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

Optimal. Leaf size=145 \[ \frac{2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac{(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-(((3*A - B)*ArcTanh[Sin[c + d*x]])/(a^3*d)) + (2*(36*A - 11*B)*Tan[c + d*x])/(15*a^3*d) - ((A - B)*Tan[c + d*
x])/(5*d*(a + a*Cos[c + d*x])^3) - ((9*A - 4*B)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((3*A - B)*Tan
[c + d*x])/(d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.469333, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2978, 2748, 3767, 8, 3770} \[ \frac{2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac{(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*A - B)*ArcTanh[Sin[c + d*x]])/(a^3*d)) + (2*(36*A - 11*B)*Tan[c + d*x])/(15*a^3*d) - ((A - B)*Tan[c + d*
x])/(5*d*(a + a*Cos[c + d*x])^3) - ((9*A - 4*B)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((3*A - B)*Tan
[c + d*x])/(d*(a^3 + a^3*Cos[c + d*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 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 8

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

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{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (6 A-B)-3 a (A-B) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(2 (36 A-11 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac{(3 A-B) \int \sec (c+d x) \, dx}{a^3}\\ &=-\frac{(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(2 (36 A-11 B)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac{(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac{(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 2.9235, size = 482, normalized size = 3.32 \[ \frac{960 (3 A-B) \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (-5 (51 A-32 B) \sin \left (\frac{d x}{2}\right )+(567 A-167 B) \sin \left (\frac{3 d x}{2}\right )-600 A \sin \left (c-\frac{d x}{2}\right )+375 A \sin \left (c+\frac{d x}{2}\right )-480 A \sin \left (2 c+\frac{d x}{2}\right )-60 A \sin \left (c+\frac{3 d x}{2}\right )+402 A \sin \left (2 c+\frac{3 d x}{2}\right )-225 A \sin \left (3 c+\frac{3 d x}{2}\right )+315 A \sin \left (c+\frac{5 d x}{2}\right )+30 A \sin \left (2 c+\frac{5 d x}{2}\right )+240 A \sin \left (3 c+\frac{5 d x}{2}\right )-45 A \sin \left (4 c+\frac{5 d x}{2}\right )+72 A \sin \left (2 c+\frac{7 d x}{2}\right )+15 A \sin \left (3 c+\frac{7 d x}{2}\right )+57 A \sin \left (4 c+\frac{7 d x}{2}\right )+170 B \sin \left (c-\frac{d x}{2}\right )-170 B \sin \left (c+\frac{d x}{2}\right )+160 B \sin \left (2 c+\frac{d x}{2}\right )+75 B \sin \left (c+\frac{3 d x}{2}\right )-167 B \sin \left (2 c+\frac{3 d x}{2}\right )+75 B \sin \left (3 c+\frac{3 d x}{2}\right )-95 B \sin \left (c+\frac{5 d x}{2}\right )+15 B \sin \left (2 c+\frac{5 d x}{2}\right )-95 B \sin \left (3 c+\frac{5 d x}{2}\right )+15 B \sin \left (4 c+\frac{5 d x}{2}\right )-22 B \sin \left (2 c+\frac{7 d x}{2}\right )-22 B \sin \left (4 c+\frac{7 d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*(3*A - B)*Cos[(c + d*x)/2]^6*(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]*(-5*(51*A - 32*B)*Sin[(d*x)/2] + (567*A - 167*B)*Sin
[(3*d*x)/2] - 600*A*Sin[c - (d*x)/2] + 170*B*Sin[c - (d*x)/2] + 375*A*Sin[c + (d*x)/2] - 170*B*Sin[c + (d*x)/2
] - 480*A*Sin[2*c + (d*x)/2] + 160*B*Sin[2*c + (d*x)/2] - 60*A*Sin[c + (3*d*x)/2] + 75*B*Sin[c + (3*d*x)/2] +
402*A*Sin[2*c + (3*d*x)/2] - 167*B*Sin[2*c + (3*d*x)/2] - 225*A*Sin[3*c + (3*d*x)/2] + 75*B*Sin[3*c + (3*d*x)/
2] + 315*A*Sin[c + (5*d*x)/2] - 95*B*Sin[c + (5*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] + 15*B*Sin[2*c + (5*d*x)/2
] + 240*A*Sin[3*c + (5*d*x)/2] - 95*B*Sin[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] + 15*B*Sin[4*c + (5*d*x
)/2] + 72*A*Sin[2*c + (7*d*x)/2] - 22*B*Sin[2*c + (7*d*x)/2] + 15*A*Sin[3*c + (7*d*x)/2] + 57*A*Sin[4*c + (7*d
*x)/2] - 22*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.108, size = 245, 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{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+3\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}}-{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-3\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}}+{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+cos(d*x+c)*a)^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+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/3/d/a^3
*B*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)-7/4/d/a^3*B*tan(1/2*d*x+1/2*c)+3/d/a^3*A*ln(tan(1/2*d*
x+1/2*c)-1)-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)-3/d/a^3*A*ln(tan(1/2*d*x+1/2*c
)+1)+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.02875, size = 386, normalized size = 2.66 \begin{align*} \frac{3 \, A{\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 )} - B{\left (\frac{\frac{105 \, \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{3 \, \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+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*A*(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) - B*((105*s
in(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 = 1.4683, size = 670, normalized size = 4.62 \begin{align*} -\frac{15 \,{\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (36 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(d*x + c)^2 + (3*A - B)*cos(
d*x + c))*log(sin(d*x + c) + 1) - 15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(
d*x + c)^2 + (3*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(36*A - 11*B)*cos(d*x + c)^3 + 3*(57*A - 17
*B)*cos(d*x + c)^2 + (117*A - 32*B)*cos(d*x + c) + 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x
+ c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(B*cos(c
+ d*x)*sec(c + d*x)**2/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3

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Giac [A]  time = 1.26267, size = 257, normalized size = 1.77 \begin{align*} -\frac{\frac{60 \,{\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \,{\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{120 \, 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^{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} + 30 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, B 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))*sec(d*x+c)^2/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^3 + 120*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*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 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*t
an(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d