\(\int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [605]

Optimal result

Integrand size = 25, antiderivative size = 85 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {x}{b^2}+\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {\sin (c+d x)}{b d (a+b \cos (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3101, 12, 2814, 2738, 211} \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}+\frac {\sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac {x}{b^2} \]

[In]

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

[Out]

-(x/b^2) + (2*a*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]*d) + Sin[c +
d*x]/(b*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 2814

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 3101

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

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {c+d x+\frac {2 a \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {b \sin (c+d x)}{a+b \cos (c+d x)}}{b^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}}{d}\)	\(110\)
default	\(\frac {-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}+\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}}{d}\)	\(110\)
risch	\(-\frac {x}{b^{2}}+\frac {2 i \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\)	\(200\)

[In]

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

[Out]

1/d*(-2/b^2*arctan(tan(1/2*d*x+1/2*c))+2/b^2*(b*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)+a/((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. 153 vs. \(2 (76) = 152\).

Time = 0.31 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.42 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {2 \, {\left (a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d\right )}}, -\frac {{\left (a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} d x - {\left (a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((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 de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.65 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b^{2}} + \frac {d x + c}{b^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} b}}{d} \]

[In]

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

[Out]

-(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)))*a/(sqrt(a^2 - b^2)*b^2) + (d*x + c)/b^2 - 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))/d

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.26 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,\left (-a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}+a^2\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}\right )}{b^2\,d\,\sqrt {b^2-a^2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )}+\frac {2\,\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {b^2-a^2}}{2}-\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}+a\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,\cos \left (c+d\,x\right )\,1{}\mathrm {i}\right )}{b\,d\,\sqrt {b^2-a^2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

(2*(a^2*atan(((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*1i - a
*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2)))/(b^2*d*(b^2 - a^2)^(1/2)*(a + b*cos(c + d*x))
) + (2*((sin(c + d*x)*(b^2 - a^2)^(1/2))/2 - cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*(b^2 - a
^2)^(1/2) + a*atan(((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*
cos(c + d*x)*1i))/(b*d*(b^2 - a^2)^(1/2)*(a + b*cos(c + d*x)))