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

Optimal. Leaf size=172 \[ \frac {(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac {(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(5 A+2 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-(5*A+2*C)*arctanh(sin(d*x+c))/a^2/d+(12*A+5*C)*tan(d*x+c)/a^2/d-(5*A+2*C)*sec(d*x+c)*tan(d*x+c)/a^2/d-2/3*(5*
A+2*C)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^2+1/3
*(12*A+5*C)*tan(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.33, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 2978, 2748, 3767, 3768, 3770} \[ \frac {(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac {(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac {(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(5 A+2 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((5*A + 2*C)*ArcTanh[Sin[c + d*x]])/(a^2*d)) + ((12*A + 5*C)*Tan[c + d*x])/(a^2*d) - ((5*A + 2*C)*Sec[c + d*
x]*Tan[c + d*x])/(a^2*d) - (2*(5*A + 2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A + C)
*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((12*A + 5*C)*Tan[c + d*x]^3)/(3*a^2*d)

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 ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(3 a (2 A+C)-a (4 A+C) \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (3 a^2 (12 A+5 C)-6 a^2 (5 A+2 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4}\\ &=-\frac {2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(2 (5 A+2 C)) \int \sec ^3(c+d x) \, dx}{a^2}+\frac {(12 A+5 C) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(5 A+2 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(5 A+2 C) \int \sec (c+d x) \, dx}{a^2}-\frac {(12 A+5 C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac {(5 A+2 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 4.78, size = 594, normalized size = 3.45 \[ \frac {192 (5 A+2 C) \cos ^4\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 )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-153 A \sin \left (c-\frac {d x}{2}\right )+21 A \sin \left (c+\frac {d x}{2}\right )-135 A \sin \left (2 c+\frac {d x}{2}\right )+25 A \sin \left (c+\frac {3 d x}{2}\right )+45 A \sin \left (2 c+\frac {3 d x}{2}\right )-85 A \sin \left (3 c+\frac {3 d x}{2}\right )+99 A \sin \left (c+\frac {5 d x}{2}\right )+21 A \sin \left (2 c+\frac {5 d x}{2}\right )+33 A \sin \left (3 c+\frac {5 d x}{2}\right )-45 A \sin \left (4 c+\frac {5 d x}{2}\right )+57 A \sin \left (2 c+\frac {7 d x}{2}\right )+18 A \sin \left (3 c+\frac {7 d x}{2}\right )+24 A \sin \left (4 c+\frac {7 d x}{2}\right )-15 A \sin \left (5 c+\frac {7 d x}{2}\right )+24 A \sin \left (3 c+\frac {9 d x}{2}\right )+11 A \sin \left (4 c+\frac {9 d x}{2}\right )+13 A \sin \left (5 c+\frac {9 d x}{2}\right )-3 (A+8 C) \sin \left (\frac {d x}{2}\right )+(155 A+66 C) \sin \left (\frac {3 d x}{2}\right )-60 C \sin \left (c-\frac {d x}{2}\right )+24 C \sin \left (c+\frac {d x}{2}\right )-60 C \sin \left (2 c+\frac {d x}{2}\right )-4 C \sin \left (c+\frac {3 d x}{2}\right )+36 C \sin \left (2 c+\frac {3 d x}{2}\right )-34 C \sin \left (3 c+\frac {3 d x}{2}\right )+42 C \sin \left (c+\frac {5 d x}{2}\right )+24 C \sin \left (3 c+\frac {5 d x}{2}\right )-18 C \sin \left (4 c+\frac {5 d x}{2}\right )+24 C \sin \left (2 c+\frac {7 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+15 C \sin \left (4 c+\frac {7 d x}{2}\right )-6 C \sin \left (5 c+\frac {7 d x}{2}\right )+10 C \sin \left (3 c+\frac {9 d x}{2}\right )+3 C \sin \left (4 c+\frac {9 d x}{2}\right )+7 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(192*(5*A + 2*C)*Cos[(c + d*x)/2]^4*(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]*Sec[c]*Sec[c + d*x]^3*(-3*(A + 8*C)*Sin[(d*x)/2] + (155*A + 66*C)*Sin[
(3*d*x)/2] - 153*A*Sin[c - (d*x)/2] - 60*C*Sin[c - (d*x)/2] + 21*A*Sin[c + (d*x)/2] + 24*C*Sin[c + (d*x)/2] -
135*A*Sin[2*c + (d*x)/2] - 60*C*Sin[2*c + (d*x)/2] + 25*A*Sin[c + (3*d*x)/2] - 4*C*Sin[c + (3*d*x)/2] + 45*A*S
in[2*c + (3*d*x)/2] + 36*C*Sin[2*c + (3*d*x)/2] - 85*A*Sin[3*c + (3*d*x)/2] - 34*C*Sin[3*c + (3*d*x)/2] + 99*A
*Sin[c + (5*d*x)/2] + 42*C*Sin[c + (5*d*x)/2] + 21*A*Sin[2*c + (5*d*x)/2] + 33*A*Sin[3*c + (5*d*x)/2] + 24*C*S
in[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] - 18*C*Sin[4*c + (5*d*x)/2] + 57*A*Sin[2*c + (7*d*x)/2] + 24*C
*Sin[2*c + (7*d*x)/2] + 18*A*Sin[3*c + (7*d*x)/2] + 3*C*Sin[3*c + (7*d*x)/2] + 24*A*Sin[4*c + (7*d*x)/2] + 15*
C*Sin[4*c + (7*d*x)/2] - 15*A*Sin[5*c + (7*d*x)/2] - 6*C*Sin[5*c + (7*d*x)/2] + 24*A*Sin[3*c + (9*d*x)/2] + 10
*C*Sin[3*c + (9*d*x)/2] + 11*A*Sin[4*c + (9*d*x)/2] + 3*C*Sin[4*c + (9*d*x)/2] + 13*A*Sin[5*c + (9*d*x)/2] + 7
*C*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 1.67, size = 237, normalized size = 1.38 \[ -\frac {3 \, {\left ({\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (12 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (33 \, A + 14 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*((5*A + 2*C)*cos(d*x + c)^5 + 2*(5*A + 2*C)*cos(d*x + c)^4 + (5*A + 2*C)*cos(d*x + c)^3)*log(sin(d*x +
c) + 1) - 3*((5*A + 2*C)*cos(d*x + c)^5 + 2*(5*A + 2*C)*cos(d*x + c)^4 + (5*A + 2*C)*cos(d*x + c)^3)*log(-sin
(d*x + c) + 1) - 2*(2*(12*A + 5*C)*cos(d*x + c)^4 + (33*A + 14*C)*cos(d*x + c)^3 + 3*(2*A + C)*cos(d*x + c)^2
- A*cos(d*x + c) + A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

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giac [A]  time = 0.48, size = 225, normalized size = 1.31 \[ -\frac {\frac {6 \, {\left (5 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (5 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {4 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(5*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(5*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 4*(15*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 - 20*A*tan(1/2*d*x + 1/2*c)^3 - 6*C*tan(1/2*
d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (
A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan(1/2
*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.22, size = 338, normalized size = 1.97 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {9 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}-\frac {A}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}}-\frac {5 A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-5/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C+5/d/a^2*A*ln(tan(1/2*d*x+1
/2*c)-1)+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^3-3/2/d/a^2*A/(tan(1/2*d*x+1/2*
c)-1)^2-5/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-5/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)
-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*C-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^3+3/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [B]  time = 0.34, size = 379, normalized size = 2.20 \[ \frac {A {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) + C*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c
)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*
sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 1.02, size = 197, normalized size = 1.15 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A+C\right )}{a^2}+\frac {5\,A+C}{2\,a^2}\right )}{d}-\frac {\left (10\,A+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {40\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (5\,A+2\,C\right )}{a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((2*(A + C))/a^2 + (5*A + C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^5*(10*A + 2*C) - tan(c/2 +
(d*x)/2)^3*((40*A)/3 + 4*C) + tan(c/2 + (d*x)/2)*(6*A + 2*C))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 - 3*a^2*tan(c/2 +
(d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2)) - (2*atanh(tan(c/2 + (d*x)/2))*(5*A + 2*C))/(a^2*d) + (tan(c/2
+ (d*x)/2)^3*(A + C))/(6*a^2*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)**4/(a+a*cos(d*x+c))**2,x)

[Out]

Timed out

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