3.616 \(\int \frac{(1-\cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=204 \[ -\frac{\left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{3 b \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

[Out]

-(((2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^
(3/2)*d)) - (3*b*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((5*a^2 - 6*b^2)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)*d) - Tan[c
+ d*x]/(2*a*d*(a + b*Cos[c + d*x])^2) - ((2*a^2 - 3*b^2)*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Cos[c + d*
x]))

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Rubi [A]  time = 0.735375, antiderivative size = 204, normalized size of antiderivative = 1., 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 = {3056, 3055, 3001, 3770, 2659, 205} \[ -\frac{\left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{3 b \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

-(((2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^
(3/2)*d)) - (3*b*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((5*a^2 - 6*b^2)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)*d) - Tan[c
+ d*x]/(2*a*d*(a + b*Cos[c + d*x])^2) - ((2*a^2 - 3*b^2)*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d*(a + b*Cos[c + d*
x]))

Rule 3056

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*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

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*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (3 \left (a^2-b^2\right )-2 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (5 a^4-11 a^2 b^2+6 b^4-a b \left (a^2-b^2\right ) \cos (c+d x)-\left (2 a^2-3 b^2\right ) \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-6 b \left (a^2-b^2\right )^2-a \left (2 a^4-5 a^2 b^2+3 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{(3 b) \int \sec (c+d x) \, dx}{a^4}-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=-\frac{3 b \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (2 a^4-9 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{3 b \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac{\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.71054, size = 200, normalized size = 0.98 \[ \frac{\frac{a b \sin (c+d x) \left (b \left (3 a^2-4 b^2\right ) \cos (c+d x)+4 a^3-5 a b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))^2}-\frac{2 \left (-9 a^2 b^2+2 a^4+6 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+2 a \tan (c+d x)+6 b \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]

[Out]

((-2*(2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 6*
b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (a*b*(4*a^3 - 5*a*b^
2 + b*(3*a^2 - 4*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + 2*a*Tan[c + d*x])
/(2*a^4*d)

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Maple [B]  time = 0.067, size = 609, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x)

[Out]

4/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3*b+1/d/a^2/(a*tan(1/2*d*
x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tan(1/2*d*x+1/2*c)^3*b^2-4/d/a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/
2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a+b)*tan(1/2*d*x+1/2*c)^3+4/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+
b)^2/(a-b)*tan(1/2*d*x+1/2*c)*b-1/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)*tan(1/2*d*
x+1/2*c)*b^2-4/d/a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a-b)*tan(1/2*d*x+1/2*c)-2/d/(a
^2-b^2)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+9/d/a^2/(a^2-b^2)/((a+b)*(a-b
))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^2-6/d/a^4/(a^2-b^2)/((a+b)*(a-b))^(1/2)*arctan
((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^4-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)+3/d*b/a^4*ln(tan(1/2*d*x+1/2
*c)-1)-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)-3/d*b/a^4*ln(tan(1/2*d*x+1/2*c)+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.45014, size = 2483, normalized size = 12.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(((2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*
a^6 - 9*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x +
c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x
+ c) + a^2)) - 6*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2
+ (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x
+ c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(-sin
(d*x + c) + 1) + 2*(2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + (5*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + (8*a^6
*b - 17*a^4*b^3 + 9*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^3 + 2
*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^2 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)), -1/2*(((2*a^4*
b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 - 9*a^4*b^
2 + 6*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + 3*
((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*
b^3 + a^2*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - 3*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^
2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - (
2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + (5*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + (8*a^6*b - 17*a^4*b^3 + 9*
a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^3 + 2*(a^9*b - 2*a^7*b^3
+ a^5*b^5)*d*cos(d*x + c)^2 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c))]

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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((1-cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.69075, size = 482, normalized size = 2.36 \begin{align*} \frac{\frac{{\left (2 \, a^{4} - 9 \, a^{2} b^{2} + 6 \, b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{4 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )}{\left (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 )^{2} + a + b\right )}^{2}} - \frac{3 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} + \frac{3 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{2 \, \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}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

((2*a^4 - 9*a^2*b^2 + 6*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*
c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^6 - a^4*b^2)*sqrt(a^2 - b^2)) + (4*a^3*b*tan(1/2*d*x + 1/2*
c)^3 - 3*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^
3*b*tan(1/2*d*x + 1/2*c) + 3*a^2*b^2*tan(1/2*d*x + 1/2*c) - 5*a*b^3*tan(1/2*d*x + 1/2*c) - 4*b^4*tan(1/2*d*x +
1/2*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) - 3*b*log(abs(tan(1
/2*d*x + 1/2*c) + 1))/a^4 + 3*b*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x
+ 1/2*c)^2 - 1)*a^3))/d