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

Optimal. Leaf size=83 \[ \frac{7 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{8 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) - (8*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (7*Sin[c + d*x]
)/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.0937042, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2758, 2750, 2648} \[ \frac{7 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{8 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) - (8*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (7*Sin[c + d*x]
)/(15*d*(a^3 + a^3*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 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))^3} \, dx &=\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{-3 a+5 a \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{7 \int \frac{1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{8 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{7 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.185515, size = 86, normalized size = 1.04 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-30 \sin \left (c+\frac{d x}{2}\right )+20 \sin \left (c+\frac{3 d x}{2}\right )-15 \sin \left (2 c+\frac{3 d x}{2}\right )+7 \sin \left (2 c+\frac{5 d x}{2}\right )+40 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{240 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(40*Sin[(d*x)/2] - 30*Sin[c + (d*x)/2] + 20*Sin[c + (3*d*x)/2] - 15*Sin[2*c + (3*
d*x)/2] + 7*Sin[2*c + (5*d*x)/2]))/(240*a^3*d)

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Maple [A]  time = 0.037, size = 45, normalized size = 0.5 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2}{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.22137, size = 90, normalized size = 1.08 \begin{align*} \frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{60 \, a^{3} 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))^3,x, algorithm="maxima")

[Out]

1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5)/(a^3*d)

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Fricas [A]  time = 1.50235, size = 186, normalized size = 2.24 \begin{align*} \frac{{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/15*(7*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a
^3*d*cos(d*x + c) + a^3*d)

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Sympy [A]  time = 5.31673, size = 68, normalized size = 0.82 \begin{align*} \begin{cases} \frac{\tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} - \frac{\tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{3} d} + \frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{3}} & \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))**3,x)

[Out]

Piecewise((tan(c/2 + d*x/2)**5/(20*a**3*d) - tan(c/2 + d*x/2)**3/(6*a**3*d) + tan(c/2 + d*x/2)/(4*a**3*d), Ne(
d, 0)), (x*cos(c)**2/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.21637, size = 62, normalized size = 0.75 \begin{align*} \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

1/60*(3*tan(1/2*d*x + 1/2*c)^5 - 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/(a^3*d)