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

Optimal. Leaf size=224 \[ -\frac {32 (54 A+5 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(21 A+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {16 (54 A+5 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(129 A+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 A \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

[Out]

1/2*(21*A+2*C)*arctanh(sin(d*x+c))/a^4/d-32/105*(54*A+5*C)*tan(d*x+c)/a^4/d+1/2*(21*A+2*C)*sec(d*x+c)*tan(d*x+
c)/a^4/d-1/105*(129*A+10*C)*sec(d*x+c)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-16/105*(54*A+5*C)*sec(d*x+c)*tan(d*x+
c)/a^4/d/(1+cos(d*x+c))-1/7*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-2/5*A*sec(d*x+c)*tan(d*x+c)/a/d/(
a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.66, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {32 (54 A+5 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {(21 A+2 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {16 (54 A+5 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(129 A+10 C) \tan (c+d x) \sec (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {2 A \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (32*(54*A + 5*C)*Tan[c + d*x])/(105*a^4*d) + ((21*A + 2*C)*Se
c[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((129*A + 10*C)*Sec[c + d*x]*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^
2) - (16*(54*A + 5*C)*Sec[c + d*x]*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Sec[c + d*x]*Tan[c
+ d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*A*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 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A+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 C)-a (5 A-2 C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \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+10 C)-56 a^2 A \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(129 A+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \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+50 C)-3 a^3 (129 A+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+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (54 A+5 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+2 C)-32 a^4 (54 A+5 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{105 a^8}\\ &=-\frac {(129 A+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (54 A+5 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(21 A+2 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac {(32 (54 A+5 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac {(21 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(129 A+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (54 A+5 C) \sec (c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(21 A+2 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac {(32 (54 A+5 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(21 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {32 (54 A+5 C) \tan (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(129 A+10 C) \sec (c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 A \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {16 (54 A+5 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 [B]  time = 6.49, size = 784, normalized size = 3.50 \[ -\frac {8 (21 A+2 C) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \cos (c+d x)+a)^4}+\frac {8 (21 A+2 C) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \cos (c+d x)+a)^4}+\frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (183162 A \sin \left (c-\frac {d x}{2}\right )-100842 A \sin \left (c+\frac {d x}{2}\right )+155526 A \sin \left (2 c+\frac {d x}{2}\right )+37380 A \sin \left (c+\frac {3 d x}{2}\right )-101148 A \sin \left (2 c+\frac {3 d x}{2}\right )+102900 A \sin \left (3 c+\frac {3 d x}{2}\right )-119364 A \sin \left (c+\frac {5 d x}{2}\right )+8820 A \sin \left (2 c+\frac {5 d x}{2}\right )-78204 A \sin \left (3 c+\frac {5 d x}{2}\right )+49980 A \sin \left (4 c+\frac {5 d x}{2}\right )-64053 A \sin \left (2 c+\frac {7 d x}{2}\right )-3885 A \sin \left (3 c+\frac {7 d x}{2}\right )-44733 A \sin \left (4 c+\frac {7 d x}{2}\right )+15435 A \sin \left (5 c+\frac {7 d x}{2}\right )-21987 A \sin \left (3 c+\frac {9 d x}{2}\right )-3675 A \sin \left (4 c+\frac {9 d x}{2}\right )-16107 A \sin \left (5 c+\frac {9 d x}{2}\right )+2205 A \sin \left (6 c+\frac {9 d x}{2}\right )-3456 A \sin \left (4 c+\frac {11 d x}{2}\right )-840 A \sin \left (5 c+\frac {11 d x}{2}\right )-2616 A \sin \left (6 c+\frac {11 d x}{2}\right )+73206 A \sin \left (\frac {d x}{2}\right )-166668 A \sin \left (\frac {3 d x}{2}\right )+17220 C \sin \left (c-\frac {d x}{2}\right )-17220 C \sin \left (c+\frac {d x}{2}\right )+14140 C \sin \left (2 c+\frac {d x}{2}\right )+9800 C \sin \left (c+\frac {3 d x}{2}\right )-15160 C \sin \left (2 c+\frac {3 d x}{2}\right )+9800 C \sin \left (3 c+\frac {3 d x}{2}\right )-10920 C \sin \left (c+\frac {5 d x}{2}\right )+4760 C \sin \left (2 c+\frac {5 d x}{2}\right )-10920 C \sin \left (3 c+\frac {5 d x}{2}\right )+4760 C \sin \left (4 c+\frac {5 d x}{2}\right )-5890 C \sin \left (2 c+\frac {7 d x}{2}\right )+1470 C \sin \left (3 c+\frac {7 d x}{2}\right )-5890 C \sin \left (4 c+\frac {7 d x}{2}\right )+1470 C \sin \left (5 c+\frac {7 d x}{2}\right )-2030 C \sin \left (3 c+\frac {9 d x}{2}\right )+210 C \sin \left (4 c+\frac {9 d x}{2}\right )-2030 C \sin \left (5 c+\frac {9 d x}{2}\right )+210 C \sin \left (6 c+\frac {9 d x}{2}\right )-320 C \sin \left (4 c+\frac {11 d x}{2}\right )-320 C \sin \left (6 c+\frac {11 d x}{2}\right )+14140 C \sin \left (\frac {d x}{2}\right )-15160 C \sin \left (\frac {3 d x}{2}\right )\right )}{6720 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(21*A + 2*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^4)
+ (8*(21*A + 2*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^
4) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(73206*A*Sin[(d*x)/2] + 14140*C*Sin[(d*x)/2] - 166668*
A*Sin[(3*d*x)/2] - 15160*C*Sin[(3*d*x)/2] + 183162*A*Sin[c - (d*x)/2] + 17220*C*Sin[c - (d*x)/2] - 100842*A*Si
n[c + (d*x)/2] - 17220*C*Sin[c + (d*x)/2] + 155526*A*Sin[2*c + (d*x)/2] + 14140*C*Sin[2*c + (d*x)/2] + 37380*A
*Sin[c + (3*d*x)/2] + 9800*C*Sin[c + (3*d*x)/2] - 101148*A*Sin[2*c + (3*d*x)/2] - 15160*C*Sin[2*c + (3*d*x)/2]
+ 102900*A*Sin[3*c + (3*d*x)/2] + 9800*C*Sin[3*c + (3*d*x)/2] - 119364*A*Sin[c + (5*d*x)/2] - 10920*C*Sin[c +
(5*d*x)/2] + 8820*A*Sin[2*c + (5*d*x)/2] + 4760*C*Sin[2*c + (5*d*x)/2] - 78204*A*Sin[3*c + (5*d*x)/2] - 10920
*C*Sin[3*c + (5*d*x)/2] + 49980*A*Sin[4*c + (5*d*x)/2] + 4760*C*Sin[4*c + (5*d*x)/2] - 64053*A*Sin[2*c + (7*d*
x)/2] - 5890*C*Sin[2*c + (7*d*x)/2] - 3885*A*Sin[3*c + (7*d*x)/2] + 1470*C*Sin[3*c + (7*d*x)/2] - 44733*A*Sin[
4*c + (7*d*x)/2] - 5890*C*Sin[4*c + (7*d*x)/2] + 15435*A*Sin[5*c + (7*d*x)/2] + 1470*C*Sin[5*c + (7*d*x)/2] -
21987*A*Sin[3*c + (9*d*x)/2] - 2030*C*Sin[3*c + (9*d*x)/2] - 3675*A*Sin[4*c + (9*d*x)/2] + 210*C*Sin[4*c + (9*
d*x)/2] - 16107*A*Sin[5*c + (9*d*x)/2] - 2030*C*Sin[5*c + (9*d*x)/2] + 2205*A*Sin[6*c + (9*d*x)/2] + 210*C*Sin
[6*c + (9*d*x)/2] - 3456*A*Sin[4*c + (11*d*x)/2] - 320*C*Sin[4*c + (11*d*x)/2] - 840*A*Sin[5*c + (11*d*x)/2] -
2616*A*Sin[6*c + (11*d*x)/2] - 320*C*Sin[6*c + (11*d*x)/2]))/(6720*d*(a + a*Cos[c + d*x])^4)

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fricas [A]  time = 1.28, size = 354, normalized size = 1.58 \[ \frac {105 \, {\left ({\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (21 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (64 \, {\left (54 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (11619 \, A + 1070 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3411 \, A + 310 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (1509 \, A + 130 \, C\right )} \cos \left (d x + c\right )^{2} + 420 \, A \cos \left (d x + c\right ) - 105 \, A\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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 + 2*C)*cos(d*x + c)^6 + 4*(21*A + 2*C)*cos(d*x + c)^5 + 6*(21*A + 2*C)*cos(d*x + c)^4 + 4*(2
1*A + 2*C)*cos(d*x + c)^3 + (21*A + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 105*((21*A + 2*C)*cos(d*x + c
)^6 + 4*(21*A + 2*C)*cos(d*x + c)^5 + 6*(21*A + 2*C)*cos(d*x + c)^4 + 4*(21*A + 2*C)*cos(d*x + c)^3 + (21*A +
2*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(64*(54*A + 5*C)*cos(d*x + c)^5 + (11619*A + 1070*C)*cos(d*x +
c)^4 + 4*(3411*A + 310*C)*cos(d*x + c)^3 + 4*(1509*A + 130*C)*cos(d*x + c)^2 + 420*A*cos(d*x + c) - 105*A)*si
n(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 = 0.60, size = 241, normalized size = 1.08 \[ \frac {\frac {420 \, {\left (21 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (21 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {840 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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 + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(21*A + 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 - 7*A*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*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 + 1
05*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 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 11
655*A*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 [A]  time = 0.23, size = 329, normalized size = 1.47 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {9 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}-\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {111 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {21 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{4}}+\frac {A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {9 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {21 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{4}}-\frac {A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {9 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+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*C*tan(1/2*d*x+1/2*c)^7-9/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/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-11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*A*t
an(1/2*d*x+1/2*c)-15/8/d/a^4*C*tan(1/2*d*x+1/2*c)-21/2/d/a^4*A*ln(tan(1/2*d*x+1/2*c)-1)-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)+21/2/d/a^4*A*ln(tan(1/2*d
*x+1/2*c)+1)+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)

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maxima [A]  time = 0.36, size = 372, normalized size = 1.66 \[ -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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) + 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*si
n(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 = 0.92, size = 260, normalized size = 1.16 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {21\,A}{2}+C\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A+C\right )}{40\,a^4}+\frac {6\,A+2\,C}{40\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{4\,a^4}+\frac {3\,\left (6\,A+2\,C\right )}{4\,a^4}+\frac {3\,\left (15\,A-C\right )}{8\,a^4}+\frac {20\,A-4\,C}{8\,a^4}\right )}{d}-\frac {7\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^4}+\frac {6\,A+2\,C}{8\,a^4}+\frac {15\,A-C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^4),x)

[Out]

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