3.371 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=248 \[ -\frac {8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(21 A-8 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B+20 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(129 A-52 B+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(2 A-B) \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

[Out]

1/2*(21*A-8*B+2*C)*arctanh(sin(d*x+c))/a^4/d-8/105*(216*A-83*B+20*C)*tan(d*x+c)/a^4/d+1/2*(21*A-8*B+2*C)*sec(d
*x+c)*tan(d*x+c)/a^4/d-1/105*(129*A-52*B+10*C)*sec(d*x+c)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-4/105*(216*A-83*B+
20*C)*sec(d*x+c)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-1/5*(2
*A-B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.76, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3041, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(21 A-8 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B+20 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(129 A-52 B+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(2 A-B) \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^4,x]

[Out]

((21*A - 8*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(216*A - 83*B + 20*C)*Tan[c + d*x])/(105*a^4*d) + ((
21*A - 8*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((129*A - 52*B + 10*C)*Sec[c + d*x]*Tan[c + d*x])/(10
5*a^4*d*(1 + Cos[c + d*x])^2) - (4*(216*A - 83*B + 20*C)*Sec[c + d*x]*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*
x])) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((2*A - B)*Sec[c + d*x]*Tan[c +
d*x])/(5*a*d*(a + a*Cos[c + d*x])^3)

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 3768

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

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 ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(a (9 A-2 B+2 C)-a (5 A-5 B-2 C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^2 (73 A-24 B+10 C)-28 a^2 (2 A-B) \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^3 (477 A-176 B+50 C)-3 a^3 (129 A-52 B+10 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (105 a^4 (21 A-8 B+2 C)-8 a^4 (216 A-83 B+20 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{105 a^8}\\ &=-\frac {(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac {(8 (216 A-83 B+20 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac {(21 A-8 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac {(8 (216 A-83 B+20 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(21 A-8 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {8 (216 A-83 B+20 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.53, size = 271, normalized size = 1.09 \[ -\frac {13440 (21 A-8 B+2 C) \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (8 (12813 A-4994 B+1130 C) \cos (c+d x)+60 (1177 A-456 B+106 C) \cos (2 (c+d x))+35928 A \cos (3 (c+d x))+11619 A \cos (4 (c+d x))+1728 A \cos (5 (c+d x))+58161 A-13864 B \cos (3 (c+d x))-4472 B \cos (4 (c+d x))-664 B \cos (5 (c+d x))-22888 B+3280 C \cos (3 (c+d x))+1070 C \cos (4 (c+d x))+160 C \cos (5 (c+d x))+5290 C)}{1680 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^4,x]

[Out]

-1/1680*(13440*(21*A - 8*B + 2*C)*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]]) + 2*Cos[(c + d*x)/2]*(58161*A - 22888*B + 5290*C + 8*(12813*A - 4994*B + 1130*C)*
Cos[c + d*x] + 60*(1177*A - 456*B + 106*C)*Cos[2*(c + d*x)] + 35928*A*Cos[3*(c + d*x)] - 13864*B*Cos[3*(c + d*
x)] + 3280*C*Cos[3*(c + d*x)] + 11619*A*Cos[4*(c + d*x)] - 4472*B*Cos[4*(c + d*x)] + 1070*C*Cos[4*(c + d*x)] +
1728*A*Cos[5*(c + d*x)] - 664*B*Cos[5*(c + d*x)] + 160*C*Cos[5*(c + d*x)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/(
a^4*d*(1 + Cos[c + d*x])^4)

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fricas [A]  time = 0.48, size = 402, normalized size = 1.62 \[ \frac {105 \, {\left ({\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (21 \, A - 8 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (216 \, A - 83 \, B + 20 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (11619 \, A - 4472 \, B + 1070 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3411 \, A - 1318 \, B + 310 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (1509 \, A - 592 \, B + 130 \, C\right )} \cos \left (d x + c\right )^{2} + 210 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right ) - 105 \, A\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

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)^3/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(105*((21*A - 8*B + 2*C)*cos(d*x + c)^6 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^5 + 6*(21*A - 8*B + 2*C)*cos
(d*x + c)^4 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^3 + (21*A - 8*B + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) -
105*((21*A - 8*B + 2*C)*cos(d*x + c)^6 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^5 + 6*(21*A - 8*B + 2*C)*cos(d*x +
c)^4 + 4*(21*A - 8*B + 2*C)*cos(d*x + c)^3 + (21*A - 8*B + 2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(1
6*(216*A - 83*B + 20*C)*cos(d*x + c)^5 + (11619*A - 4472*B + 1070*C)*cos(d*x + c)^4 + 4*(3411*A - 1318*B + 310
*C)*cos(d*x + c)^3 + 4*(1509*A - 592*B + 130*C)*cos(d*x + c)^2 + 210*(2*A - B)*cos(d*x + c) - 105*A)*sin(d*x +
c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*
cos(d*x + c)^2)

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giac [A]  time = 0.75, size = 339, normalized size = 1.37 \[ \frac {\frac {420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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)^3/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(420*(21*A - 8*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(21*A - 8*B + 2*C)*log(abs(tan(1/2*
d*x + 1/2*c) - 1))/a^4 + 840*(9*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 7*A*tan(1/2*d*x + 1/2*
c) + 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B
*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 189*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*B*a^
24*tan(1/2*d*x + 1/2*c)^5 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 1365*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 805*B*a^2
4*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 11655*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*a^24
*tan(1/2*d*x + 1/2*c) + 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [B]  time = 0.25, size = 493, normalized size = 1.99 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {9 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {7 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}+\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {111 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {9 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {21 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{4}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{4}}+\frac {A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {9 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {21 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{4}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{4}}-\frac {A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

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)^3/(a+a*cos(d*x+c))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A+1/56/d/a^4*B*tan(1/2*d*x+1/2*c)^7-1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7-9/40/d/
a^4*A*tan(1/2*d*x+1/2*c)^5+7/40/d/a^4*B*tan(1/2*d*x+1/2*c)^5-1/8/d/a^4*C*tan(1/2*d*x+1/2*c)^5-13/8/d/a^4*tan(1
/2*d*x+1/2*c)^3*A+23/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3-11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*A*tan(1/2*
d*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)-15/8/d/a^4*C*tan(1/2*d*x+1/2*c)+9/2/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)-
1/d/a^4/(tan(1/2*d*x+1/2*c)-1)*B-21/2/d/a^4*A*ln(tan(1/2*d*x+1/2*c)-1)+4/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/
a^4*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)^2+9/2/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)-1/d/a^4
/(tan(1/2*d*x+1/2*c)+1)*B+21/2/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-4/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*B+1/d/a^4*ln(
tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [B]  time = 0.37, size = 556, normalized size = 2.24 \[ -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \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 {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, 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)^3/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(3*A*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x +
c) + 1) - 1)/a^4) - B*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))
+ (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^
4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) + 5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7
)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/
d

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mupad [B]  time = 1.23, size = 318, normalized size = 1.28 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (9\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,A-2\,B\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (6\,A-4\,B+2\,C\right )}{4\,a^4}-\frac {3\,\left (5\,B-15\,A+C\right )}{8\,a^4}+\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {20\,A-4\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {6\,A-4\,B+2\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {6\,A-4\,B+2\,C}{8\,a^4}-\frac {5\,B-15\,A+C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}+\frac {2\,\mathrm {atanh}\left (\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {21\,A}{2}-4\,B+C\right )}{21\,A-8\,B+2\,C}\right )\,\left (\frac {21\,A}{2}-4\,B+C\right )}{a^4\,d} \]

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)^3*(a + a*cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(9*A - 2*B) - tan(c/2 + (d*x)/2)*(7*A - 2*B))/(d*(a^4*tan(c/2 + (d*x)/2)^4 - 2*a^4*tan(c
/2 + (d*x)/2)^2 + a^4)) - (tan(c/2 + (d*x)/2)*((3*(6*A - 4*B + 2*C))/(4*a^4) - (3*(5*B - 15*A + C))/(8*a^4) +
(5*(A - B + C))/(4*a^4) + (20*A - 4*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((6*A - 4*B + 2*C)/(40*a^4) + (3*(A
- B + C))/(40*a^4)))/d - (tan(c/2 + (d*x)/2)^3*((6*A - 4*B + 2*C)/(8*a^4) - (5*B - 15*A + C)/(24*a^4) + (A -
B + C)/(4*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) + (2*atanh((2*tan(c/2 + (d*x)/2)*((21*A)/2
- 4*B + C))/(21*A - 8*B + 2*C))*((21*A)/2 - 4*B + C))/(a^4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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