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

Optimal. Leaf size=155 \[ \frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 a x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.256047, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2792, 3023, 2735, 2659, 205} \[ \frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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

Mathematica [A]  time = 0.664714, size = 113, normalized size = 0.73 \[ \frac{\frac{2 a^2 \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\sin (c+d x) \left (\frac{a^3 b}{(a-b) (a+b) (a+b \cos (c+d x))}+b\right )-2 a (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.096, size = 238, normalized size = 1.5 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{3}}}+2\,{\frac{{a}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{{b}^{2}d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+4\,{\frac{{a}^{4}}{d{b}^{3} \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }-6\,{\frac{{a}^{2}}{bd \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \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)^3/(a+b*cos(d*x+c))^2,x)

[Out]

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+2/d*a^3/b^2/(a^2-b^2)
*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)+4/d*a^4/b^3/(a-b)/(a+b)/((a-b)*(a+b))^
(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-6/d*a^2/b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(ta
n(1/2*d*x+1/2*c)*(a-b)/((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)^3/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.17531, size = 1220, normalized size = 7.87 \begin{align*} \left [-\frac{4 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x +{\left (2 \, a^{5} - 3 \, a^{3} b^{2} +{\left (2 \, a^{4} b - 3 \, 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 (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} +{\left (a^{4} b^{2} - 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^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d\right )}}, -\frac{2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x -{\left (2 \, a^{5} - 3 \, a^{3} b^{2} +{\left (2 \, a^{4} b - 3 \, 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 (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} +{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.42021, size = 412, normalized size = 2.66 \begin{align*} -\frac{2 \,{\left (\frac{{\left (2 \, a^{4} - 3 \, a^{2} 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} - 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} + 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 ) + 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 ) - b^{3} \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 )^{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 )}{\left (a^{2} b^{2} - b^{4}\right )}} + \frac{{\left (d x + c\right )} a}{b^{3}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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