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

Optimal. Leaf size=210 \[ -\frac{2 (76 A-36 B+11 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 A-6 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{(13 A-6 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

((13*A - 6*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(76*A - 36*B + 11*C)*Tan[c + d*x])/(15*a^3*d) + ((13
*A - 6*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos
[c + d*x])^3) - ((11*A - 6*B + C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((76*A - 36*B +
11*C)*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.556505, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3041, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{2 (76 A-36 B+11 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 A-6 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{(13 A-6 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{(76 A-36 B+11 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(11 A-6 B+C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((13*A - 6*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(76*A - 36*B + 11*C)*Tan[c + d*x])/(15*a^3*d) + ((13
*A - 6*B + 2*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos
[c + d*x])^3) - ((11*A - 6*B + C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((76*A - 36*B +
11*C)*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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 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 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]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (7 A-2 B+2 C)-a (4 A-4 B-C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (a^2 (43 A-18 B+8 C)-3 a^2 (11 A-6 B+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(76 A-36 B+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (15 a^3 (13 A-6 B+2 C)-2 a^3 (76 A-36 B+11 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(76 A-36 B+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(13 A-6 B+2 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac{(2 (76 A-36 B+11 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac{(13 A-6 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(76 A-36 B+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(13 A-6 B+2 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac{(2 (76 A-36 B+11 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{(13 A-6 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{2 (76 A-36 B+11 C) \tan (c+d x)}{15 a^3 d}+\frac{(13 A-6 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(11 A-6 B+C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(76 A-36 B+11 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.45282, size = 206, normalized size = 0.98 \[ \frac{2 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (-4 (107 A-57 B+22 C) \tan \left (\frac{1}{2} (c+d x)\right )-96 (A-B+C) \sin ^6\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)-16 (17 A-12 B+7 C) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-30 (13 A-6 B+2 C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-60 (3 A-B) \tan (c+d x)+30 A \tan (c+d x) \sec (c+d x)\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^6*(-30*(13*A - 6*B + 2*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) - 16*(17*A - 12*B + 7*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 96*(A - B + C)*Csc[c + d*x]
^5*Sin[(c + d*x)/2]^6 - 4*(107*A - 57*B + 22*C)*Tan[(c + d*x)/2] - 60*(3*A - B)*Tan[c + d*x] + 30*A*Sec[c + d*
x]*Tan[c + d*x]))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [B]  time = 0.08, size = 433, normalized size = 2.1 \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{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{2\,A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{13\,A}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{B\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}}-{\frac{C}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7\,A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{13\,A}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{B\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}}+{\frac{C}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}} \end{align*}

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)^3/(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*B*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-2/3/d/a
^3*tan(1/2*d*x+1/2*c)^3*A+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*B-1/3/d/a^3*C*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*A*tan(1
/2*d*x+1/2*c)+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)+7/2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1
)-1/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)-13/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)+3/d/a^3*B*ln(tan(1/2*d*x+1/2*c)-1)-1/
d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)^2+7/2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)-1/d/a
^3*B/(tan(1/2*d*x+1/2*c)+1)+13/2/d/a^3*A*ln(tan(1/2*d*x+1/2*c)+1)-3/d/a^3*B*ln(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*l
n(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)^2

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Maxima [B]  time = 1.09989, size = 666, normalized size = 3.17 \begin{align*} \text{result too large to display} \end{align*}

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)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(A*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*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 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(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(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20
*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 - 60*log(sin(d*x + c)/(cos(d
*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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Fricas [A]  time = 2.32358, size = 849, normalized size = 4.04 \begin{align*} \frac{15 \,{\left ({\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (13 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (76 \, A - 36 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (239 \, A - 114 \, B + 34 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (479 \, A - 234 \, B + 64 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

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)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((13*A - 6*B + 2*C)*cos(d*x + c)^5 + 3*(13*A - 6*B + 2*C)*cos(d*x + c)^4 + 3*(13*A - 6*B + 2*C)*cos(d
*x + c)^3 + (13*A - 6*B + 2*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*A - 6*B + 2*C)*cos(d*x + c)^5 +
3*(13*A - 6*B + 2*C)*cos(d*x + c)^4 + 3*(13*A - 6*B + 2*C)*cos(d*x + c)^3 + (13*A - 6*B + 2*C)*cos(d*x + c)^2
)*log(-sin(d*x + c) + 1) - 2*(4*(76*A - 36*B + 11*C)*cos(d*x + c)^4 + 3*(239*A - 114*B + 34*C)*cos(d*x + c)^3
+ (479*A - 234*B + 64*C)*cos(d*x + c)^2 + 15*(3*A - 2*B)*cos(d*x + c) - 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c
)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33398, size = 389, normalized size = 1.85 \begin{align*} \frac{\frac{30 \,{\left (13 \, A - 6 \, B + 2 \, 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 (13 \, A - 6 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{60 \,{\left (7 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B \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 )}^{2} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B 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} + 40 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 105 \, 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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(13*A - 6*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*A - 6*B + 2*C)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1))/a^3 + 60*(7*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c) +
2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*
tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/
2*d*x + 1/2*c)^3 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x
+ 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d