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

Optimal. Leaf size=112 \[ -\frac {2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {2 a x}{b^3}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}-\frac {\sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3032, 3023, 2735, 2659, 205} \[ -\frac {2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {a \sin (c+d x)}{b^2 d (a+b \cos (c+d x))}+\frac {2 a x}{b^3}-\frac {\sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*x)/b^3 - (2*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sqrt[a + b
]*d) - Sin[c + d*x]/(b^2*d) - (a*Sin[c + d*x])/(b^2*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 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1
))/(b^2*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b
*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f
*x] + b*C*d*(m + 1)*(a^2 - b^2)*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] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 1.07, size = 132, normalized size = 1.18 \[ \frac {\frac {4 a^2 c+4 a^2 d x-4 a b \sin (c+d x)+4 a b (c+d x) \cos (c+d x)-b^2 \sin (2 (c+d x))}{a+b \cos (c+d x)}+\frac {4 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 463, normalized size = 4.13 \[ \left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (2 \, a^{3} b - 2 \, a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d\right )}}, \frac {2 \, {\left (a^{3} b - a b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} - a^{2} b^{2}\right )} d x - {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{3} b - 2 \, a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 3.93, size = 422, normalized size = 3.77 \[ -\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} {\left (2 \, a - b\right )} {\left | -a + b \right |} {\left | b \right |} + {\left (4 \, a^{2} - 2 \, a b - b^{2}\right )} \sqrt {a^{2} - b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a b^{2} + \sqrt {a^{2} b^{4} - {\left (a b^{2} + b^{3}\right )} {\left (a b^{2} - b^{3}\right )}}}{a b^{2} - b^{3}}}}\right )\right )}}{{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} b^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} {\left | b \right |}} + \frac {{\left (4 \, a^{2} - 2 \, a b - b^{2} - 2 \, a {\left | b \right |} + b {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {a b^{2} - \sqrt {a^{2} b^{4} - {\left (a b^{2} + b^{3}\right )} {\left (a b^{2} - b^{3}\right )}}}{a b^{2} - b^{3}}}}\right )\right )}}{b^{4} - a b^{2} {\left | b \right |}} + \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((sqrt(a^2 - b^2)*(2*a - b)*abs(-a + b)*abs(b) + (4*a^2 - 2*a*b - b^2)*sqrt(a^2 - b^2)*abs(-a + b))*(pi*floor
(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt((a*b^2 + sqrt(a^2*b^4 - (a*b^2 + b^3)*(a*b^2 - b^3
)))/(a*b^2 - b^3))))/((a^2*b^2 - 2*a*b^3 + b^4)*b^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*abs(b)) + (4*a^2 - 2*a*b -
b^2 - 2*a*abs(b) + b*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt((a*b^2 - sq
rt(a^2*b^4 - (a*b^2 + b^3)*(a*b^2 - b^3)))/(a*b^2 - b^3))))/(b^4 - a*b^2*abs(b)) + 2*(2*a*tan(1/2*d*x + 1/2*c)
^3 - b*tan(1/2*d*x + 1/2*c)^3 + 2*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)^4
- b*tan(1/2*d*x + 1/2*c)^4 + 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*b^2))/d

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maple [A]  time = 0.09, size = 198, normalized size = 1.77 \[ -\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {4 a^{2} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(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 = 1.76, size = 314, normalized size = 2.80 \[ \frac {4\,a\,\mathrm {atan}\left (\frac {128\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,a-\frac {128\,a^2}{b}}-\frac {128\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,a\,b-128\,a^2}\right )}{b^3\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a-b\right )}{b^2}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a+b\right )}{b^2}}{d\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}-\frac {\ln \left (b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-b^2\right )}{d\,\left (b^5-a^2\,b^3\right )}-\frac {\ln \left (a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (2\,a^2\,\sqrt {b^2-a^2}-b^2\,\sqrt {b^2-a^2}\right )}{b^3\,d\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*atan((128*a*tan(c/2 + (d*x)/2))/(128*a - (128*a^2)/b) - (128*a^2*tan(c/2 + (d*x)/2))/(128*a*b - 128*a^2))
)/(b^3*d) - ((2*tan(c/2 + (d*x)/2)^3*(2*a - b))/b^2 + (2*tan(c/2 + (d*x)/2)*(2*a + b))/b^2)/(d*(a + b + tan(c/
2 + (d*x)/2)^4*(a - b) + 2*a*tan(c/2 + (d*x)/2)^2)) - (log(b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2) + (b^2
- a^2)^(1/2))*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2))/(d*(b^5 - a^2*b^3)) - (log(a*tan(c/2 + (d*x)/2) - b*tan(
c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(2*a^2*(b^2 - a^2)^(1/2) - b^2*(b^2 - a^2)^(1/2)))/(b^3*d*(a^2 - b^2))

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

[Out]

Timed out

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