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

Optimal. Leaf size=157 \[ -\frac {2 (80 A-3 B-4 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

A*arctanh(sin(d*x+c))/a^4/d-1/105*(55*A-6*B-8*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-2/105*(80*A-3*B-4*C)*sin(d*
x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*(10*A-3*B-4*C)*sin(d*x+c)/a/d/(a+a*
cos(d*x+c))^3

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Rubi [A]  time = 0.49, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3041, 2978, 12, 3770} \[ -\frac {2 (80 A-3 B-4 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((55*A - 6*B - 8*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - (2*(8
0*A - 3*B - 4*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Sin[c + d*x])/(7*d*(a + a*Cos[c +
d*x])^4) - ((10*A - 3*B - 4*C)*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 (c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(7 a A-a (3 A-3 B-4 C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (35 a^2 A-2 a^2 (10 A-3 B-4 C) \cos (c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (105 a^3 A-a^3 (55 A-6 B-8 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {2 (80 A-3 B-4 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int 105 a^4 A \sec (c+d x) \, dx}{105 a^8}\\ &=-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {2 (80 A-3 B-4 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {A \int \sec (c+d x) \, dx}{a^4}\\ &=\frac {A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(55 A-6 B-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(10 A-3 B-4 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {2 (80 A-3 B-4 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 2.62, size = 334, normalized size = 2.13 \[ -\frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (70 (49 A-3 B-2 C) \sin \left (\frac {d x}{2}\right )-70 (31 A-2 C) \sin \left (c+\frac {d x}{2}\right )+2625 A \sin \left (c+\frac {3 d x}{2}\right )-735 A \sin \left (2 c+\frac {3 d x}{2}\right )+1015 A \sin \left (2 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {5 d x}{2}\right )+160 A \sin \left (3 c+\frac {7 d x}{2}\right )-126 B \sin \left (c+\frac {3 d x}{2}\right )-42 B \sin \left (2 c+\frac {5 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )-168 C \sin \left (c+\frac {3 d x}{2}\right )-56 C \sin \left (2 c+\frac {5 d x}{2}\right )-8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )+6720 A \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{210 a^4 d (\cos (c+d x)+1)^4 (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/210*((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*(6720*A*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]]) + Cos[(c + d*x)/2]*Sec[c/2]*(70*(49*A - 3*B - 2*C)*Sin[(d*x
)/2] - 70*(31*A - 2*C)*Sin[c + (d*x)/2] + 2625*A*Sin[c + (3*d*x)/2] - 126*B*Sin[c + (3*d*x)/2] - 168*C*Sin[c +
(3*d*x)/2] - 735*A*Sin[2*c + (3*d*x)/2] + 1015*A*Sin[2*c + (5*d*x)/2] - 42*B*Sin[2*c + (5*d*x)/2] - 56*C*Sin[
2*c + (5*d*x)/2] - 105*A*Sin[3*c + (5*d*x)/2] + 160*A*Sin[3*c + (7*d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] - 8*C*Si
n[3*c + (7*d*x)/2])))/(a^4*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.47, size = 248, normalized size = 1.58 \[ \frac {105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (80 \, A - 3 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (535 \, A - 24 \, B - 32 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (620 \, A - 39 \, B - 52 \, C\right )} \cos \left (d x + c\right ) + 260 \, A - 36 \, B - 13 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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

[Out]

1/210*(105*(A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 6*A*cos(d*x + c)^2 + 4*A*cos(d*x + c) + A)*log(sin(d*x + c
) + 1) - 105*(A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 6*A*cos(d*x + c)^2 + 4*A*cos(d*x + c) + A)*log(-sin(d*x
+ c) + 1) - 2*(2*(80*A - 3*B - 4*C)*cos(d*x + c)^3 + (535*A - 24*B - 32*C)*cos(d*x + c)^2 + (620*A - 39*B - 52
*C)*cos(d*x + c) + 260*A - 36*B - 13*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d
*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [A]  time = 1.39, size = 248, normalized size = 1.58 \[ \frac {\frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 63 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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

[Out]

1/840*(840*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (15*A*a^2
4*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 + 105*A*a^24*ta
n(1/2*d*x + 1/2*c)^5 - 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 21*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/
2*d*x + 1/2*c)^3 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^3 - 35*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*
d*x + 1/2*c) - 105*B*a^24*tan(1/2*d*x + 1/2*c) - 105*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.22, size = 277, normalized size = 1.76 \[ \frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {3 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}} \]

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

[Out]

1/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-11/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A-15/8/d/a^4*A*tan(1/2*d*x+1/2*c)+1/8/d/a^
4*C*tan(1/2*d*x+1/2*c)-1/8/d/a^4*A*tan(1/2*d*x+1/2*c)^5+3/40/d/a^4*B*tan(1/2*d*x+1/2*c)^5-1/40/d/a^4*C*tan(1/2
*d*x+1/2*c)^5-1/56/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+1/8/d/a^4*B*tan(1/2*d*x+1/2*c)+1/8/d/a^4*B*tan(1/2*d*x+1/2*c)^3+1/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3-1/d/a^
4*A*ln(tan(1/2*d*x+1/2*c)-1)

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maxima [B]  time = 0.37, size = 313, normalized size = 1.99 \[ -\frac {5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} - \frac {3 \, B {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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

[Out]

-1/840*(5*A*((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) - C*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(
d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*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 - 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

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mupad [B]  time = 1.14, size = 199, normalized size = 1.27 \[ \frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B+C}{8\,a^4}+\frac {4\,A-2\,B}{8\,a^4}+\frac {4\,A+2\,B}{8\,a^4}+\frac {6\,A-2\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{24\,a^4}+\frac {4\,A-2\,B}{24\,a^4}+\frac {6\,A-2\,C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-B+C}{40\,a^4}+\frac {4\,A-2\,B}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \]

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)*(a + a*cos(c + d*x))^4),x)

[Out]

(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (tan(c/2 + (d*x)/2)*((A - B + C)/(8*a^4) + (4*A - 2*B)/(8*a^4) + (4*
A + 2*B)/(8*a^4) + (6*A - 2*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^3*((A - B + C)/(24*a^4) + (4*A - 2*B)/(24*a^4
) + (6*A - 2*C)/(24*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((A - B + C)/(40*a^4) + (4*A - 2*B)/(40*a^4)))/d - (tan(c
/2 + (d*x)/2)^7*(A - B + 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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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