3.76 \(\int \frac{\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=112 \[ \frac{13 \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac{13 \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}-\frac{11 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}+\frac{\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) - (11*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3) + (13*Sin[c + d*
x])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + (13*Sin[c + d*x])/(105*d*(a^4 + a^4*Cos[c + d*x]))

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Rubi [A]  time = 0.11202, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2758, 2750, 2650, 2648} \[ \frac{13 \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac{13 \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}-\frac{11 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}+\frac{\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^4,x]

[Out]

Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) - (11*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3) + (13*Sin[c + d*
x])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + (13*Sin[c + d*x])/(105*d*(a^4 + a^4*Cos[c + d*x]))

Rule 2758

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b
*(2*m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=\frac{\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{-4 a+7 a \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=\frac{\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{11 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{13 \int \frac{1}{(a+a \cos (c+d x))^2} \, dx}{35 a^2}\\ &=\frac{\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{11 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{13 \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac{13 \int \frac{1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac{\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{11 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{13 \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac{13 \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.248377, size = 99, normalized size = 0.88 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-175 \sin \left (c+\frac{d x}{2}\right )+168 \sin \left (c+\frac{3 d x}{2}\right )-105 \sin \left (2 c+\frac{3 d x}{2}\right )+91 \sin \left (2 c+\frac{5 d x}{2}\right )+13 \sin \left (3 c+\frac{7 d x}{2}\right )+280 \sin \left (\frac{d x}{2}\right )\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right )}{6720 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(280*Sin[(d*x)/2] - 175*Sin[c + (d*x)/2] + 168*Sin[c + (3*d*x)/2] - 105*Sin[2*c +
(3*d*x)/2] + 91*Sin[2*c + (5*d*x)/2] + 13*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)

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Maple [A]  time = 0.037, size = 58, normalized size = 0.5 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{1}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+cos(d*x+c)*a)^4,x)

[Out]

1/8/d/a^4*(1/7*tan(1/2*d*x+1/2*c)^7-1/5*tan(1/2*d*x+1/2*c)^5-1/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.15492, size = 117, normalized size = 1.04 \begin{align*} \frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*d)

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Fricas [A]  time = 1.54048, size = 251, normalized size = 2.24 \begin{align*} \frac{{\left (13 \, \cos \left (d x + c\right )^{3} + 52 \, \cos \left (d x + c\right )^{2} + 32 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(13*cos(d*x + c)^3 + 52*cos(d*x + c)^2 + 32*cos(d*x + c) + 8)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4
*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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Sympy [A]  time = 10.9928, size = 87, normalized size = 0.78 \begin{align*} \begin{cases} \frac{\tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} - \frac{\tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{4} d} - \frac{\tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{4} d} + \frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((tan(c/2 + d*x/2)**7/(56*a**4*d) - tan(c/2 + d*x/2)**5/(40*a**4*d) - tan(c/2 + d*x/2)**3/(24*a**4*d)
+ tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.3442, size = 80, normalized size = 0.71 \begin{align*} \frac{15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{840 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(15*tan(1/2*d*x + 1/2*c)^7 - 21*tan(1/2*d*x + 1/2*c)^5 - 35*tan(1/2*d*x + 1/2*c)^3 + 105*tan(1/2*d*x + 1
/2*c))/(a^4*d)