3.87 \(\int \frac{\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx\)

Optimal. Leaf size=139 \[ \frac{2 \sin (c+d x)}{45 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{2 \sin (c+d x)}{45 a^3 d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{15 a^2 d (a \cos (c+d x)+a)^3}-\frac{2 \sin (c+d x)}{9 a d (a \cos (c+d x)+a)^4}+\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

[Out]

Sin[c + d*x]/(9*d*(a + a*Cos[c + d*x])^5) - (2*Sin[c + d*x])/(9*a*d*(a + a*Cos[c + d*x])^4) + Sin[c + d*x]/(15
*a^2*d*(a + a*Cos[c + d*x])^3) + (2*Sin[c + d*x])/(45*a^3*d*(a + a*Cos[c + d*x])^2) + (2*Sin[c + d*x])/(45*d*(
a^5 + a^5*Cos[c + d*x]))

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Rubi [A]  time = 0.144441, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2758, 2750, 2650, 2648} \[ \frac{2 \sin (c+d x)}{45 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{2 \sin (c+d x)}{45 a^3 d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{15 a^2 d (a \cos (c+d x)+a)^3}-\frac{2 \sin (c+d x)}{9 a d (a \cos (c+d x)+a)^4}+\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^5,x]

[Out]

Sin[c + d*x]/(9*d*(a + a*Cos[c + d*x])^5) - (2*Sin[c + d*x])/(9*a*d*(a + a*Cos[c + d*x])^4) + Sin[c + d*x]/(15
*a^2*d*(a + a*Cos[c + d*x])^3) + (2*Sin[c + d*x])/(45*a^3*d*(a + a*Cos[c + d*x])^2) + (2*Sin[c + d*x])/(45*d*(
a^5 + a^5*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))^5} \, dx &=\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\int \frac{-5 a+9 a \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx}{9 a^2}\\ &=\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac{\int \frac{1}{(a+a \cos (c+d x))^3} \, dx}{3 a^2}\\ &=\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac{\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac{2 \int \frac{1}{(a+a \cos (c+d x))^2} \, dx}{15 a^3}\\ &=\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac{\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac{2 \sin (c+d x)}{45 a^3 d (a+a \cos (c+d x))^2}+\frac{2 \int \frac{1}{a+a \cos (c+d x)} \, dx}{45 a^4}\\ &=\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac{\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac{2 \sin (c+d x)}{45 a^3 d (a+a \cos (c+d x))^2}+\frac{2 \sin (c+d x)}{45 d \left (a^5+a^5 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.223414, size = 110, normalized size = 0.79 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-45 \sin \left (c+\frac{d x}{2}\right )+54 \sin \left (c+\frac{3 d x}{2}\right )-30 \sin \left (2 c+\frac{3 d x}{2}\right )+36 \sin \left (2 c+\frac{5 d x}{2}\right )+9 \sin \left (3 c+\frac{7 d x}{2}\right )+\sin \left (4 c+\frac{9 d x}{2}\right )+81 \sin \left (\frac{d x}{2}\right )\right ) \sec ^9\left (\frac{1}{2} (c+d x)\right )}{5760 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^5,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(81*Sin[(d*x)/2] - 45*Sin[c + (d*x)/2] + 54*Sin[c + (3*d*x)/2] - 30*Sin[2*c + (3*
d*x)/2] + 36*Sin[2*c + (5*d*x)/2] + 9*Sin[3*c + (7*d*x)/2] + Sin[4*c + (9*d*x)/2]))/(5760*a^5*d)

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Maple [A]  time = 0.039, size = 45, normalized size = 0.3 \begin{align*}{\frac{1}{16\,d{a}^{5}} \left ({\frac{1}{9} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}-{\frac{2}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+\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)^5,x)

[Out]

1/16/d/a^5*(1/9*tan(1/2*d*x+1/2*c)^9-2/5*tan(1/2*d*x+1/2*c)^5+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.17754, size = 90, normalized size = 0.65 \begin{align*} \frac{\frac{45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{18 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{720 \, a^{5} 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))^5,x, algorithm="maxima")

[Out]

1/720*(45*sin(d*x + c)/(cos(d*x + c) + 1) - 18*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^9/(cos(d*x
+ c) + 1)^9)/(a^5*d)

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Fricas [A]  time = 1.53705, size = 312, normalized size = 2.24 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{45 \,{\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} 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))^5,x, algorithm="fricas")

[Out]

1/45*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d*x + c) + 2)*sin(d*x + c)/(a^5*d*cos(
d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c)
+ a^5*d)

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Sympy [A]  time = 23.3924, size = 68, normalized size = 0.49 \begin{align*} \begin{cases} \frac{\tan ^{9}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{144 a^{5} d} - \frac{\tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{5} d} + \frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{16 a^{5} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{5}} & \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))**5,x)

[Out]

Piecewise((tan(c/2 + d*x/2)**9/(144*a**5*d) - tan(c/2 + d*x/2)**5/(40*a**5*d) + tan(c/2 + d*x/2)/(16*a**5*d),
Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**5, True))

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Giac [A]  time = 1.30468, size = 62, normalized size = 0.45 \begin{align*} \frac{5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{720 \, a^{5} 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))^5,x, algorithm="giac")

[Out]

1/720*(5*tan(1/2*d*x + 1/2*c)^9 - 18*tan(1/2*d*x + 1/2*c)^5 + 45*tan(1/2*d*x + 1/2*c))/(a^5*d)