3.10 \(\int \cos ^3(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=191 \[ -\frac {a^2 (9 A+8 B) \sin ^3(c+d x)}{15 d}+\frac {a^2 (9 A+8 B) \sin (c+d x)}{5 d}+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^4(c+d x)}{30 d}+\frac {a^2 (12 A+11 B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a^2 (12 A+11 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^2 x (12 A+11 B)+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d} \]

[Out]

1/16*a^2*(12*A+11*B)*x+1/5*a^2*(9*A+8*B)*sin(d*x+c)/d+1/16*a^2*(12*A+11*B)*cos(d*x+c)*sin(d*x+c)/d+1/24*a^2*(1
2*A+11*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/30*a^2*(6*A+7*B)*cos(d*x+c)^4*sin(d*x+c)/d+1/6*B*cos(d*x+c)^4*(a^2+a^2*c
os(d*x+c))*sin(d*x+c)/d-1/15*a^2*(9*A+8*B)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.31, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2976, 2968, 3023, 2748, 2633, 2635, 8} \[ -\frac {a^2 (9 A+8 B) \sin ^3(c+d x)}{15 d}+\frac {a^2 (9 A+8 B) \sin (c+d x)}{5 d}+\frac {a^2 (6 A+7 B) \sin (c+d x) \cos ^4(c+d x)}{30 d}+\frac {a^2 (12 A+11 B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a^2 (12 A+11 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a^2 x (12 A+11 B)+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A + 11*B)*x)/16 + (a^2*(9*A + 8*B)*Sin[c + d*x])/(5*d) + (a^2*(12*A + 11*B)*Cos[c + d*x]*Sin[c + d*x]
)/(16*d) + (a^2*(12*A + 11*B)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (a^2*(6*A + 7*B)*Cos[c + d*x]^4*Sin[c + d*
x])/(30*d) + (B*Cos[c + d*x]^4*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(6*d) - (a^2*(9*A + 8*B)*Sin[c + d*x]^3)
/(15*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 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 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]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx &=\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^3(c+d x) (a+a \cos (c+d x)) (2 a (3 A+2 B)+a (6 A+7 B) \cos (c+d x)) \, dx\\ &=\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^3(c+d x) \left (2 a^2 (3 A+2 B)+\left (2 a^2 (3 A+2 B)+a^2 (6 A+7 B)\right ) \cos (c+d x)+a^2 (6 A+7 B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (6 A+7 B) \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{30} \int \cos ^3(c+d x) \left (6 a^2 (9 A+8 B)+5 a^2 (12 A+11 B) \cos (c+d x)\right ) \, dx\\ &=\frac {a^2 (6 A+7 B) \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{5} \left (a^2 (9 A+8 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{6} \left (a^2 (12 A+11 B)\right ) \int \cos ^4(c+d x) \, dx\\ &=\frac {a^2 (12 A+11 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^2 (6 A+7 B) \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{8} \left (a^2 (12 A+11 B)\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (a^2 (9 A+8 B)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {a^2 (9 A+8 B) \sin (c+d x)}{5 d}+\frac {a^2 (12 A+11 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (12 A+11 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^2 (6 A+7 B) \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {a^2 (9 A+8 B) \sin ^3(c+d x)}{15 d}+\frac {1}{16} \left (a^2 (12 A+11 B)\right ) \int 1 \, dx\\ &=\frac {1}{16} a^2 (12 A+11 B) x+\frac {a^2 (9 A+8 B) \sin (c+d x)}{5 d}+\frac {a^2 (12 A+11 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (12 A+11 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^2 (6 A+7 B) \cos ^4(c+d x) \sin (c+d x)}{30 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {a^2 (9 A+8 B) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 134, normalized size = 0.70 \[ \frac {a^2 (120 (11 A+10 B) \sin (c+d x)+15 (32 A+31 B) \sin (2 (c+d x))+180 A \sin (3 (c+d x))+60 A \sin (4 (c+d x))+12 A \sin (5 (c+d x))+720 A d x+200 B \sin (3 (c+d x))+75 B \sin (4 (c+d x))+24 B \sin (5 (c+d x))+5 B \sin (6 (c+d x))+660 B c+660 B d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(660*B*c + 720*A*d*x + 660*B*d*x + 120*(11*A + 10*B)*Sin[c + d*x] + 15*(32*A + 31*B)*Sin[2*(c + d*x)] + 1
80*A*Sin[3*(c + d*x)] + 200*B*Sin[3*(c + d*x)] + 60*A*Sin[4*(c + d*x)] + 75*B*Sin[4*(c + d*x)] + 12*A*Sin[5*(c
+ d*x)] + 24*B*Sin[5*(c + d*x)] + 5*B*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.56, size = 130, normalized size = 0.68 \[ \frac {15 \, {\left (12 \, A + 11 \, B\right )} a^{2} d x + {\left (40 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (12 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 16 \, {\left (9 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (12 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 32 \, {\left (9 \, A + 8 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(12*A + 11*B)*a^2*d*x + (40*B*a^2*cos(d*x + c)^5 + 48*(A + 2*B)*a^2*cos(d*x + c)^4 + 10*(12*A + 11*B
)*a^2*cos(d*x + c)^3 + 16*(9*A + 8*B)*a^2*cos(d*x + c)^2 + 15*(12*A + 11*B)*a^2*cos(d*x + c) + 32*(9*A + 8*B)*
a^2)*sin(d*x + c))/d

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giac [A]  time = 0.41, size = 166, normalized size = 0.87 \[ \frac {B a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (12 \, A a^{2} + 11 \, B a^{2}\right )} x + \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (4 \, A a^{2} + 5 \, B a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (9 \, A a^{2} + 10 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (32 \, A a^{2} + 31 \, B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (11 \, A a^{2} + 10 \, B a^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*B*a^2*sin(6*d*x + 6*c)/d + 1/16*(12*A*a^2 + 11*B*a^2)*x + 1/80*(A*a^2 + 2*B*a^2)*sin(5*d*x + 5*c)/d + 1/
64*(4*A*a^2 + 5*B*a^2)*sin(4*d*x + 4*c)/d + 1/48*(9*A*a^2 + 10*B*a^2)*sin(3*d*x + 3*c)/d + 1/64*(32*A*a^2 + 31
*B*a^2)*sin(2*d*x + 2*c)/d + 1/8*(11*A*a^2 + 10*B*a^2)*sin(d*x + c)/d

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maple [A]  time = 0.08, size = 217, normalized size = 1.14 \[ \frac {\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B \,a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)),x)

[Out]

1/d*(1/5*a^2*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+B*a^2*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*c
os(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+2*a^2*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2/
5*B*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/3*a^2*A*(2+cos(d*x+c)^2)*sin(d*x+c)+B*a^2*(1/4*(cos(d
*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

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maxima [A]  time = 0.65, size = 216, normalized size = 1.13 \[ \frac {64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^2 + 60*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 + 128*(3*sin(d*x + c)^5 - 10*sin(d
*x + c)^3 + 15*sin(d*x + c))*B*a^2 - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d
*x + 2*c))*B*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2)/d

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mupad [B]  time = 1.59, size = 315, normalized size = 1.65 \[ \frac {\left (\frac {3\,A\,a^2}{2}+\frac {11\,B\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {17\,A\,a^2}{2}+\frac {187\,B\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {107\,A\,a^2}{5}+\frac {331\,B\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {117\,A\,a^2}{5}+\frac {501\,B\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {31\,A\,a^2}{2}+\frac {87\,B\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+\frac {53\,B\,a^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\left (12\,A+11\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,A+11\,B\right )}{8\,\left (\frac {3\,A\,a^2}{2}+\frac {11\,B\,a^2}{8}\right )}\right )\,\left (12\,A+11\,B\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B*cos(c + d*x))*(a + a*cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a^2)/2 + (53*B*a^2)/8) + tan(c/2 + (d*x)/2)^11*((3*A*a^2)/2 + (11*B*a^2)/8) + tan(c
/2 + (d*x)/2)^3*((31*A*a^2)/2 + (87*B*a^2)/8) + tan(c/2 + (d*x)/2)^9*((17*A*a^2)/2 + (187*B*a^2)/24) + tan(c/2
+ (d*x)/2)^7*((107*A*a^2)/5 + (331*B*a^2)/20) + tan(c/2 + (d*x)/2)^5*((117*A*a^2)/5 + (501*B*a^2)/20))/(d*(6*
tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2
+ (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) - (a^2*(12*A + 11*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*d)
+ (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(12*A + 11*B))/(8*((3*A*a^2)/2 + (11*B*a^2)/8)))*(12*A + 11*B))/(8*d)

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sympy [A]  time = 4.34, size = 600, normalized size = 3.14 \[ \begin {cases} \frac {3 A a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 A a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {5 B a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 B a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {15 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {5 B a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {16 B a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {8 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {11 B a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {2 B a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{2} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((3*A*a**2*x*sin(c + d*x)**4/4 + 3*A*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 3*A*a**2*x*cos(c + d*
x)**4/4 + 8*A*a**2*sin(c + d*x)**5/(15*d) + 4*A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*A*a**2*sin(c +
d*x)**3*cos(c + d*x)/(4*d) + 2*A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)*cos(c + d*x)**4/d + 5*A*a**2
*sin(c + d*x)*cos(c + d*x)**3/(4*d) + A*a**2*sin(c + d*x)*cos(c + d*x)**2/d + 5*B*a**2*x*sin(c + d*x)**6/16 +
15*B*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*B*a**2*x*sin(c + d*x)**4/8 + 15*B*a**2*x*sin(c + d*x)**2*co
s(c + d*x)**4/16 + 3*B*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 5*B*a**2*x*cos(c + d*x)**6/16 + 3*B*a**2*x*c
os(c + d*x)**4/8 + 5*B*a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 16*B*a**2*sin(c + d*x)**5/(15*d) + 5*B*a**2*
sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 8*B*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*B*a**2*sin(c + d*x)
**3*cos(c + d*x)/(8*d) + 11*B*a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 2*B*a**2*sin(c + d*x)*cos(c + d*x)**4
/d + 5*B*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)**2*cos(c)**3, Tr
ue))

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