3.333 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=206 \[ -\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}+\frac{a^4 (13 A+8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(15 A+6 B-2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}-\frac{(18 A+3 B-8 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (8 A+13 B+12 C)+\frac{a (2 A+B) \tan (c+d x) (a \cos (c+d x)+a)^3}{d}+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d} \]

[Out]

(a^4*(8*A + 13*B + 12*C)*x)/2 + (a^4*(13*A + 8*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A - B - 2*C)*Si
n[c + d*x])/(2*d) - ((15*A + 6*B - 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) - ((18*A + 3*B - 8*C)*(
a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*d) + (a*(2*A + B)*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d + (A*(a + a*
Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.685534, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3043, 2975, 2976, 2968, 3023, 2735, 3770} \[ -\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}+\frac{a^4 (13 A+8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(15 A+6 B-2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}-\frac{(18 A+3 B-8 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (8 A+13 B+12 C)+\frac{a (2 A+B) \tan (c+d x) (a \cos (c+d x)+a)^3}{d}+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d} \]

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]^3,x]

[Out]

(a^4*(8*A + 13*B + 12*C)*x)/2 + (a^4*(13*A + 8*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A - B - 2*C)*Si
n[c + d*x])/(2*d) - ((15*A + 6*B - 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) - ((18*A + 3*B - 8*C)*(
a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*d) + (a*(2*A + B)*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d + (A*(a + a*
Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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 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 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 ^3(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x))^4 (2 a (2 A+B)-a (3 A-2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x))^3 \left (a^2 (13 A+8 B+2 C)-a^2 (15 A+6 B-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x))^2 \left (3 a^3 (13 A+8 B+2 C)-2 a^3 (18 A+3 B-8 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac{(18 A+3 B-8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x)) \left (6 a^4 (13 A+8 B+2 C)-30 a^4 (A-B-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac{(18 A+3 B-8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (6 a^5 (13 A+8 B+2 C)+\left (-30 a^5 (A-B-2 C)+6 a^5 (13 A+8 B+2 C)\right ) \cos (c+d x)-30 a^5 (A-B-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac{(18 A+3 B-8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (6 a^5 (13 A+8 B+2 C)+6 a^5 (8 A+13 B+12 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=\frac{1}{2} a^4 (8 A+13 B+12 C) x-\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac{(18 A+3 B-8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^4 (13 A+8 B+2 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (8 A+13 B+12 C) x+\frac{a^4 (13 A+8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{(15 A+6 B-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac{(18 A+3 B-8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{a (2 A+B) (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 3.66794, size = 299, normalized size = 1.45 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (6 (8 A+13 B+12 C) (c+d x)+3 (4 A+16 B+27 C) \sin (c+d x)-6 (13 A+8 B+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 (13 A+8 B+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 (4 A+B) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 (4 A+B) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{3 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{3 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+3 (B+4 C) \sin (2 (c+d x))+C \sin (3 (c+d x))\right )}{192 d} \]

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]^3,x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(6*(8*A + 13*B + 12*C)*(c + d*x) - 6*(13*A + 8*B + 2*C)*Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(13*A + 8*B + 2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*A)/(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*(4*A + B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (3
*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (12*(4*A + B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]) + 3*(4*A + 16*B + 27*C)*Sin[c + d*x] + 3*(B + 4*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(192*d)

________________________________________________________________________________________

Maple [A]  time = 0.095, size = 280, normalized size = 1.4 \begin{align*}{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{13\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{a}^{4}Cx+6\,{\frac{{a}^{4}Cc}{d}}+{\frac{13\,{a}^{4}Bx}{2}}+{\frac{13\,{a}^{4}Bc}{2\,d}}+{\frac{20\,{a}^{4}C\sin \left ( dx+c \right ) }{3\,d}}+4\,A{a}^{4}x+4\,{\frac{A{a}^{4}c}{d}}+4\,{\frac{{a}^{4}B\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{4}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,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)^3,x)

[Out]

1/2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+13/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^4*B*tan(d*x+c)+1/d*a^4*C*ln(sec
(d*x+c)+tan(d*x+c))+4/d*A*a^4*tan(d*x+c)+4/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+6*a^4*C*x+6/d*a^4*C*c+13/2*a^4*B*
x+13/2/d*a^4*B*c+20/3/d*a^4*C*sin(d*x+c)+4*A*a^4*x+4/d*A*a^4*c+4/d*a^4*B*sin(d*x+c)+2/d*a^4*C*cos(d*x+c)*sin(d
*x+c)+1/d*A*a^4*sin(d*x+c)+1/2/d*a^4*B*cos(d*x+c)*sin(d*x+c)+1/3/d*a^4*C*sin(d*x+c)*cos(d*x+c)^2

________________________________________________________________________________________

Maxima [A]  time = 1.10959, size = 400, normalized size = 1.94 \begin{align*} \frac{48 \,{\left (d x + c\right )} A a^{4} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 72 \,{\left (d x + c\right )} B a^{4} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 48 \,{\left (d x + c\right )} C a^{4} - 3 \, A a^{4}{\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 )} + 36 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 48 \, B a^{4} \sin \left (d x + c\right ) + 72 \, C a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right ) + 12 \, B a^{4} \tan \left (d x + c\right )}{12 \, 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)^3,x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*A*a^4 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 72*(d*x + c)*B*a^4 - 4*(sin(d*x + c)^3 -
3*sin(d*x + c))*C*a^4 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 48*(d*x + c)*C*a^4 - 3*A*a^4*(2*sin(d*x +
c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*A*a^4*(log(sin(d*x + c) + 1) -
log(sin(d*x + c) - 1)) + 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^4*(log(sin(d*x + c)
+ 1) - log(sin(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 48*B*a^4*sin(d*x + c) + 72*C*a^4*sin(d*x + c) + 48*A*a
^4*tan(d*x + c) + 12*B*a^4*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 2.18024, size = 486, normalized size = 2.36 \begin{align*} \frac{6 \,{\left (8 \, A + 13 \, B + 12 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \,{\left (13 \, A + 8 \, B + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (13 \, A + 8 \, B + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C a^{4} \cos \left (d x + c\right )^{4} + 3 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A + 12 \, B + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 6 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 3 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \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)^3,x, algorithm="fricas")

[Out]

1/12*(6*(8*A + 13*B + 12*C)*a^4*d*x*cos(d*x + c)^2 + 3*(13*A + 8*B + 2*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c)
+ 1) - 3*(13*A + 8*B + 2*C)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*a^4*cos(d*x + c)^4 + 3*(B + 4*C
)*a^4*cos(d*x + c)^3 + 2*(3*A + 12*B + 20*C)*a^4*cos(d*x + c)^2 + 6*(4*A + B)*a^4*cos(d*x + c) + 3*A*a^4)*sin(
d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.2468, size = 468, normalized size = 2.27 \begin{align*} \frac{3 \,{\left (8 \, A a^{4} + 13 \, B a^{4} + 12 \, C a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (13 \, A a^{4} + 8 \, B a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (13 \, A a^{4} + 8 \, B a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (7 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B 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 )}^{2}} + \frac{2 \,{\left (6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 48 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 76 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 54 \, 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 )}^{3}}}{6 \, 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)^3,x, algorithm="giac")

[Out]

1/6*(3*(8*A*a^4 + 13*B*a^4 + 12*C*a^4)*(d*x + c) + 3*(13*A*a^4 + 8*B*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) + 1)) - 3*(13*A*a^4 + 8*B*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(7*A*a^4*tan(1/2*d*x + 1/2*
c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/2*c) - 2*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*
d*x + 1/2*c)^2 - 1)^2 + 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*C*a^4*tan(1/2
*d*x + 1/2*c)^5 + 12*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 76*C*a^4*tan(1/2*d*x + 1
/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 54*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 + 1)^3)/d