\(\int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [622]

Optimal result

Integrand size = 30, antiderivative size = 140 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {2 a \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]

[Out]

2*a*(a^2+2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-a*b*sin(d*x+c)/(a
^2-b^2)/d/(a+b*cos(d*x+c))^2-b*(2*a^2+b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3095, 2833, 12, 2738, 211} \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {2 a \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

[In]

Int[(a^2 - b^2*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^4,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 2738

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 2833

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

Rule 3095

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[-a + b*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {-a+b \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx \\ & = -\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\int \frac {2 \left (a^2+b^2\right )-2 a b \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )} \\ & = -\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\int -\frac {2 a \left (a^2+2 b^2\right )}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = -\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+2 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2} \\ & = -\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (2 a \left (a^2+2 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d} \\ & = \frac {2 a \left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {2 a \left (a^2+2 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {b \left (3 a^3+b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{d} \]

[In]

Integrate[(a^2 - b^2*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^4,x]

[Out]

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

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 \left (3 a^{2}+a b +b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \left (3 a^{2}-a b +b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {2 a \left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(197\)
default	\(\frac {\frac {-\frac {2 \left (3 a^{2}+a b +b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \left (3 a^{2}-a b +b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {2 a \left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(197\)
risch	\(-\frac {2 i \left (b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}+4 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+7 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+2 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a +2 a^{2} b^{2}+b^{4}\right )}{\left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(498\)

[In]

int((-b^2*cos(d*x+c)^2+a^2)/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (131) = 262\).

Time = 0.31 (sec) , antiderivative size = 598, normalized size of antiderivative = 4.27 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\left [-\frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d}\right ] \]

[In]

integrate((a^2-b^2*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

[-1/2*((a^5 + 2*a^3*b^2 + (a^3*b^2 + 2*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b + 2*a^2*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*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)) + 2*(3*a^5*b - 3*a^3*b^3 + (2*a^4*b^2 -
a^2*b^4 - b^6)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*
b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d), ((a^5 + 2*a^3*
b^2 + (a^3*b^2 + 2*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b + 2*a^2*b^3)*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^5*b - 3*a^3*b^3 + (2*a^4*b^2 - a^2*b^4 - b^6)*cos(d*x + c
))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5
- a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate((a**2-b**2*cos(d*x+c)**2)/(a+b*cos(d*x+c))**4,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a^2-b^2*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,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 de

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.82 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {2 \, {\left (\frac {{\left (a^{3} + 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}\right )}}{d} \]

[In]

integrate((a^2-b^2*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.55 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.73 \[ \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {2\,a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}\,\left (a^3+2\,a\,b^2\right )}\right )\,\left (a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]

[In]

int((a^2 - b^2*cos(c + d*x)^2)/(a + b*cos(c + d*x))^4,x)

[Out]

(2*a*atan((a*tan(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1/2)*(a - b)^(5/2)*
(2*a*b^2 + a^3)))*(a^2 + 2*b^2))/(d*(a + b)^(5/2)*(a - b)^(5/2)) - ((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + 3*a^2*b +
b^3))/((a + b)^2*(a - b)) + (2*tan(c/2 + (d*x)/2)*(3*a^2*b - a*b^2 + b^3))/((a + b)*(a^2 - 2*a*b + b^2)))/(d*
(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2))