3.64 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=225 \[ \frac{4 (34 A+9 C) \tan ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(23 A+6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(23 A+6 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(13 A+3 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-((23*A + 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (4*(34*A + 9*C)*Tan[c + d*x])/(5*a^3*d) - ((23*A + 6*C)*Sec[
c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((13*A
+ 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((23*A + 6*C)*Sec[c + d*x]^2*Tan[c + d*
x])/(3*d*(a^3 + a^3*Cos[c + d*x])) + (4*(34*A + 9*C)*Tan[c + d*x]^3)/(15*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.554484, antiderivative size = 225, normalized size of antiderivative = 1., 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, 3768, 3770} \[ \frac{4 (34 A+9 C) \tan ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{(23 A+6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(23 A+6 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(13 A+3 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

-((23*A + 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (4*(34*A + 9*C)*Tan[c + d*x])/(5*a^3*d) - ((23*A + 6*C)*Sec[
c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((13*A
+ 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((23*A + 6*C)*Sec[c + d*x]^2*Tan[c + d*
x])/(3*d*(a^3 + a^3*Cos[c + d*x])) + (4*(34*A + 9*C)*Tan[c + d*x]^3)/(15*a^3*d)

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 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 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 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))^3} \, dx &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (8 A+3 C)-5 a A \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (9 a^2 (7 A+2 C)-4 a^2 (13 A+3 C) \cos (c+d x)\right ) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (12 a^3 (34 A+9 C)-15 a^3 (23 A+6 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{15 a^6}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(23 A+6 C) \int \sec ^3(c+d x) \, dx}{a^3}+\frac{(4 (34 A+9 C)) \int \sec ^4(c+d x) \, dx}{5 a^3}\\ &=-\frac{(23 A+6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(23 A+6 C) \int \sec (c+d x) \, dx}{2 a^3}-\frac{(4 (34 A+9 C)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 a^3 d}\\ &=-\frac{(23 A+6 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{4 (34 A+9 C) \tan (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(13 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(23 A+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{4 (34 A+9 C) \tan ^3(c+d x)}{15 a^3 d}\\ \end{align*}

Mathematica [B]  time = 6.45378, size = 798, normalized size = 3.55 \[ \frac{4 (23 A+6 C) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (c+d x) a+a)^3}-\frac{4 (23 A+6 C) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (\cos (c+d x) a+a)^3}+\frac{\sec \left (\frac{c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (-2484 A \sin \left (\frac{d x}{2}\right )-1764 C \sin \left (\frac{d x}{2}\right )+12622 A \sin \left (\frac{3 d x}{2}\right )+3372 C \sin \left (\frac{3 d x}{2}\right )-13340 A \sin \left (c-\frac{d x}{2}\right )-3480 C \sin \left (c-\frac{d x}{2}\right )+4140 A \sin \left (c+\frac{d x}{2}\right )+2100 C \sin \left (c+\frac{d x}{2}\right )-11684 A \sin \left (2 c+\frac{d x}{2}\right )-3144 C \sin \left (2 c+\frac{d x}{2}\right )-450 A \sin \left (c+\frac{3 d x}{2}\right )-960 C \sin \left (c+\frac{3 d x}{2}\right )+5022 A \sin \left (2 c+\frac{3 d x}{2}\right )+2232 C \sin \left (2 c+\frac{3 d x}{2}\right )-8050 A \sin \left (3 c+\frac{3 d x}{2}\right )-2100 C \sin \left (3 c+\frac{3 d x}{2}\right )+9230 A \sin \left (c+\frac{5 d x}{2}\right )+2460 C \sin \left (c+\frac{5 d x}{2}\right )+630 A \sin \left (2 c+\frac{5 d x}{2}\right )-390 C \sin \left (2 c+\frac{5 d x}{2}\right )+4230 A \sin \left (3 c+\frac{5 d x}{2}\right )+1710 C \sin \left (3 c+\frac{5 d x}{2}\right )-4370 A \sin \left (4 c+\frac{5 d x}{2}\right )-1140 C \sin \left (4 c+\frac{5 d x}{2}\right )+5347 A \sin \left (2 c+\frac{7 d x}{2}\right )+1422 C \sin \left (2 c+\frac{7 d x}{2}\right )+875 A \sin \left (3 c+\frac{7 d x}{2}\right )-60 C \sin \left (3 c+\frac{7 d x}{2}\right )+2747 A \sin \left (4 c+\frac{7 d x}{2}\right )+1032 C \sin \left (4 c+\frac{7 d x}{2}\right )-1725 A \sin \left (5 c+\frac{7 d x}{2}\right )-450 C \sin \left (5 c+\frac{7 d x}{2}\right )+2375 A \sin \left (3 c+\frac{9 d x}{2}\right )+630 C \sin \left (3 c+\frac{9 d x}{2}\right )+655 A \sin \left (4 c+\frac{9 d x}{2}\right )+60 C \sin \left (4 c+\frac{9 d x}{2}\right )+1375 A \sin \left (5 c+\frac{9 d x}{2}\right )+480 C \sin \left (5 c+\frac{9 d x}{2}\right )-345 A \sin \left (6 c+\frac{9 d x}{2}\right )-90 C \sin \left (6 c+\frac{9 d x}{2}\right )+544 A \sin \left (4 c+\frac{11 d x}{2}\right )+144 C \sin \left (4 c+\frac{11 d x}{2}\right )+200 A \sin \left (5 c+\frac{11 d x}{2}\right )+30 C \sin \left (5 c+\frac{11 d x}{2}\right )+344 A \sin \left (6 c+\frac{11 d x}{2}\right )+114 C \sin \left (6 c+\frac{11 d x}{2}\right )\right ) \cos \left (\frac{c}{2}+\frac{d x}{2}\right )}{960 d (\cos (c+d x) a+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

(4*(23*A + 6*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^3)
- (4*(23*A + 6*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^3
) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(-2484*A*Sin[(d*x)/2] - 1764*C*Sin[(d*x)/2] + 12622*A*S
in[(3*d*x)/2] + 3372*C*Sin[(3*d*x)/2] - 13340*A*Sin[c - (d*x)/2] - 3480*C*Sin[c - (d*x)/2] + 4140*A*Sin[c + (d
*x)/2] + 2100*C*Sin[c + (d*x)/2] - 11684*A*Sin[2*c + (d*x)/2] - 3144*C*Sin[2*c + (d*x)/2] - 450*A*Sin[c + (3*d
*x)/2] - 960*C*Sin[c + (3*d*x)/2] + 5022*A*Sin[2*c + (3*d*x)/2] + 2232*C*Sin[2*c + (3*d*x)/2] - 8050*A*Sin[3*c
+ (3*d*x)/2] - 2100*C*Sin[3*c + (3*d*x)/2] + 9230*A*Sin[c + (5*d*x)/2] + 2460*C*Sin[c + (5*d*x)/2] + 630*A*Si
n[2*c + (5*d*x)/2] - 390*C*Sin[2*c + (5*d*x)/2] + 4230*A*Sin[3*c + (5*d*x)/2] + 1710*C*Sin[3*c + (5*d*x)/2] -
4370*A*Sin[4*c + (5*d*x)/2] - 1140*C*Sin[4*c + (5*d*x)/2] + 5347*A*Sin[2*c + (7*d*x)/2] + 1422*C*Sin[2*c + (7*
d*x)/2] + 875*A*Sin[3*c + (7*d*x)/2] - 60*C*Sin[3*c + (7*d*x)/2] + 2747*A*Sin[4*c + (7*d*x)/2] + 1032*C*Sin[4*
c + (7*d*x)/2] - 1725*A*Sin[5*c + (7*d*x)/2] - 450*C*Sin[5*c + (7*d*x)/2] + 2375*A*Sin[3*c + (9*d*x)/2] + 630*
C*Sin[3*c + (9*d*x)/2] + 655*A*Sin[4*c + (9*d*x)/2] + 60*C*Sin[4*c + (9*d*x)/2] + 1375*A*Sin[5*c + (9*d*x)/2]
+ 480*C*Sin[5*c + (9*d*x)/2] - 345*A*Sin[6*c + (9*d*x)/2] - 90*C*Sin[6*c + (9*d*x)/2] + 544*A*Sin[4*c + (11*d*
x)/2] + 144*C*Sin[4*c + (11*d*x)/2] + 200*A*Sin[5*c + (11*d*x)/2] + 30*C*Sin[5*c + (11*d*x)/2] + 344*A*Sin[6*c
+ (11*d*x)/2] + 114*C*Sin[6*c + (11*d*x)/2]))/(960*d*(a + a*Cos[c + d*x])^3)

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Maple [A]  time = 0.072, size = 378, normalized size = 1.7 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{5\,A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{49\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{17\,A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{23\,A}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{3}}}-{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-2\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}-{\frac{23\,A}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{3}}}-{\frac{17\,A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}} \end{align*}

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))^3,x)

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+5/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/2/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+49/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)-17/2/d/a^3*A/(tan(1/2*
d*x+1/2*c)-1)-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*C+23/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)+3/d/a^3*ln(tan(1/2*d*x+1/
2*c)-1)*C-1/3/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^3-2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^2-23/2/d/a^3*A*ln(tan(1/2*d*x+
1/2*c)+1)-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-17/2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)-1/d/a^3/(tan(1/2*d*x+1/2*c)+1
)*C-1/3/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)^3+2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)^2

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Maxima [A]  time = 1.06271, size = 568, normalized size = 2.52 \begin{align*} \frac{A{\left (\frac{20 \,{\left (\frac{33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{51 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3} - \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{50 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{690 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{690 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + 3 \, C{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \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{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \end{align*}

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))^3,x, algorithm="maxima")

[Out]

1/60*(A*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) - 76*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 51*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5)/(a^3 - 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x + c) + 1) + 50*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 690*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a
^3 + 690*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + 3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(si
n(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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Fricas [A]  time = 1.53899, size = 786, normalized size = 3.49 \begin{align*} -\frac{15 \,{\left ({\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (23 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (34 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{5} + 9 \,{\left (143 \, A + 38 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (869 \, A + 234 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \,{\left (19 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \, A \cos \left (d x + c\right ) + 10 \, A\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{6} + 3 \, a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + a^{3} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

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))^3,x, algorithm="fricas")

[Out]

-1/60*(15*((23*A + 6*C)*cos(d*x + c)^6 + 3*(23*A + 6*C)*cos(d*x + c)^5 + 3*(23*A + 6*C)*cos(d*x + c)^4 + (23*A
+ 6*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 15*((23*A + 6*C)*cos(d*x + c)^6 + 3*(23*A + 6*C)*cos(d*x + c)^
5 + 3*(23*A + 6*C)*cos(d*x + c)^4 + (23*A + 6*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(16*(34*A + 9*C)*c
os(d*x + c)^5 + 9*(143*A + 38*C)*cos(d*x + c)^4 + (869*A + 234*C)*cos(d*x + c)^3 + 5*(19*A + 6*C)*cos(d*x + c)
^2 - 15*A*cos(d*x + c) + 10*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^6 + 3*a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x
+ c)^4 + a^3*d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25775, size = 352, normalized size = 1.56 \begin{align*} -\frac{\frac{30 \,{\left (23 \, A + 6 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{30 \,{\left (23 \, A + 6 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{20 \,{\left (51 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 76 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 50 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 735 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

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))^3,x, algorithm="giac")

[Out]

-1/60*(30*(23*A + 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(23*A + 6*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 20*(51*A*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 76*A*tan(1/2*d*x + 1/2*c)^3 - 12*C*t
an(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*
a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 50*A*a^12*tan(1/2*d*x + 1/2*c)^3 +
30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d