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

Optimal. Leaf size=149 \[ \frac{a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}-\frac{x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.291669, antiderivative size = 149, normalized size of antiderivative = 1., 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, 3021, 2735, 2659, 205} \[ \frac{a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}-\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + 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[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && 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 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{\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}-\frac{\int \frac{-2 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)+2 b \left (a^2-b^2\right ) \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{-a b^2 \left (a^2-b^2\right )-2 b \left (a^2-b^2\right )^2 \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{x}{b^3}-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{x}{b^3}-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a \left (2 a^2-3 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 \left (a^2-b^2\right ) d}\\ &=-\frac{x}{b^3}+\frac{a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}-\frac{a \sin (c+d x)}{2 b^2 d (a+b \cos (c+d x))^2}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.59163, size = 291, normalized size = 1.95 \[ \frac{\frac{\frac{\sin (c+d x) \left (b \left (a^2+2 b^2\right ) \cos (c+d x)+a \left (2 a^2+b^2\right )\right )}{(a+b \cos (c+d x))^2}+\frac{6 a b \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{(a-b)^2 (a+b)^2}-\frac{-\frac{3 b \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{a b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac{2 a \left (-20 a^2 b^2+8 a^4+15 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+8 (c+d x)}{b^3}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a
^2 + b^2)^(5/2) + (a*b*(4*a^2 - 3*b^2)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) - (3*b*(4*a^4 -
7*a^2*b^2 + 2*b^4)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/b^3) + ((6*a*b*ArcTanh[((a - b)*T
an[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + ((a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Cos[c + d*x])*Sin[c
+ d*x])/(a + b*Cos[c + d*x])^2)/((a - b)^2*(a + b)^2))/(8*d)

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Maple [B]  time = 0.033, size = 475, normalized size = 3.2 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{3}}}+2\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{b}^{2} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{2} \left ( a+b \right ) }}-{\frac{a}{db \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}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{2} \left ( a+b \right ) }}+2\,{\frac{{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{2} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{2} \left ( a-b \right ) }}+{\frac{a}{db \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}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{2} \left ( a-b \right ) }}+2\,{\frac{{a}^{3}}{d{b}^{3} \left ({a}^{2}-{b}^{2} \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-3\,{\frac{a}{db \left ({a}^{2}-{b}^{2} \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

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))^3,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.66012, size = 1628, normalized size = 10.93 \begin{align*} \text{result too large to display} \end{align*}

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))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 + 8*(a^5*b - 2*a^3*b^3 + a*b^5)*d*x*cos(d*x + c) + 4*(
a^6 - 2*a^4*b^2 + a^2*b^4)*d*x + (2*a^5 - 3*a^3*b^2 + (2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(2*a^4*b - 3*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*(2*a
^5*b - 3*a^3*b^3 + a*b^5 + (3*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 +
b^9)*d*cos(d*x + c)^2 + 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*d),
-1/2*(2*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 + 4*(a^5*b - 2*a^3*b^3 + a*b^5)*d*x*cos(d*x + c) + 2*(a
^6 - 2*a^4*b^2 + a^2*b^4)*d*x - (2*a^5 - 3*a^3*b^2 + (2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(2*a^4*b - 3*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))) - (2*a^5*b -
3*a^3*b^3 + a*b^5 + (3*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d
*cos(d*x + c)^2 + 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*d)]

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

[Out]

Timed out

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Giac [B]  time = 1.68489, size = 392, normalized size = 2.63 \begin{align*} -\frac{\frac{{\left (2 \, a^{3} - 3 \, 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^{2} b^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (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}} + \frac{d x + c}{b^{3}}}{d} \end{align*}

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))^3,x, algorithm="giac")

[Out]

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