3.361 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=150 \[ \frac {2 (36 A-11 B+C) \tan (c+d x)}{15 a^3 d}-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-(3*A-B)*arctanh(sin(d*x+c))/a^3/d+2/15*(36*A-11*B+C)*tan(d*x+c)/a^3/d-1/5*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c
))^3-1/15*(9*A-4*B-C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^2-(3*A-B)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.52, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3041, 2978, 2748, 3767, 8, 3770} \[ \frac {2 (36 A-11 B+C) \tan (c+d x)}{15 a^3 d}-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*A - B)*ArcTanh[Sin[c + d*x]])/(a^3*d)) + (2*(36*A - 11*B + C)*Tan[c + d*x])/(15*a^3*d) - ((A - B + C)*Ta
n[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((9*A - 4*B - C)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((
3*A - B)*Tan[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

Rule 8

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

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 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3041

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[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (6 A-B+C)-a (3 A-3 B-2 C) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (27 A-7 B+2 C)-2 a^2 (9 A-4 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (2 a^3 (36 A-11 B+C)-15 a^3 (3 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(3 A-B) \int \sec (c+d x) \, dx}{a^3}+\frac {(2 (36 A-11 B+C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(2 (36 A-11 B+C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B+C) \tan (c+d x)}{15 a^3 d}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.37, size = 839, normalized size = 5.59 \[ \frac {\frac {16 (3 A-B) \cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (c+d x)+1)^3 (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x))}-\frac {16 (3 A-B) \cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (c+d x)+1)^3 (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x))}+\frac {\cos (c+d x) \sec \left (\frac {c}{2}\right ) \sec (c) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (-255 A \sin \left (\frac {d x}{2}\right )+160 B \sin \left (\frac {d x}{2}\right )-20 C \sin \left (\frac {d x}{2}\right )+567 A \sin \left (\frac {3 d x}{2}\right )-167 B \sin \left (\frac {3 d x}{2}\right )+22 C \sin \left (\frac {3 d x}{2}\right )-600 A \sin \left (c-\frac {d x}{2}\right )+170 B \sin \left (c-\frac {d x}{2}\right )-10 C \sin \left (c-\frac {d x}{2}\right )+375 A \sin \left (c+\frac {d x}{2}\right )-170 B \sin \left (c+\frac {d x}{2}\right )+10 C \sin \left (c+\frac {d x}{2}\right )-480 A \sin \left (2 c+\frac {d x}{2}\right )+160 B \sin \left (2 c+\frac {d x}{2}\right )-20 C \sin \left (2 c+\frac {d x}{2}\right )-60 A \sin \left (c+\frac {3 d x}{2}\right )+75 B \sin \left (c+\frac {3 d x}{2}\right )+402 A \sin \left (2 c+\frac {3 d x}{2}\right )-167 B \sin \left (2 c+\frac {3 d x}{2}\right )+22 C \sin \left (2 c+\frac {3 d x}{2}\right )-225 A \sin \left (3 c+\frac {3 d x}{2}\right )+75 B \sin \left (3 c+\frac {3 d x}{2}\right )+315 A \sin \left (c+\frac {5 d x}{2}\right )-95 B \sin \left (c+\frac {5 d x}{2}\right )+10 C \sin \left (c+\frac {5 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+15 B \sin \left (2 c+\frac {5 d x}{2}\right )+240 A \sin \left (3 c+\frac {5 d x}{2}\right )-95 B \sin \left (3 c+\frac {5 d x}{2}\right )+10 C \sin \left (3 c+\frac {5 d x}{2}\right )-45 A \sin \left (4 c+\frac {5 d x}{2}\right )+15 B \sin \left (4 c+\frac {5 d x}{2}\right )+72 A \sin \left (2 c+\frac {7 d x}{2}\right )-22 B \sin \left (2 c+\frac {7 d x}{2}\right )+2 C \sin \left (2 c+\frac {7 d x}{2}\right )+15 A \sin \left (3 c+\frac {7 d x}{2}\right )+57 A \sin \left (4 c+\frac {7 d x}{2}\right )-22 B \sin \left (4 c+\frac {7 d x}{2}\right )+2 C \sin \left (4 c+\frac {7 d x}{2}\right )\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{60 d (\cos (c+d x)+1)^3 (2 A+C+2 B \cos (c+d x)+C \cos (2 c+2 d x))}}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*(3*A - B)*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(C + B*Sec[c +
d*x] + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (16*(3
*A - B)*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + B*Sec[c + d*x] +
A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (Cos[c/2 + (d
*x)/2]*Cos[c + d*x]*Sec[c/2]*Sec[c]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(-255*A*Sin[(d*x)/2] + 160*B*Sin[(
d*x)/2] - 20*C*Sin[(d*x)/2] + 567*A*Sin[(3*d*x)/2] - 167*B*Sin[(3*d*x)/2] + 22*C*Sin[(3*d*x)/2] - 600*A*Sin[c
- (d*x)/2] + 170*B*Sin[c - (d*x)/2] - 10*C*Sin[c - (d*x)/2] + 375*A*Sin[c + (d*x)/2] - 170*B*Sin[c + (d*x)/2]
+ 10*C*Sin[c + (d*x)/2] - 480*A*Sin[2*c + (d*x)/2] + 160*B*Sin[2*c + (d*x)/2] - 20*C*Sin[2*c + (d*x)/2] - 60*A
*Sin[c + (3*d*x)/2] + 75*B*Sin[c + (3*d*x)/2] + 402*A*Sin[2*c + (3*d*x)/2] - 167*B*Sin[2*c + (3*d*x)/2] + 22*C
*Sin[2*c + (3*d*x)/2] - 225*A*Sin[3*c + (3*d*x)/2] + 75*B*Sin[3*c + (3*d*x)/2] + 315*A*Sin[c + (5*d*x)/2] - 95
*B*Sin[c + (5*d*x)/2] + 10*C*Sin[c + (5*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] + 15*B*Sin[2*c + (5*d*x)/2] + 240*
A*Sin[3*c + (5*d*x)/2] - 95*B*Sin[3*c + (5*d*x)/2] + 10*C*Sin[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] + 1
5*B*Sin[4*c + (5*d*x)/2] + 72*A*Sin[2*c + (7*d*x)/2] - 22*B*Sin[2*c + (7*d*x)/2] + 2*C*Sin[2*c + (7*d*x)/2] +
15*A*Sin[3*c + (7*d*x)/2] + 57*A*Sin[4*c + (7*d*x)/2] - 22*B*Sin[4*c + (7*d*x)/2] + 2*C*Sin[4*c + (7*d*x)/2]))
/(60*d*(1 + Cos[c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])))/a^3

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fricas [A]  time = 0.52, size = 279, normalized size = 1.86 \[ -\frac {15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, A - 11 \, B + C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, A - 17 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, A - 32 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(d*x + c)^2 + (3*A - B)*cos(
d*x + c))*log(sin(d*x + c) + 1) - 15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(
d*x + c)^2 + (3*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(36*A - 11*B + C)*cos(d*x + c)^3 + 3*(57*A
- 17*B + 2*C)*cos(d*x + c)^2 + (117*A - 32*B + 7*C)*cos(d*x + c) + 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 +
3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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giac [A]  time = 0.76, size = 239, normalized size = 1.59 \[ -\frac {\frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^3 + 120*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^
12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*B*a^12*tan
(1/2*d*x + 1/2*c)^3 + 10*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*
d*x + 1/2*c) + 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [B]  time = 0.23, size = 303, normalized size = 2.02 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/2/d/a^
3*tan(1/2*d*x+1/2*c)^3*A-1/3/d/a^3*B*tan(1/2*d*x+1/2*c)^3+1/6/d/a^3*C*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/
2*d*x+1/2*c)-7/4/d/a^3*B*tan(1/2*d*x+1/2*c)+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)+3/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)-
1/d/a^3*B*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)-3/d/a^3*A*ln(tan(1/2*d*x+1/2*c)+1)+1/d/a^3
*B*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.37, size = 350, normalized size = 2.33 \[ \frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*A*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - B*((105*s
in(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^
5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) +
C*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c
) + 1)^5)/a^3)/d

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mupad [B]  time = 1.22, size = 177, normalized size = 1.18 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}+\frac {6\,A-2\,C}{4\,a^3}\right )}{d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A-B\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*((A - B + C)/(6*a^3) + (4*A - 2*B)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)*((3*(A - B + C))/(
4*a^3) + (4*A - 2*B)/(2*a^3) + (6*A - 2*C)/(4*a^3)))/d - (2*A*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/2)^2
- a^3)) + (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d) - (2*atanh(tan(c/2 + (d*x)/2))*(3*A - B))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(B*cos(c
+ d*x)*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(C*cos(c + d*x
)**2*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3

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