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

Optimal. Leaf size=179 \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{x}{b^3} \]

[Out]

x/b^3 - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^3*
(a + b)^(5/2)*d) - (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a^2*(2*a^2 -
5*b^2)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.303562, antiderivative size = 179, 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, 3021, 2735, 2659, 205} \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/b^3 - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^3*
(a + b)^(5/2)*d) - (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a^2*(2*a^2 -
5*b^2)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^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 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 ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{a^2-2 a b \cos (c+d x)-2 \left (a^2-b^2\right ) \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{a b \left (a^2-4 b^2\right )+2 \left (a^2-b^2\right )^2 \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}-\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}-\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\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 )^2 d}\\ &=\frac{x}{b^3}-\frac{a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.09943, size = 149, normalized size = 0.83 \[ \frac{-\frac{a^2 b \sin (c+d x) \left (3 b \left (a^2-2 b^2\right ) \cos (c+d x)+2 a^3-5 a b^2\right )}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}+\frac{2 a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+2 (c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.094, size = 639, normalized size = 3.6 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

2/d/b^3*arctan(tan(1/2*d*x+1/2*c))-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^
2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2
*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+
b^2)*tan(1/2*d*x+1/2*c)^3-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b
^2)*tan(1/2*d*x+1/2*c)-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*t
an(1/2*d*x+1/2*c)+6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*
d*x+1/2*c)-2/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(
1/2))+5/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((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*b/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.21593, size = 1944, normalized size = 10.86 \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)^3/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(4*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*x*cos(d*x + c)^2 + 8*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)
*d*x*cos(d*x + c) + 4*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x - (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4 + (2*a^5*b^
2 - 5*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*l
og((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^7*b - 7*a^5*b^3 + 5*a^3*b^5 + 3*(a^6*b
^2 - 3*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x +
c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^
9)*d), 1/2*(2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*x*cos(d*x + c)^2 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 -
a*b^7)*d*x*cos(d*x + c) + 2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x - (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4 + (2*
a^5*b^2 - 5*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*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^7*b - 7*a^5*b^3 + 5*a^3*b^5 + 3*(a^6*b
^2 - 3*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x +
c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^
9)*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.46025, size = 431, normalized size = 2.41 \begin{align*} \frac{\frac{{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4}\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^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\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)^3/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

((2*a^5 - 5*a^3*b^2 + 6*a*b^4)*(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^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(a^2 - b^2)) - (2*a^5*tan(1/
2*d*x + 1/2*c)^3 - 3*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^3*tan(1/2*d*x +
1/2*c)^3 + 2*a^5*tan(1/2*d*x + 1/2*c) + 3*a^4*b*tan(1/2*d*x + 1/2*c) - 5*a^3*b^2*tan(1/2*d*x + 1/2*c) - 6*a^2
*b^3*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(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