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

Optimal. Leaf size=161 \[ \frac {2 (332 A+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {4 A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {4 A \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

-4*A*arctanh(sin(d*x+c))/a^4/d+2/105*(332*A+3*C)*tan(d*x+c)/a^4/d-1/105*(88*A-3*C)*tan(d*x+c)/a^4/d/(1+cos(d*x
+c))^2-4*A*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-2/35*(6*A-C)*tan(d*x+c)/a
/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.62, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 2978, 2748, 3767, 8, 3770} \[ \frac {2 (332 A+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {4 A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {4 A \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*A*ArcTanh[Sin[c + d*x]])/(a^4*d) + (2*(332*A + 3*C)*Tan[c + d*x])/(105*a^4*d) - ((88*A - 3*C)*Tan[c + d*x]
)/(105*a^4*d*(1 + Cos[c + d*x])^2) - (4*A*Tan[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Tan[c + d*x])/(7
*d*(a + a*Cos[c + d*x])^4) - (2*(6*A - 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 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 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 ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(a (8 A+C)-a (4 A-3 C) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^2 (52 A+3 C)-6 a^2 (6 A-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-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (2 a^3 (122 A+3 C)-2 a^3 (88 A-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-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 A \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (2 a^4 (332 A+3 C)-420 a^4 A \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{105 a^8}\\ &=-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 A \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(4 A) \int \sec (c+d x) \, dx}{a^4}+\frac {(2 (332 A+3 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=-\frac {4 A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 A \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(2 (332 A+3 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=-\frac {4 A \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {2 (332 A+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (6 A-C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 A \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.43, size = 680, normalized size = 4.22 \[ \frac {\frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos (c+d x) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-20524 A \sin \left (c-\frac {d x}{2}\right )+14644 A \sin \left (c+\frac {d x}{2}\right )-16660 A \sin \left (2 c+\frac {d x}{2}\right )-4690 A \sin \left (c+\frac {3 d x}{2}\right )+14378 A \sin \left (2 c+\frac {3 d x}{2}\right )-9100 A \sin \left (3 c+\frac {3 d x}{2}\right )+11668 A \sin \left (c+\frac {5 d x}{2}\right )-630 A \sin \left (2 c+\frac {5 d x}{2}\right )+9358 A \sin \left (3 c+\frac {5 d x}{2}\right )-2940 A \sin \left (4 c+\frac {5 d x}{2}\right )+4228 A \sin \left (2 c+\frac {7 d x}{2}\right )+315 A \sin \left (3 c+\frac {7 d x}{2}\right )+3493 A \sin \left (4 c+\frac {7 d x}{2}\right )-420 A \sin \left (5 c+\frac {7 d x}{2}\right )+664 A \sin \left (3 c+\frac {9 d x}{2}\right )+105 A \sin \left (4 c+\frac {9 d x}{2}\right )+559 A \sin \left (5 c+\frac {9 d x}{2}\right )-10780 A \sin \left (\frac {d x}{2}\right )+18788 A \sin \left (\frac {3 d x}{2}\right )-126 C \sin \left (c-\frac {d x}{2}\right )+126 C \sin \left (c+\frac {d x}{2}\right )-210 C \sin \left (2 c+\frac {d x}{2}\right )+252 C \sin \left (2 c+\frac {3 d x}{2}\right )+132 C \sin \left (c+\frac {5 d x}{2}\right )+132 C \sin \left (3 c+\frac {5 d x}{2}\right )+42 C \sin \left (2 c+\frac {7 d x}{2}\right )+42 C \sin \left (4 c+\frac {7 d x}{2}\right )+6 C \sin \left (3 c+\frac {9 d x}{2}\right )+6 C \sin \left (5 c+\frac {9 d x}{2}\right )-210 C \sin \left (\frac {d x}{2}\right )+252 C \sin \left (\frac {3 d x}{2}\right )\right ) \left (A \sec ^2(c+d x)+C\right )}{840 d (\cos (c+d x)+1)^4 (2 A+C \cos (2 c+2 d x)+C)}+\frac {128 A \cos ^2(c+d x) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sec ^2(c+d x)+C\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 (\cos (c+d x)+1)^4 (2 A+C \cos (2 c+2 d x)+C)}-\frac {128 A \cos ^2(c+d x) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sec ^2(c+d x)+C\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 (\cos (c+d x)+1)^4 (2 A+C \cos (2 c+2 d x)+C)}}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((128*A*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 + A*Sec[c + d*x]^2
))/(d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2*c + 2*d*x])) - (128*A*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]^2*Log[Co
s[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2*c +
2*d*x])) + (Cos[c/2 + (d*x)/2]*Cos[c + d*x]*Sec[c/2]*Sec[c]*(C + A*Sec[c + d*x]^2)*(-10780*A*Sin[(d*x)/2] - 21
0*C*Sin[(d*x)/2] + 18788*A*Sin[(3*d*x)/2] + 252*C*Sin[(3*d*x)/2] - 20524*A*Sin[c - (d*x)/2] - 126*C*Sin[c - (d
*x)/2] + 14644*A*Sin[c + (d*x)/2] + 126*C*Sin[c + (d*x)/2] - 16660*A*Sin[2*c + (d*x)/2] - 210*C*Sin[2*c + (d*x
)/2] - 4690*A*Sin[c + (3*d*x)/2] + 14378*A*Sin[2*c + (3*d*x)/2] + 252*C*Sin[2*c + (3*d*x)/2] - 9100*A*Sin[3*c
+ (3*d*x)/2] + 11668*A*Sin[c + (5*d*x)/2] + 132*C*Sin[c + (5*d*x)/2] - 630*A*Sin[2*c + (5*d*x)/2] + 9358*A*Sin
[3*c + (5*d*x)/2] + 132*C*Sin[3*c + (5*d*x)/2] - 2940*A*Sin[4*c + (5*d*x)/2] + 4228*A*Sin[2*c + (7*d*x)/2] + 4
2*C*Sin[2*c + (7*d*x)/2] + 315*A*Sin[3*c + (7*d*x)/2] + 3493*A*Sin[4*c + (7*d*x)/2] + 42*C*Sin[4*c + (7*d*x)/2
] - 420*A*Sin[5*c + (7*d*x)/2] + 664*A*Sin[3*c + (9*d*x)/2] + 6*C*Sin[3*c + (9*d*x)/2] + 105*A*Sin[4*c + (9*d*
x)/2] + 559*A*Sin[5*c + (9*d*x)/2] + 6*C*Sin[5*c + (9*d*x)/2]))/(840*d*(1 + Cos[c + d*x])^4*(2*A + C + C*Cos[2
*c + 2*d*x])))/a^4

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fricas [A]  time = 0.53, size = 277, normalized size = 1.72 \[ -\frac {210 \, {\left (A \cos \left (d x + c\right )^{5} + 4 \, A \cos \left (d x + c\right )^{4} + 6 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 210 \, {\left (A \cos \left (d x + c\right )^{5} + 4 \, A \cos \left (d x + c\right )^{4} + 6 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, {\left (332 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (559 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2636 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (296 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 105 \, A\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/105*(210*(A*cos(d*x + c)^5 + 4*A*cos(d*x + c)^4 + 6*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + A*cos(d*x + c))
*log(sin(d*x + c) + 1) - 210*(A*cos(d*x + c)^5 + 4*A*cos(d*x + c)^4 + 6*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2
+ A*cos(d*x + c))*log(-sin(d*x + c) + 1) - (2*(332*A + 3*C)*cos(d*x + c)^4 + 4*(559*A + 6*C)*cos(d*x + c)^3 +
(2636*A + 39*C)*cos(d*x + c)^2 + 4*(296*A + 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 = 0.52, size = 212, normalized size = 1.32 \[ -\frac {\frac {3360 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3360 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(3360*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3360*A*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 + 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 + 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 + 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) + 105*C*a^24*tan(1/2*d*x
+ 1/2*c))/a^28)/d

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maple [A]  time = 0.20, size = 244, normalized size = 1.52 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {3 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}}-\frac {A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+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*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+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+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)+1/8/d/a^4*C*tan(1/2*d*x+1/2*c)-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*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)

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maxima [A]  time = 0.35, size = 274, normalized size = 1.70 \[ \frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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) + 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 = 0.89, size = 204, normalized size = 1.27 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{20\,a^4}+\frac {5\,A+C}{40\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{2\,a^4}+\frac {3\,\left (5\,A+C\right )}{8\,a^4}+\frac {3\,\left (10\,A-2\,C\right )}{8\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{8\,a^4}+\frac {5\,A+C}{12\,a^4}+\frac {10\,A-2\,C}{24\,a^4}\right )}{d}-\frac {8\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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