3.29 \(\int \cos (c+d x) (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=185 \[ -\frac {2 a^4 (8 A+7 B) \sin ^3(c+d x)}{15 d}+\frac {4 a^4 (8 A+7 B) \sin (c+d x)}{5 d}+\frac {a^4 (8 A+7 B) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {27 a^4 (8 A+7 B) \sin (c+d x) \cos (c+d x)}{80 d}+\frac {7}{16} a^4 x (8 A+7 B)+\frac {(6 A-B) \sin (c+d x) (a \cos (c+d x)+a)^4}{30 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d} \]

[Out]

7/16*a^4*(8*A+7*B)*x+4/5*a^4*(8*A+7*B)*sin(d*x+c)/d+27/80*a^4*(8*A+7*B)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^4*(8*A+
7*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/30*(6*A-B)*(a+a*cos(d*x+c))^4*sin(d*x+c)/d+1/6*B*(a+a*cos(d*x+c))^5*sin(d*x+c
)/a/d-2/15*a^4*(8*A+7*B)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.30, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2968, 3023, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac {2 a^4 (8 A+7 B) \sin ^3(c+d x)}{15 d}+\frac {4 a^4 (8 A+7 B) \sin (c+d x)}{5 d}+\frac {a^4 (8 A+7 B) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {27 a^4 (8 A+7 B) \sin (c+d x) \cos (c+d x)}{80 d}+\frac {7}{16} a^4 x (8 A+7 B)+\frac {(6 A-B) \sin (c+d x) (a \cos (c+d x)+a)^4}{30 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*(8*A + 7*B)*x)/16 + (4*a^4*(8*A + 7*B)*Sin[c + d*x])/(5*d) + (27*a^4*(8*A + 7*B)*Cos[c + d*x]*Sin[c + d
*x])/(80*d) + (a^4*(8*A + 7*B)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + ((6*A - B)*(a + a*Cos[c + d*x])^4*Sin[c +
d*x])/(30*d) + (B*(a + a*Cos[c + d*x])^5*Sin[c + d*x])/(6*a*d) - (2*a^4*(8*A + 7*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

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]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx &=\int (a+a \cos (c+d x))^4 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}+\frac {\int (a+a \cos (c+d x))^4 (5 a B+a (6 A-B) \cos (c+d x)) \, dx}{6 a}\\ &=\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}+\frac {1}{10} (8 A+7 B) \int (a+a \cos (c+d x))^4 \, dx\\ &=\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}+\frac {1}{10} (8 A+7 B) \int \left (a^4+4 a^4 \cos (c+d x)+6 a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+a^4 \cos ^4(c+d x)\right ) \, dx\\ &=\frac {1}{10} a^4 (8 A+7 B) x+\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}+\frac {1}{10} \left (a^4 (8 A+7 B)\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{5} \left (2 a^4 (8 A+7 B)\right ) \int \cos (c+d x) \, dx+\frac {1}{5} \left (2 a^4 (8 A+7 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{5} \left (3 a^4 (8 A+7 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{10} a^4 (8 A+7 B) x+\frac {2 a^4 (8 A+7 B) \sin (c+d x)}{5 d}+\frac {3 a^4 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{10 d}+\frac {a^4 (8 A+7 B) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}+\frac {1}{40} \left (3 a^4 (8 A+7 B)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{10} \left (3 a^4 (8 A+7 B)\right ) \int 1 \, dx-\frac {\left (2 a^4 (8 A+7 B)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {2}{5} a^4 (8 A+7 B) x+\frac {4 a^4 (8 A+7 B) \sin (c+d x)}{5 d}+\frac {27 a^4 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{80 d}+\frac {a^4 (8 A+7 B) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}-\frac {2 a^4 (8 A+7 B) \sin ^3(c+d x)}{15 d}+\frac {1}{80} \left (3 a^4 (8 A+7 B)\right ) \int 1 \, dx\\ &=\frac {7}{16} a^4 (8 A+7 B) x+\frac {4 a^4 (8 A+7 B) \sin (c+d x)}{5 d}+\frac {27 a^4 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{80 d}+\frac {a^4 (8 A+7 B) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {(6 A-B) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {B (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}-\frac {2 a^4 (8 A+7 B) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 134, normalized size = 0.72 \[ \frac {a^4 (120 (49 A+44 B) \sin (c+d x)+15 (128 A+127 B) \sin (2 (c+d x))+580 A \sin (3 (c+d x))+120 A \sin (4 (c+d x))+12 A \sin (5 (c+d x))+3360 A d x+720 B \sin (3 (c+d x))+225 B \sin (4 (c+d x))+48 B \sin (5 (c+d x))+5 B \sin (6 (c+d x))+2940 B c+2940 B d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(2940*B*c + 3360*A*d*x + 2940*B*d*x + 120*(49*A + 44*B)*Sin[c + d*x] + 15*(128*A + 127*B)*Sin[2*(c + d*x)
] + 580*A*Sin[3*(c + d*x)] + 720*B*Sin[3*(c + d*x)] + 120*A*Sin[4*(c + d*x)] + 225*B*Sin[4*(c + d*x)] + 12*A*S
in[5*(c + d*x)] + 48*B*Sin[5*(c + d*x)] + 5*B*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.70, size = 130, normalized size = 0.70 \[ \frac {105 \, {\left (8 \, A + 7 \, B\right )} a^{4} d x + {\left (40 \, B a^{4} \cos \left (d x + c\right )^{5} + 48 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 10 \, {\left (24 \, A + 41 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 32 \, {\left (17 \, A + 18 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 105 \, {\left (8 \, A + 7 \, B\right )} a^{4} \cos \left (d x + c\right ) + 16 \, {\left (83 \, A + 72 \, B\right )} a^{4}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(105*(8*A + 7*B)*a^4*d*x + (40*B*a^4*cos(d*x + c)^5 + 48*(A + 4*B)*a^4*cos(d*x + c)^4 + 10*(24*A + 41*B)
*a^4*cos(d*x + c)^3 + 32*(17*A + 18*B)*a^4*cos(d*x + c)^2 + 105*(8*A + 7*B)*a^4*cos(d*x + c) + 16*(83*A + 72*B
)*a^4)*sin(d*x + c))/d

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giac [A]  time = 1.00, size = 166, normalized size = 0.90 \[ \frac {B a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {7}{16} \, {\left (8 \, A a^{4} + 7 \, B a^{4}\right )} x + \frac {{\left (A a^{4} + 4 \, B a^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (8 \, A a^{4} + 15 \, B a^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (29 \, A a^{4} + 36 \, B a^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (128 \, A a^{4} + 127 \, B a^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (49 \, A a^{4} + 44 \, B a^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*B*a^4*sin(6*d*x + 6*c)/d + 7/16*(8*A*a^4 + 7*B*a^4)*x + 1/80*(A*a^4 + 4*B*a^4)*sin(5*d*x + 5*c)/d + 1/64
*(8*A*a^4 + 15*B*a^4)*sin(4*d*x + 4*c)/d + 1/48*(29*A*a^4 + 36*B*a^4)*sin(3*d*x + 3*c)/d + 1/64*(128*A*a^4 + 1
27*B*a^4)*sin(2*d*x + 2*c)/d + 1/8*(49*A*a^4 + 44*B*a^4)*sin(d*x + c)/d

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maple [A]  time = 0.07, size = 306, normalized size = 1.65 \[ \frac {\frac {A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{4} B \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+4 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a^{4} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 a^{4} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 a^{4} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{4} \sin \left (d x +c \right )+a^{4} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 297, normalized size = 1.61 \[ \frac {64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{4} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 960 \, A a^{4} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^4*(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^4 - 1920*(sin(d*x + c)^3 - 3*sin(d*x +
c))*A*a^4 + 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 960*(2*d*x + 2*c + sin(2*d*x +
2*c))*A*a^4 + 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^4 - 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^4 - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4
+ 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*
a^4 + 960*A*a^4*sin(d*x + c))/d

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mupad [B]  time = 1.62, size = 316, normalized size = 1.71 \[ \frac {\left (7\,A\,a^4+\frac {49\,B\,a^4}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {119\,A\,a^4}{3}+\frac {833\,B\,a^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {462\,A\,a^4}{5}+\frac {1617\,B\,a^4}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {562\,A\,a^4}{5}+\frac {1967\,B\,a^4}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {233\,A\,a^4}{3}+\frac {1471\,B\,a^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (25\,A\,a^4+\frac {207\,B\,a^4}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {7\,a^4\,\left (8\,A+7\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,A+7\,B\right )}{8\,\left (7\,A\,a^4+\frac {49\,B\,a^4}{8}\right )}\right )\,\left (8\,A+7\,B\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(25*A*a^4 + (207*B*a^4)/8) + tan(c/2 + (d*x)/2)^11*(7*A*a^4 + (49*B*a^4)/8) + tan(c/2 + (d
*x)/2)^9*((119*A*a^4)/3 + (833*B*a^4)/24) + tan(c/2 + (d*x)/2)^3*((233*A*a^4)/3 + (1471*B*a^4)/24) + tan(c/2 +
(d*x)/2)^7*((462*A*a^4)/5 + (1617*B*a^4)/20) + tan(c/2 + (d*x)/2)^5*((562*A*a^4)/5 + (1967*B*a^4)/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)) - (7*a^4*(8*A + 7*B)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*d)
+ (7*a^4*atan((7*a^4*tan(c/2 + (d*x)/2)*(8*A + 7*B))/(8*(7*A*a^4 + (49*B*a^4)/8)))*(8*A + 7*B))/(8*d)

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sympy [A]  time = 4.83, size = 765, normalized size = 4.14 \[ \begin {cases} \frac {3 A a^{4} x \sin ^{4}{\left (c + d x \right )}}{2} + 3 A a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + 2 A a^{4} x \sin ^{2}{\left (c + d x \right )} + \frac {3 A a^{4} x \cos ^{4}{\left (c + d x \right )}}{2} + 2 A a^{4} x \cos ^{2}{\left (c + d x \right )} + \frac {8 A a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 A a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 A a^{4} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {A a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 A a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {6 A a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 A a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {A a^{4} \sin {\left (c + d x \right )}}{d} + \frac {5 B a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 B a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 B a^{4} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {15 B a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 B a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {5 B a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 B a^{4} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {B a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {5 B a^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {32 B a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 B a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {16 B a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 B a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {8 B a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {11 B a^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {4 B a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 B a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {4 B a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{4} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))**4*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((3*A*a**4*x*sin(c + d*x)**4/2 + 3*A*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2 + 2*A*a**4*x*sin(c + d*x)
**2 + 3*A*a**4*x*cos(c + d*x)**4/2 + 2*A*a**4*x*cos(c + d*x)**2 + 8*A*a**4*sin(c + d*x)**5/(15*d) + 4*A*a**4*s
in(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*A*a**4*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 4*A*a**4*sin(c + d*x)**3/
d + A*a**4*sin(c + d*x)*cos(c + d*x)**4/d + 5*A*a**4*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 6*A*a**4*sin(c + d*x
)*cos(c + d*x)**2/d + 2*A*a**4*sin(c + d*x)*cos(c + d*x)/d + A*a**4*sin(c + d*x)/d + 5*B*a**4*x*sin(c + d*x)**
6/16 + 15*B*a**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*B*a**4*x*sin(c + d*x)**4/4 + 15*B*a**4*x*sin(c + d*x
)**2*cos(c + d*x)**4/16 + 9*B*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + B*a**4*x*sin(c + d*x)**2/2 + 5*B*a**4
*x*cos(c + d*x)**6/16 + 9*B*a**4*x*cos(c + d*x)**4/4 + B*a**4*x*cos(c + d*x)**2/2 + 5*B*a**4*sin(c + d*x)**5*c
os(c + d*x)/(16*d) + 32*B*a**4*sin(c + d*x)**5/(15*d) + 5*B*a**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 16*B*
a**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 9*B*a**4*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 8*B*a**4*sin(c + d*
x)**3/(3*d) + 11*B*a**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 4*B*a**4*sin(c + d*x)*cos(c + d*x)**4/d + 15*B*a
**4*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 4*B*a**4*sin(c + d*x)*cos(c + d*x)**2/d + B*a**4*sin(c + d*x)*cos(c +
d*x)/(2*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)**4*cos(c), True))

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