*c + 2*d*x] + 570*C*S
in[3*c + 2*d*x] + 1520*B*Sin[2*c + 3*d*x] + 1680*C*Sin[2*c + 3*d*x] - 120*C*Sin[4*c + 3*d*x] + 195*B*Sin[3*c +
4*d*x] + 225*C*Sin[3*c + 4*d*x] + 195*B*Sin[5*c + 4*d*x] + 225*C*Sin[5*c + 4*d*x] + 304*B*Sin[4*c + 5*d*x] +
360*C*Sin[4*c + 5*d*x])))/(15360*d)

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Maple [A]  time = 0.08, size = 234, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{38\,{a}^{3}B\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{19\,{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{13\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.33351, size = 455, normalized size = 2.46 \begin{align*} \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 )} B a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, 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 )} - 15 \, 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 )} - 60 \, 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 )} + 240 \, C a^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c
))*B*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 45*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)) - 15*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)) - 60*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)) + 240*C*a^3*ta
n(d*x + c))/d

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Fricas [A]  time = 1.72745, size = 431, normalized size = 2.33 \begin{align*} \frac{15 \,{\left (13 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (13 \, B + 15 \, 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 \, B + 45 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (13 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (19 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \,{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, B a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.60367, size = 332, normalized size = 1.79 \begin{align*} \frac{15 \,{\left (13 \, B a^{3} + 15 \, 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 \, B a^{3} + 15 \, 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 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 225 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 910 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1050 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1664 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1920 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1330 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1830 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 765 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 735 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(13*B*a^3 + 15*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(13*B*a^3 + 15*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(195*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 225*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 910*B*a^3*tan(1/
2*d*x + 1/2*c)^7 - 1050*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1920*C*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1330*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 1830*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*B*a^3*tan(1/2*d*x
+ 1/2*c) + 735*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d