3.38 \(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=153 \[ -\frac {4 (A-B) \sin ^3(c+d x)}{3 a d}+\frac {4 (A-B) \sin (c+d x)}{a d}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(4 A-5 B) \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {3 (4 A-5 B) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {3 x (4 A-5 B)}{8 a} \]

[Out]

-3/8*(4*A-5*B)*x/a+4*(A-B)*sin(d*x+c)/a/d-3/8*(4*A-5*B)*cos(d*x+c)*sin(d*x+c)/a/d-1/4*(4*A-5*B)*cos(d*x+c)^3*s
in(d*x+c)/a/d+(A-B)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))-4/3*(A-B)*sin(d*x+c)^3/a/d

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Rubi [A]  time = 0.21, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2977, 2748, 2633, 2635, 8} \[ -\frac {4 (A-B) \sin ^3(c+d x)}{3 a d}+\frac {4 (A-B) \sin (c+d x)}{a d}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(4 A-5 B) \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {3 (4 A-5 B) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {3 x (4 A-5 B)}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*A - 5*B)*x)/(8*a) + (4*(A - B)*Sin[c + d*x])/(a*d) - (3*(4*A - 5*B)*Cos[c + d*x]*Sin[c + d*x])/(8*a*d)
- ((4*A - 5*B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) + ((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(d*(a + a*Cos[c +
d*x])) - (4*(A - B)*Sin[c + d*x]^3)/(3*a*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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps

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

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Mathematica [B]  time = 0.70, size = 311, normalized size = 2.03 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-72 d x (4 A-5 B) \cos \left (c+\frac {d x}{2}\right )-72 d x (4 A-5 B) \cos \left (\frac {d x}{2}\right )+168 A \sin \left (c+\frac {d x}{2}\right )+144 A \sin \left (c+\frac {3 d x}{2}\right )+144 A \sin \left (2 c+\frac {3 d x}{2}\right )-16 A \sin \left (2 c+\frac {5 d x}{2}\right )-16 A \sin \left (3 c+\frac {5 d x}{2}\right )+8 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 A \sin \left (4 c+\frac {7 d x}{2}\right )+552 A \sin \left (\frac {d x}{2}\right )-168 B \sin \left (c+\frac {d x}{2}\right )-120 B \sin \left (c+\frac {3 d x}{2}\right )-120 B \sin \left (2 c+\frac {3 d x}{2}\right )+40 B \sin \left (2 c+\frac {5 d x}{2}\right )+40 B \sin \left (3 c+\frac {5 d x}{2}\right )-5 B \sin \left (3 c+\frac {7 d x}{2}\right )-5 B \sin \left (4 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {9 d x}{2}\right )+3 B \sin \left (5 c+\frac {9 d x}{2}\right )-552 B \sin \left (\frac {d x}{2}\right )\right )}{192 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-72*(4*A - 5*B)*d*x*Cos[(d*x)/2] - 72*(4*A - 5*B)*d*x*Cos[c + (d*x)/2] + 552*A*Sin
[(d*x)/2] - 552*B*Sin[(d*x)/2] + 168*A*Sin[c + (d*x)/2] - 168*B*Sin[c + (d*x)/2] + 144*A*Sin[c + (3*d*x)/2] -
120*B*Sin[c + (3*d*x)/2] + 144*A*Sin[2*c + (3*d*x)/2] - 120*B*Sin[2*c + (3*d*x)/2] - 16*A*Sin[2*c + (5*d*x)/2]
+ 40*B*Sin[2*c + (5*d*x)/2] - 16*A*Sin[3*c + (5*d*x)/2] + 40*B*Sin[3*c + (5*d*x)/2] + 8*A*Sin[3*c + (7*d*x)/2
] - 5*B*Sin[3*c + (7*d*x)/2] + 8*A*Sin[4*c + (7*d*x)/2] - 5*B*Sin[4*c + (7*d*x)/2] + 3*B*Sin[4*c + (9*d*x)/2]
+ 3*B*Sin[5*c + (9*d*x)/2]))/(192*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.62, size = 120, normalized size = 0.78 \[ -\frac {9 \, {\left (4 \, A - 5 \, B\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (4 \, A - 5 \, B\right )} d x - {\left (6 \, B \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (28 \, A - 19 \, B\right )} \cos \left (d x + c\right ) + 64 \, A - 64 \, B\right )} \sin \left (d x + c\right )}{24 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(9*(4*A - 5*B)*d*x*cos(d*x + c) + 9*(4*A - 5*B)*d*x - (6*B*cos(d*x + c)^4 + 2*(4*A - B)*cos(d*x + c)^3 -
(4*A - 13*B)*cos(d*x + c)^2 + (28*A - 19*B)*cos(d*x + c) + 64*A - 64*B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d
)

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giac [A]  time = 0.85, size = 181, normalized size = 1.18 \[ -\frac {\frac {9 \, {\left (d x + c\right )} {\left (4 \, A - 5 \, B\right )}}{a} - \frac {24 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (60 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 124 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 115 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 100 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 109 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(9*(d*x + c)*(4*A - 5*B)/a - 24*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a - 2*(60*A*tan(1/2*d*
x + 1/2*c)^7 - 75*B*tan(1/2*d*x + 1/2*c)^7 + 124*A*tan(1/2*d*x + 1/2*c)^5 - 115*B*tan(1/2*d*x + 1/2*c)^5 + 100
*A*tan(1/2*d*x + 1/2*c)^3 - 109*B*tan(1/2*d*x + 1/2*c)^3 + 36*A*tan(1/2*d*x + 1/2*c) - 21*B*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d

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maple [B]  time = 0.09, size = 351, normalized size = 2.29 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {115 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{12 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {31 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {109 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{12 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{a d}+\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{4 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*B*tan(1/2*d*x+1/2*c)-25/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7
*B+5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*A-115/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1
/2*c)^5*B+31/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A-109/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan
(1/2*d*x+1/2*c)^3*B+25/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A-7/4/a/d/(1+tan(1/2*d*x+1/2*c)^2
)^4*B*tan(1/2*d*x+1/2*c)+3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*A*tan(1/2*d*x+1/2*c)-3/a/d*arctan(tan(1/2*d*x+1/2*c)
)*A+15/4/a/d*arctan(tan(1/2*d*x+1/2*c))*B

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maxima [B]  time = 0.56, size = 394, normalized size = 2.58 \[ -\frac {B {\left (\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {12 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 4 \, A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(B*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6
*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8) - 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 12*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 4*A*((9
*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) +
1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))
)/d

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mupad [B]  time = 0.38, size = 170, normalized size = 1.11 \[ \frac {15\,B\,x}{8\,a}-\frac {3\,A\,x}{2\,a}+\frac {7\,A\,\sin \left (c+d\,x\right )}{4\,a\,d}-\frac {7\,B\,\sin \left (c+d\,x\right )}{4\,a\,d}-\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,a\,d}+\frac {A\,\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{2\,a\,d}-\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {B\,\sin \left (4\,c+4\,d\,x\right )}{32\,a\,d}-\frac {B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*B*x)/(8*a) - (3*A*x)/(2*a) + (7*A*sin(c + d*x))/(4*a*d) - (7*B*sin(c + d*x))/(4*a*d) - (A*sin(2*c + 2*d*x)
)/(4*a*d) + (A*sin(3*c + 3*d*x))/(12*a*d) + (A*tan(c/2 + (d*x)/2))/(a*d) + (B*sin(2*c + 2*d*x))/(2*a*d) - (B*s
in(3*c + 3*d*x))/(12*a*d) + (B*sin(4*c + 4*d*x))/(32*a*d) - (B*tan(c/2 + (d*x)/2))/(a*d)

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sympy [A]  time = 7.46, size = 1794, normalized size = 11.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-36*A*d*x*tan(c/2 + d*x/2)**8/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*ta
n(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 144*A*d*x*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x
/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 216
*A*d*x*tan(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)
**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 144*A*d*x*tan(c/2 + d*x/2)**2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*
d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 36*A*d*x/(24*a*d*
tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 +
24*a*d) + 24*A*tan(c/2 + d*x/2)**9/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2
+ d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 216*A*tan(c/2 + d*x/2)**7/(24*a*d*tan(c/2 + d*x/2)**8 + 9
6*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 392*A*tan(c/2
+ d*x/2)**5/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*t
an(c/2 + d*x/2)**2 + 24*a*d) + 296*A*tan(c/2 + d*x/2)**3/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)
**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 96*A*tan(c/2 + d*x/2)/(24*a*d*tan(c
/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*
d) + 45*B*d*x*tan(c/2 + d*x/2)**8/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 +
d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 180*B*d*x*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x/2)**8
+ 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 270*B*d*x*
tan(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 9
6*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 180*B*d*x*tan(c/2 + d*x/2)**2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c
/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 45*B*d*x/(24*a*d*tan(c/2
+ d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d)
- 24*B*tan(c/2 + d*x/2)**9/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2
)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 246*B*tan(c/2 + d*x/2)**7/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*t
an(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 374*B*tan(c/2 + d*x/
2)**5/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2
+ d*x/2)**2 + 24*a*d) - 314*B*tan(c/2 + d*x/2)**3/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 1
44*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 66*B*tan(c/2 + d*x/2)/(24*a*d*tan(c/2 + d*
x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d), Ne(
d, 0)), (x*(A + B*cos(c))*cos(c)**4/(a*cos(c) + a), True))

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