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

Optimal. Leaf size=185 \[ \frac {2 (332 A-80 B+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(4 A-B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(4 A-B) \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

-(4*A-B)*arctanh(sin(d*x+c))/a^4/d+2/105*(332*A-80*B+3*C)*tan(d*x+c)/a^4/d-1/105*(88*A-25*B-3*C)*tan(d*x+c)/a^
4/d/(1+cos(d*x+c))^2-(4*A-B)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*
(12*A-5*B-2*C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.72, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {2 (332 A-80 B+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(4 A-B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(4 A-B) \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((4*A - B)*ArcTanh[Sin[c + d*x]])/(a^4*d)) + (2*(332*A - 80*B + 3*C)*Tan[c + d*x])/(105*a^4*d) - ((88*A - 25
*B - 3*C)*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - ((4*A - B)*Tan[c + d*x])/(a^4*d*(1 + Cos[c + d*x]))
- ((A - B + C)*Tan[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((12*A - 5*B - 2*C)*Tan[c + d*x])/(35*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 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))^4} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(a (8 A-B+C)-a (4 A-4 B-3 C) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^2 (52 A-10 B+3 C)-3 a^2 (12 A-5 B-2 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^3 (244 A-55 B+6 C)-2 a^3 (88 A-25 B-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (2 a^4 (332 A-80 B+3 C)-105 a^4 (4 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{105 a^8}\\ &=-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(4 A-B) \int \sec (c+d x) \, dx}{a^4}+\frac {(2 (332 A-80 B+3 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=-\frac {(4 A-B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(2 (332 A-80 B+3 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=-\frac {(4 A-B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {2 (332 A-80 B+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.40, size = 1190, normalized size = 6.43 \[ \text {result too large to display} \]

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])^4,x]

[Out]

((32*(4*A - B)*Cos[c/2 + (d*x)/2]^8*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])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (32*(4
*A - B)*Cos[c/2 + (d*x)/2]^8*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])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (4*Cos[c/2 +
(d*x)/2]^2*Cos[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(A*Sin[c/2] - B*Sin[c/2] + C*Sin[c/
2]))/(7*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (8*Cos[c/2 + (d*x)/2]^4*Co
s[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(17*A*Sin[c/2] - 10*B*Sin[c/2] + 3*C*Sin[c/2]))/
(35*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (16*Cos[c/2 + (d*x)/2]^6*Cos[c
+ d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(139*A*Sin[c/2] - 55*B*Sin[c/2] + 6*C*Sin[c/2]))/(1
05*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (4*Cos[c/2 + (d*x)/2]*Cos[c + d
*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(7*
d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (8*Cos[c/2 + (d*x)/2]^3*Cos[c + d*
x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(17*A*Sin[(d*x)/2] - 10*B*Sin[(d*x)/2] + 3*C*Sin[(d*x)/2
]))/(35*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (32*Cos[c/2 + (d*x)/2]^7*C
os[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(559*A*Sin[(d*x)/2] - 160*B*Sin[(d*x)/2] + 6*C*
Sin[(d*x)/2]))/(105*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (16*Cos[c/2 +
(d*x)/2]^5*Cos[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(139*A*Sin[(d*x)/2] - 55*B*Sin[(d*x
)/2] + 6*C*Sin[(d*x)/2]))/(105*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (32
*A*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]*Sec[c]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*Sin[d*x])/(d*(1 + Cos[c +
d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])))/a^4

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fricas [A]  time = 0.49, size = 348, normalized size = 1.88 \[ -\frac {105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (332 \, A - 80 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2236 \, A - 535 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2636 \, A - 620 \, B + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (296 \, A - 65 \, B + 9 \, C\right )} \cos \left (d x + c\right ) + 105 \, A\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

-1/210*(105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*cos(d*x + c)^4 + 6*(4*A - B)*cos(d*x + c)^3 + 4*(4*A - B)*
cos(d*x + c)^2 + (4*A - B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*c
os(d*x + c)^4 + 6*(4*A - B)*cos(d*x + c)^3 + 4*(4*A - B)*cos(d*x + c)^2 + (4*A - B)*cos(d*x + c))*log(-sin(d*x
+ c) + 1) - 2*(2*(332*A - 80*B + 3*C)*cos(d*x + c)^4 + (2236*A - 535*B + 24*C)*cos(d*x + c)^3 + (2636*A - 620
*B + 39*C)*cos(d*x + c)^2 + 4*(296*A - 65*B + 9*C)*cos(d*x + c) + 105*A)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 +
4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

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giac [A]  time = 1.15, size = 290, normalized size = 1.57 \[ -\frac {\frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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)^2/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1)
)/a^4 + 1680*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 1
5*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 + 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*B
*a^24*tan(1/2*d*x + 1/2*c)^5 + 63*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^
24*tan(1/2*d*x + 1/2*c)^3 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*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 [B]  time = 0.23, size = 363, normalized size = 1.96 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {3 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}-\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{4}}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{4}}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(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+7/40/d/a
^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a^4*B*tan(1/2*d*x+1/2*c)^5+3/40/d/a^4*C*tan(1/2*d*x+1/2*c)^5+23/24/d/a^4*tan(1
/2*d*x+1/2*c)^3*A-11/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3+1/8/d/a^4*C*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*A*tan(1/2*d*x
+1/2*c)-15/8/d/a^4*B*tan(1/2*d*x+1/2*c)+1/8/d/a^4*C*tan(1/2*d*x+1/2*c)+4/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)*B-1/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)-4/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)*B-1/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.36, size = 411, normalized size = 2.22 \[ \frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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)^2/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(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*B*((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*l
og(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) + 3*C*(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 = 1.21, size = 252, normalized size = 1.36 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {5\,A-3\,B+C}{40\,a^4}+\frac {A-B+C}{20\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,A-3\,B+C}{12\,a^4}-\frac {2\,B-10\,A+2\,C}{24\,a^4}+\frac {A-B+C}{8\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (5\,A-3\,B+C\right )}{8\,a^4}-\frac {2\,B-10\,A+2\,C}{4\,a^4}+\frac {10\,A+2\,B-2\,C}{8\,a^4}+\frac {A-B+C}{2\,a^4}\right )}{d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-B\right )}{a^4\,d} \]

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)^2*(a + a*cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)^5*((5*A - 3*B + C)/(40*a^4) + (A - B + C)/(20*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((5*A - 3*B
+ C)/(12*a^4) - (2*B - 10*A + 2*C)/(24*a^4) + (A - B + C)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)*((3*(5*A - 3*B +
C))/(8*a^4) - (2*B - 10*A + 2*C)/(4*a^4) + (10*A + 2*B - 2*C)/(8*a^4) + (A - B + C)/(2*a^4)))/d - (2*A*tan(c/2
+ (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) + (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) - (2*atanh(t
an(c/2 + (d*x)/2))*(4*A - B))/(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)**2/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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