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

Optimal. Leaf size=195 \[ \frac{a^4 (40 A+35 B+28 C) \sin (c+d x)}{8 d}+\frac{(20 A+35 B+28 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{60 d}+\frac{(32 A+35 B+28 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (48 A+35 B+28 C)+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (5 B+4 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

[Out]

(a^4*(48*A + 35*B + 28*C)*x)/8 + (a^4*A*ArcTanh[Sin[c + d*x]])/d + (a^4*(40*A + 35*B + 28*C)*Sin[c + d*x])/(8*
d) + (a*(5*B + 4*C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d
) + ((20*A + 35*B + 28*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(60*d) + ((32*A + 35*B + 28*C)*(a^4 + a^4*C
os[c + d*x])*Sin[c + d*x])/(24*d)

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Rubi [A]  time = 0.603097, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3045, 2976, 2968, 3023, 2735, 3770} \[ \frac{a^4 (40 A+35 B+28 C) \sin (c+d x)}{8 d}+\frac{(20 A+35 B+28 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{60 d}+\frac{(32 A+35 B+28 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (48 A+35 B+28 C)+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (5 B+4 C) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(48*A + 35*B + 28*C)*x)/8 + (a^4*A*ArcTanh[Sin[c + d*x]])/d + (a^4*(40*A + 35*B + 28*C)*Sin[c + d*x])/(8*
d) + (a*(5*B + 4*C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d
) + ((20*A + 35*B + 28*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(60*d) + ((32*A + 35*B + 28*C)*(a^4 + a^4*C
os[c + d*x])*Sin[c + d*x])/(24*d)

Rule 3045

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3023

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^4 (5 a A+a (5 B+4 C) \cos (c+d x)) \sec (c+d x) \, dx}{5 a}\\ &=\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^3 \left (20 a^2 A+a^2 (20 A+35 B+28 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{20 a}\\ &=\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{\int (a+a \cos (c+d x))^2 \left (60 a^3 A+5 a^3 (32 A+35 B+28 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{60 a}\\ &=\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(32 A+35 B+28 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int (a+a \cos (c+d x)) \left (120 a^4 A+15 a^4 (40 A+35 B+28 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a}\\ &=\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(32 A+35 B+28 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \left (120 a^5 A+\left (120 a^5 A+15 a^5 (40 A+35 B+28 C)\right ) \cos (c+d x)+15 a^5 (40 A+35 B+28 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{120 a}\\ &=\frac{a^4 (40 A+35 B+28 C) \sin (c+d x)}{8 d}+\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(32 A+35 B+28 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \left (120 a^5 A+15 a^5 (48 A+35 B+28 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a}\\ &=\frac{1}{8} a^4 (48 A+35 B+28 C) x+\frac{a^4 (40 A+35 B+28 C) \sin (c+d x)}{8 d}+\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(32 A+35 B+28 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^4 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^4 (48 A+35 B+28 C) x+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^4 (40 A+35 B+28 C) \sin (c+d x)}{8 d}+\frac{a (5 B+4 C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac{C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(20 A+35 B+28 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(32 A+35 B+28 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end{align*}

Mathematica [A]  time = 0.744236, size = 182, normalized size = 0.93 \[ \frac{a^4 \left (60 (54 A+56 B+49 C) \sin (c+d x)+120 (4 A+7 B+8 C) \sin (2 (c+d x))+40 A \sin (3 (c+d x))-480 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+480 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2880 A d x+160 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+2100 B d x+290 C \sin (3 (c+d x))+60 C \sin (4 (c+d x))+6 C \sin (5 (c+d x))+1680 C d x\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(2880*A*d*x + 2100*B*d*x + 1680*C*d*x - 480*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 480*A*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]] + 60*(54*A + 56*B + 49*C)*Sin[c + d*x] + 120*(4*A + 7*B + 8*C)*Sin[2*(c + d*x)]
+ 40*A*Sin[3*(c + d*x)] + 160*B*Sin[3*(c + d*x)] + 290*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 60*C*Sin[
4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d)

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Maple [A]  time = 0.075, size = 320, normalized size = 1.6 \begin{align*}{\frac{7\,{a}^{4}Cx}{2}}+{\frac{7\,{a}^{4}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{35\,{a}^{4}Bx}{8}}+{\frac{27\,{a}^{4}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+6\,A{a}^{4}x+2\,{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{{a}^{4}B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{4\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,{a}^{4}B\sin \left ( dx+c \right ) }{3\,d}}+{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{20\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{83\,{a}^{4}C\sin \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{34\,{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{7\,{a}^{4}Cc}{2\,d}}+{\frac{35\,{a}^{4}Bc}{8\,d}}+6\,{\frac{A{a}^{4}c}{d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/2*a^4*C*x+7/2/d*a^4*C*cos(d*x+c)*sin(d*x+c)+35/8*a^4*B*x+27/8/d*a^4*B*cos(d*x+c)*sin(d*x+c)+6*A*a^4*x+2/d*A*
a^4*cos(d*x+c)*sin(d*x+c)+1/d*a^4*C*sin(d*x+c)*cos(d*x+c)^3+1/4/d*a^4*B*sin(d*x+c)*cos(d*x+c)^3+4/3/d*B*cos(d*
x+c)^2*sin(d*x+c)*a^4+20/3/d*a^4*B*sin(d*x+c)+1/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*A*a^4*sin(d*x+c)+83/1
5/d*a^4*C*sin(d*x+c)+1/5/d*a^4*C*sin(d*x+c)*cos(d*x+c)^4+34/15/d*a^4*C*sin(d*x+c)*cos(d*x+c)^2+7/2/d*a^4*C*c+3
5/8/d*a^4*B*c+6/d*A*a^4*c+1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.03721, size = 439, normalized size = 2.25 \begin{align*} -\frac{160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 480 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 1920 \,{\left (d x + c\right )} A a^{4} + 640 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 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^{4} - 720 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 480 \,{\left (d x + c\right )} B a^{4} - 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 )} C a^{4} + 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 60 \,{\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} - 480 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 480 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 2880 \, A a^{4} \sin \left (d x + c\right ) - 1920 \, B a^{4} \sin \left (d x + c\right ) - 480 \, C a^{4} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 1920*(d*x +
c)*A*a^4 + 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x +
2*c))*B*a^4 - 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 480*(d*x + c)*B*a^4 - 32*(3*sin(d*x + c)^5 - 10*si
n(d*x + c)^3 + 15*sin(d*x + c))*C*a^4 + 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 - 60*(12*d*x + 12*c + sin(
4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^4 - 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 480*A*a^4*log(sec(d*x
+ c) + tan(d*x + c)) - 2880*A*a^4*sin(d*x + c) - 1920*B*a^4*sin(d*x + c) - 480*C*a^4*sin(d*x + c))/d

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Fricas [A]  time = 2.12322, size = 408, normalized size = 2.09 \begin{align*} \frac{15 \,{\left (48 \, A + 35 \, B + 28 \, C\right )} a^{4} d x + 60 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 60 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (24 \, C a^{4} \cos \left (d x + c\right )^{4} + 30 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 20 \, B + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \,{\left (16 \, A + 27 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right ) + 8 \,{\left (100 \, A + 100 \, B + 83 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(48*A + 35*B + 28*C)*a^4*d*x + 60*A*a^4*log(sin(d*x + c) + 1) - 60*A*a^4*log(-sin(d*x + c) + 1) + (2
4*C*a^4*cos(d*x + c)^4 + 30*(B + 4*C)*a^4*cos(d*x + c)^3 + 8*(5*A + 20*B + 34*C)*a^4*cos(d*x + c)^2 + 15*(16*A
+ 27*B + 28*C)*a^4*cos(d*x + c) + 8*(100*A + 100*B + 83*C)*a^4)*sin(d*x + c))/d

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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))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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Giac [A]  time = 1.25558, size = 455, normalized size = 2.33 \begin{align*} \frac{120 \, A a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 120 \, A a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 15 \,{\left (48 \, A a^{4} + 35 \, B a^{4} + 28 \, C a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (600 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 420 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 2720 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1960 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 4720 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3584 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3680 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3160 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1080 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1500 \, 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(120*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 120*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(48*A*
a^4 + 35*B*a^4 + 28*C*a^4)*(d*x + c) + 2*(600*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c)^9
+ 420*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 2720*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 1
960*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 4720*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 358
4*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 3680*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 3160*
C*a^4*tan(1/2*d*x + 1/2*c)^3 + 1080*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*d*x + 1/2*c) + 1500*C*a^4*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d