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

Optimal. Leaf size=219 \[ -\frac{5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac{a^4 (12 A+13 B+8 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(22 A+18 B+3 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac{(16 A+15 B+6 C) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}+\frac{1}{2} a^4 x (2 A+8 B+13 C)+\frac{a (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d} \]

[Out]

(a^4*(2*A + 8*B + 13*C)*x)/2 + (a^4*(12*A + 13*B + 8*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(2*A + B - C)*Si
n[c + d*x])/(2*d) - ((22*A + 18*B + 3*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*d) + ((16*A + 15*B + 6*C)*(
a^2 + a^2*Cos[c + d*x])^2*Tan[c + d*x])/(6*d) + (a*(4*A + 3*B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x
])/(6*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.713735, antiderivative size = 219, 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} \[ -\frac{5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac{a^4 (12 A+13 B+8 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(22 A+18 B+3 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac{(16 A+15 B+6 C) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}+\frac{1}{2} a^4 x (2 A+8 B+13 C)+\frac{a (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d} \]

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]^4,x]

[Out]

(a^4*(2*A + 8*B + 13*C)*x)/2 + (a^4*(12*A + 13*B + 8*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(2*A + B - C)*Si
n[c + d*x])/(2*d) - ((22*A + 18*B + 3*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*d) + ((16*A + 15*B + 6*C)*(
a^2 + a^2*Cos[c + d*x])^2*Tan[c + d*x])/(6*d) + (a*(4*A + 3*B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x
])/(6*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 ^4(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x))^4 (a (4 A+3 B)-a (2 A-3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x))^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x))^2 \left (3 a^3 (12 A+13 B+8 C)-2 a^3 (22 A+18 B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac{(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x)) \left (6 a^4 (12 A+13 B+8 C)-30 a^4 (2 A+B-C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac{(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \left (6 a^5 (12 A+13 B+8 C)+\left (-30 a^5 (2 A+B-C)+6 a^5 (12 A+13 B+8 C)\right ) \cos (c+d x)-30 a^5 (2 A+B-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac{5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}-\frac{(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \left (6 a^5 (12 A+13 B+8 C)+6 a^5 (2 A+8 B+13 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=\frac{1}{2} a^4 (2 A+8 B+13 C) x-\frac{5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}-\frac{(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (a^4 (12 A+13 B+8 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (2 A+8 B+13 C) x+\frac{a^4 (12 A+13 B+8 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}-\frac{(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac{a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 5.69424, size = 354, normalized size = 1.62 \[ \frac{a^4 \left (6 (2 A+8 B+13 C) (c+d x)+\frac{4 (20 A+3 (4 B+C)) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 (20 A+3 (4 B+C)) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-6 (12 A+13 B+8 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 (12 A+13 B+8 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{-13 A-3 B}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{13 A+3 B}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+12 (B+4 C) \sin (c+d x)+3 C \sin (2 (c+d x))\right )}{12 d} \]

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]^4,x]

[Out]

(a^4*(6*(2*A + 8*B + 13*C)*(c + d*x) - 6*(12*A + 13*B + 8*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(12*
A + 13*B + 8*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (13*A + 3*B)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^
2 + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*(20*A + 3*(4*B + C))*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (-13
*A - 3*B)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*(20*A + 3*(4*B + C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]) + 12*(B + 4*C)*Sin[c + d*x] + 3*C*Sin[2*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.099, size = 279, normalized size = 1.3 \begin{align*}{\frac{20\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{4}B\tan \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{13\,{a}^{4}Cx}{2}}+{\frac{13\,{a}^{4}Cc}{2\,d}}+4\,{a}^{4}Bx+4\,{\frac{B{a}^{4}c}{d}}+4\,{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}}+A{a}^{4}x+{\frac{A{a}^{4}c}{d}}+{\frac{{a}^{4}B\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,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)^4,x)

[Out]

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

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Maxima [A]  time = 1.02482, size = 432, normalized size = 1.97 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \,{\left (d x + c\right )} A a^{4} + 48 \,{\left (d x + c\right )} B a^{4} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 72 \,{\left (d x + c\right )} C a^{4} - 12 \, 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 )} - 3 \, B 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 )} + 24 \, 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 )} + 36 \, 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 )} + 24 \, 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 \, B a^{4} \sin \left (d x + c\right ) + 48 \, C a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right ) + 12 \, C 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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.29112, size = 487, normalized size = 2.22 \begin{align*} \frac{6 \,{\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, C a^{4} \cos \left (d x + c\right )^{4} + 6 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (20 \, A + 12 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

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

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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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.28613, size = 468, normalized size = 2.14 \begin{align*} \frac{3 \,{\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, 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 (2 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, C 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 ) + 9 \, 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 )}^{2}} - \frac{2 \,{\left (30 \, 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} + 6 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 76 \, 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} - 12 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 54 \, 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 ) + 6 \, 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)^4,x, algorithm="giac")

[Out]

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