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

Optimal. Leaf size=257 \[ \frac {4 (454 A+83 C) \tan ^3(c+d x)}{105 a^4 d}+\frac {4 (454 A+83 C) \tan (c+d x)}{35 a^4 d}-\frac {2 (11 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {2 (11 A+2 C) \tan (c+d x) \sec (c+d x)}{a^4 d}-\frac {4 (11 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^4 d (\cos (c+d x)+1)}-\frac {(178 A+31 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {2 (8 A+C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

-2*(11*A+2*C)*arctanh(sin(d*x+c))/a^4/d+4/35*(454*A+83*C)*tan(d*x+c)/a^4/d-2*(11*A+2*C)*sec(d*x+c)*tan(d*x+c)/
a^4/d-1/105*(178*A+31*C)*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-4/3*(11*A+2*C)*sec(d*x+c)^2*tan(d*x+c)
/a^4/d/(1+cos(d*x+c))-1/7*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-2/35*(8*A+C)*sec(d*x+c)^2*tan(d*x
+c)/a/d/(a+a*cos(d*x+c))^3+4/105*(454*A+83*C)*tan(d*x+c)^3/a^4/d

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Rubi [A]  time = 0.71, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {4 (454 A+83 C) \tan ^3(c+d x)}{105 a^4 d}+\frac {4 (454 A+83 C) \tan (c+d x)}{35 a^4 d}-\frac {2 (11 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {2 (11 A+2 C) \tan (c+d x) \sec (c+d x)}{a^4 d}-\frac {4 (11 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^4 d (\cos (c+d x)+1)}-\frac {(178 A+31 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {2 (8 A+C) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(11*A + 2*C)*ArcTanh[Sin[c + d*x]])/(a^4*d) + (4*(454*A + 83*C)*Tan[c + d*x])/(35*a^4*d) - (2*(11*A + 2*C)
*Sec[c + d*x]*Tan[c + d*x])/(a^4*d) - ((178*A + 31*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x
])^2) - (4*(11*A + 2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Sec[c + d*x]^2*Ta
n[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*(8*A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Cos[c + d*
x])^3) + (4*(454*A + 83*C)*Tan[c + d*x]^3)/(105*a^4*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))^4} \, dx &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(a (10 A+3 C)-a (6 A-C) \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (7 a^2 (14 A+3 C)-10 a^2 (8 A+C) \cos (c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(178 A+31 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (12 a^3 (69 A+13 C)-4 a^3 (178 A+31 C) \cos (c+d x)\right ) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(178 A+31 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 (11 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (12 a^4 (454 A+83 C)-420 a^4 (11 A+2 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{105 a^8}\\ &=-\frac {(178 A+31 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 (11 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(4 (11 A+2 C)) \int \sec ^3(c+d x) \, dx}{a^4}+\frac {(4 (454 A+83 C)) \int \sec ^4(c+d x) \, dx}{35 a^4}\\ &=-\frac {2 (11 A+2 C) \sec (c+d x) \tan (c+d x)}{a^4 d}-\frac {(178 A+31 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 (11 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(2 (11 A+2 C)) \int \sec (c+d x) \, dx}{a^4}-\frac {(4 (454 A+83 C)) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 a^4 d}\\ &=-\frac {2 (11 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {4 (454 A+83 C) \tan (c+d x)}{35 a^4 d}-\frac {2 (11 A+2 C) \sec (c+d x) \tan (c+d x)}{a^4 d}-\frac {(178 A+31 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (8 A+C) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 (11 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {4 (454 A+83 C) \tan ^3(c+d x)}{105 a^4 d}\\ \end {align*}

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Mathematica [A]  time = 4.59, size = 361, normalized size = 1.40 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (4 (412 A+139 C) \tan \left (\frac {c}{2}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+6 (31 A+17 C) \tan \left (\frac {c}{2}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+15 (A+C) \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+15 (A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 (2512 A+559 C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right )+4 (412 A+139 C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+6 (31 A+17 C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+280 \cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (\sec (c) \sin (d x) \sec (c+d x) \left (A \sec ^2(c+d x)-6 A \sec (c+d x)+32 A+3 C\right )+6 (11 A+2 C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+A \tan (c) \sec ^2(c+d x)-6 A \tan (c) \sec (c+d x)\right )\right )}{105 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(15*(A + C)*Sec[c/2]*Sin[(d*x)/2] + 6*(31*A + 17*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/
2] + 4*(412*A + 139*C)*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 8*(2512*A + 559*C)*Cos[(c + d*x)/2]^6*Sec[c/
2]*Sin[(d*x)/2] + 15*(A + C)*Cos[(c + d*x)/2]*Tan[c/2] + 6*(31*A + 17*C)*Cos[(c + d*x)/2]^3*Tan[c/2] + 4*(412*
A + 139*C)*Cos[(c + d*x)/2]^5*Tan[c/2] + 280*Cos[(c + d*x)/2]^7*(6*(11*A + 2*C)*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*Sec[c + d*x]*(32*A + 3*C - 6*A*Sec[c + d*x] +
A*Sec[c + d*x]^2)*Sin[d*x] - 6*A*Sec[c + d*x]*Tan[c] + A*Sec[c + d*x]^2*Tan[c])))/(105*a^4*d*(1 + Cos[c + d*x
])^4)

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fricas [A]  time = 1.76, size = 372, normalized size = 1.45 \[ -\frac {105 \, {\left ({\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{7} + 4 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{7} + 4 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (11 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, {\left (454 \, A + 83 \, C\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (6109 \, A + 1118 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (3592 \, A + 659 \, C\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (799 \, A + 148 \, C\right )} \cos \left (d x + c\right )^{3} + 35 \, {\left (14 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} - 70 \, A \cos \left (d x + c\right ) + 35 \, A\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{7} + 4 \, a^{4} d \cos \left (d x + c\right )^{6} + 6 \, a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + a^{4} d \cos \left (d x + c\right )^{3}\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

-1/105*(105*((11*A + 2*C)*cos(d*x + c)^7 + 4*(11*A + 2*C)*cos(d*x + c)^6 + 6*(11*A + 2*C)*cos(d*x + c)^5 + 4*(
11*A + 2*C)*cos(d*x + c)^4 + (11*A + 2*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 105*((11*A + 2*C)*cos(d*x +
c)^7 + 4*(11*A + 2*C)*cos(d*x + c)^6 + 6*(11*A + 2*C)*cos(d*x + c)^5 + 4*(11*A + 2*C)*cos(d*x + c)^4 + (11*A +
2*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - (8*(454*A + 83*C)*cos(d*x + c)^6 + 2*(6109*A + 1118*C)*cos(d*x
+ c)^5 + 4*(3592*A + 659*C)*cos(d*x + c)^4 + 8*(799*A + 148*C)*cos(d*x + c)^3 + 35*(14*A + 3*C)*cos(d*x + c)^2
- 70*A*cos(d*x + c) + 35*A)*sin(d*x + c))/(a^4*d*cos(d*x + c)^7 + 4*a^4*d*cos(d*x + c)^6 + 6*a^4*d*cos(d*x +
c)^5 + 4*a^4*d*cos(d*x + c)^4 + a^4*d*cos(d*x + c)^3)

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giac [A]  time = 0.58, size = 295, normalized size = 1.15 \[ -\frac {\frac {1680 \, {\left (11 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {1680 \, {\left (11 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {560 \, {\left (39 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 62 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 231 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2065 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21945 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(1680*(11*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 1680*(11*A + 2*C)*log(abs(tan(1/2*d*x + 1/2
*c) - 1))/a^4 + 560*(39*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 - 62*A*tan(1/2*d*x + 1/2*c)^3 -
6*C*tan(1/2*d*x + 1/2*c)^3 + 27*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^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 + 231*A*a^24*tan(1/2*d*x + 1/
2*c)^5 + 147*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 2065*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 805*C*a^24*tan(1/2*d*x + 1/2
*c)^3 + 21945*A*a^24*tan(1/2*d*x + 1/2*c) + 5145*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.24, size = 418, normalized size = 1.63 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {11 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {7 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {59 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {23 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {209 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {49 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {13 A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {22 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{4}}-\frac {A}{3 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {5 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {22 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{4}}-\frac {13 A}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A}{3 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5 A}{2 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

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))^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+11/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5+7/40/d/
a^4*C*tan(1/2*d*x+1/2*c)^5+59/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A+23/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3+209/8/d/a^4*A
*tan(1/2*d*x+1/2*c)+49/8/d/a^4*C*tan(1/2*d*x+1/2*c)-13/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)-1/d/a^4/(tan(1/2*d*x+1/2
*c)-1)*C+22/d/a^4*A*ln(tan(1/2*d*x+1/2*c)-1)+4/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*C-1/3/d/a^4*A/(tan(1/2*d*x+1/2*c
)-1)^3-5/2/d/a^4*A/(tan(1/2*d*x+1/2*c)-1)^2-22/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-4/d/a^4*ln(tan(1/2*d*x+1/2*c)+
1)*C-13/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)-1/d/a^4/(tan(1/2*d*x+1/2*c)+1)*C-1/3/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)^3+5
/2/d/a^4*A/(tan(1/2*d*x+1/2*c)+1)^2

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maxima [A]  time = 0.35, size = 461, normalized size = 1.79 \[ \frac {A {\left (\frac {560 \, {\left (\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {62 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {39 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{4} - \frac {3 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {21945 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2065 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {231 \, \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 {18480 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {18480 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + C {\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 )}}{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)^4/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(560*(27*sin(d*x + c)/(cos(d*x + c) + 1) - 62*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 39*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5)/(a^4 - 3*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^4*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 - a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (21945*sin(d*x + c)/(cos(d*x + c) + 1) + 2065*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 + 231*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
- 18480*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 18480*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) +
C*(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*s
in(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))/d

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mupad [B]  time = 0.94, size = 303, normalized size = 1.18 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{2\,a^4}+\frac {21\,A+C}{2\,a^4}+\frac {5\,\left (7\,A+3\,C\right )}{4\,a^4}+\frac {35\,A-5\,C}{8\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{10\,a^4}+\frac {7\,A+3\,C}{40\,a^4}\right )}{d}-\frac {\left (26\,A+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {124\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (18\,A+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,\left (A+C\right )}{12\,a^4}+\frac {21\,A+C}{24\,a^4}+\frac {7\,A+3\,C}{6\,a^4}\right )}{d}-\frac {4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (11\,A+2\,C\right )}{a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((5*(A + C))/(2*a^4) + (21*A + C)/(2*a^4) + (5*(7*A + 3*C))/(4*a^4) + (35*A - 5*C)/(8*a^4)
))/d + (tan(c/2 + (d*x)/2)^5*((A + C)/(10*a^4) + (7*A + 3*C)/(40*a^4)))/d - (tan(c/2 + (d*x)/2)^5*(26*A + 2*C)
- tan(c/2 + (d*x)/2)^3*((124*A)/3 + 4*C) + tan(c/2 + (d*x)/2)*(18*A + 2*C))/(d*(3*a^4*tan(c/2 + (d*x)/2)^2 -
3*a^4*tan(c/2 + (d*x)/2)^4 + a^4*tan(c/2 + (d*x)/2)^6 - a^4)) + (tan(c/2 + (d*x)/2)^3*((5*(A + C))/(12*a^4) +
(21*A + C)/(24*a^4) + (7*A + 3*C)/(6*a^4)))/d - (4*atanh(tan(c/2 + (d*x)/2))*(11*A + 2*C))/(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)**4/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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