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

Optimal. Leaf size=117 \[ -\frac{a \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12988, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3022, 12, 2754, 2659, 205} \[ -\frac{a \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3022

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]

Rule 12

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

Rule 2754

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

Mathematica [A]  time = 0.279818, size = 94, normalized size = 0.8 \[ -\frac{\frac{2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+\frac{\sin (c+d x) (a \cos (c+d x)+b)}{(a+b \cos (c+d x))^2}}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.021, size = 160, normalized size = 1.4 \begin{align*}{\frac{1}{d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( a \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+a+b \right ) ^{-2}}-{\frac{1}{d \left ( a-b \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( a \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+a+b \right ) ^{-2}}+{\frac{1}{d \left ({a}^{2}-{b}^{2} \right ) }\arctan \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(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)^3-1/d/(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)+1/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))

________________________________________________________________________________________

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)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.24778, size = 1015, normalized size = 8.68 \begin{align*} \left [\frac{{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, 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} +{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}, \frac{{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, 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} +{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.08321, size = 239, normalized size = 2.04 \begin{align*} -\frac{\frac{\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 )}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} - \frac{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} - 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 )}{{\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}{\left (a^{2} - b^{2}\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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