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

Optimal. Leaf size=154 \[ \frac {2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (6 a^2-b^2\right )}{2 b^4}+\frac {3 a \sin (c+d x)}{b^3 d}+\frac {\sin (c+d x) \cos ^2(c+d x)}{b d (a+b \cos (c+d x))}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

[Out]

-1/2*(6*a^2-b^2)*x/b^4+3*a*sin(d*x+c)/b^3/d-3/2*cos(d*x+c)*sin(d*x+c)/b^2/d+cos(d*x+c)^2*sin(d*x+c)/b/d/(a+b*c
os(d*x+c))+2*a*(3*a^2-2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b^4/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3048, 3050, 3023, 2735, 2659, 205} \[ \frac {2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (6 a^2-b^2\right )}{2 b^4}+\frac {3 a \sin (c+d x)}{b^3 d}+\frac {\sin (c+d x) \cos ^2(c+d x)}{b d (a+b \cos (c+d x))}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2735

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3048

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[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 131, normalized size = 0.85 \[ \frac {2 \left (b^2-6 a^2\right ) (c+d x)-\frac {8 a \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {4 a^2 b \sin (c+d x)}{a+b \cos (c+d x)}+8 a b \sin (c+d x)-b^2 \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-6*a^2 + b^2)*(c + d*x) - (8*a*(3*a^2 - 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[
-a^2 + b^2] + 8*a*b*Sin[c + d*x] + (4*a^2*b*Sin[c + d*x])/(a + b*Cos[c + d*x]) - b^2*Sin[2*(c + d*x)])/(4*b^4*
d)

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fricas [A]  time = 0.50, size = 547, normalized size = 3.55 \[ \left [-\frac {{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x - {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, a^{4} b - 6 \, a^{2} b^{3} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}, -\frac {{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x - 2 \, {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, a^{4} b - 6 \, a^{2} b^{3} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((6*a^4*b - 7*a^2*b^3 + b^5)*d*x*cos(d*x + c) + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - (3*a^4 - 2*a^2*b^2 + (
3*a^3*b - 2*a*b^3)*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*s
qrt(-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 - 6*a^2*b^3 - (a^2*b^3 - b^5)*cos(d*x + c)^2 + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/
((a^2*b^5 - b^7)*d*cos(d*x + c) + (a^3*b^4 - a*b^6)*d), -1/2*((6*a^4*b - 7*a^2*b^3 + b^5)*d*x*cos(d*x + c) + (
6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - 2*(3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arct
an(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*a^4*b - 6*a^2*b^3 - (a^2*b^3 - b^5)*cos(d*x + c)
^2 + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*cos(d*x + c) + (a^3*b^4 - a*b^6)*d)]

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giac [A]  time = 2.86, size = 238, normalized size = 1.55 \[ \frac {\frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} b^{3}} - \frac {{\left (6 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2 \, {\left (4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 321, normalized size = 2.08 \[ \frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{3} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {6 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{4}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.01, size = 664, normalized size = 4.31 \[ \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^2+b^2\right )}{b^3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2+3\,a\,b-b^2\right )}{b^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-6\,a^2+3\,a\,b+b^2\right )}{b^3}}{d\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (3\,a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (3\,a+b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {\ln \left (b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (3\,a^3\,\sqrt {b^2-a^2}-2\,a\,b^2\,\sqrt {b^2-a^2}\right )}{b^4\,d\,\left (a^2-b^2\right )}-\frac {a\,\ln \left (a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (3\,a^2-2\,b^2\right )}{d\,\left (b^6-a^2\,b^4\right )}-\frac {\mathrm {atan}\left (\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {8\,a}{b}+\frac {24\,a^2}{b^2}-\frac {24\,a^3}{b^3}+\frac {144\,a^4}{b^4}-\frac {144\,a^5}{b^5}-8}-\frac {8\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a-8\,b+\frac {24\,a^2}{b}-\frac {24\,a^3}{b^2}+\frac {144\,a^4}{b^3}-\frac {144\,a^5}{b^4}}-\frac {24\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,b+24\,a^2-8\,b^2-\frac {24\,a^3}{b}+\frac {144\,a^4}{b^2}-\frac {144\,a^5}{b^3}}+\frac {24\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,b^2+24\,a^2\,b-24\,a^3-8\,b^3+\frac {144\,a^4}{b}-\frac {144\,a^5}{b^2}}-\frac {144\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,b^3-24\,a^3\,b+144\,a^4-8\,b^4+24\,a^2\,b^2-\frac {144\,a^5}{b}}+\frac {144\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{-144\,a^5+144\,a^4\,b-24\,a^3\,b^2+24\,a^2\,b^3+8\,a\,b^4-8\,b^5}\right )\,\left (a^2\,6{}\mathrm {i}-b^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3*(6*a^2 + b^2))/b^3 + (tan(c/2 + (d*x)/2)*(3*a*b + 6*a^2 - b^2))/b^3 - (tan(c/2 + (d*x
)/2)^5*(3*a*b - 6*a^2 + b^2))/b^3)/(d*(a + b + tan(c/2 + (d*x)/2)^2*(3*a + b) + tan(c/2 + (d*x)/2)^6*(a - b) +
tan(c/2 + (d*x)/2)^4*(3*a - b))) - (atan((8*tan(c/2 + (d*x)/2))/((8*a)/b + (24*a^2)/b^2 - (24*a^3)/b^3 + (144
*a^4)/b^4 - (144*a^5)/b^5 - 8) - (8*a*tan(c/2 + (d*x)/2))/(8*a - 8*b + (24*a^2)/b - (24*a^3)/b^2 + (144*a^4)/b
^3 - (144*a^5)/b^4) - (24*a^2*tan(c/2 + (d*x)/2))/(8*a*b + 24*a^2 - 8*b^2 - (24*a^3)/b + (144*a^4)/b^2 - (144*
a^5)/b^3) + (24*a^3*tan(c/2 + (d*x)/2))/(8*a*b^2 + 24*a^2*b - 24*a^3 - 8*b^3 + (144*a^4)/b - (144*a^5)/b^2) -
(144*a^4*tan(c/2 + (d*x)/2))/(8*a*b^3 - 24*a^3*b + 144*a^4 - 8*b^4 + 24*a^2*b^2 - (144*a^5)/b) + (144*a^5*tan(
c/2 + (d*x)/2))/(8*a*b^4 + 144*a^4*b - 144*a^5 - 8*b^5 + 24*a^2*b^3 - 24*a^3*b^2))*(a^2*6i - b^2*1i)*1i)/(b^4*
d) - (log(b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(3*a^3*(b^2 - a^2)^(1/2) - 2*a*b^2*
(b^2 - a^2)^(1/2)))/(b^4*d*(a^2 - b^2)) - (a*log(a*tan(c/2 + (d*x)/2) - b*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/
2))*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 2*b^2))/(d*(b^6 - a^2*b^4))

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

[Out]

Timed out

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