3.75 \(\int \frac{\cos ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=114 \[ \frac{12 \sin (c+d x)}{35 a^4 d (\cos (c+d x)+1)}-\frac{18 \sin (c+d x)}{35 a^4 d (\cos (c+d x)+1)^2}-\frac{\sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{8 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(-18*Sin[c + d*x])/(35*a^4*d*(1 + Cos[c + d*x])^2) + (12*Sin[c + d*x])/(35*a^4*d*(1 + Cos[c + d*x])) - (Cos[c
+ d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + (8*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.19936, antiderivative size = 114, 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 = {2765, 2968, 3019, 2750, 2648} \[ \frac{12 \sin (c+d x)}{35 a^4 d (\cos (c+d x)+1)}-\frac{18 \sin (c+d x)}{35 a^4 d (\cos (c+d x)+1)^2}-\frac{\sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{8 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^4,x]

[Out]

(-18*Sin[c + d*x])/(35*a^4*d*(1 + Cos[c + d*x])^2) + (12*Sin[c + d*x])/(35*a^4*d*(1 + Cos[c + d*x])) - (Cos[c
+ d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + (8*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 2765

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3019

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 - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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 ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{\int \frac{\cos (c+d x) (2 a-6 a \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac{\int \frac{2 a \cos (c+d x)-6 a \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{8 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{-24 a^2+30 a^2 \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{18 \sin (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{8 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{12 \int \frac{1}{a+a \cos (c+d x)} \, dx}{35 a^3}\\ &=-\frac{18 \sin (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac{\cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{8 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{12 \sin (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.257397, size = 112, normalized size = 0.98 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-210 \sin \left (c+\frac{d x}{2}\right )+147 \sin \left (c+\frac{3 d x}{2}\right )-105 \sin \left (2 c+\frac{3 d x}{2}\right )+49 \sin \left (2 c+\frac{5 d x}{2}\right )-35 \sin \left (3 c+\frac{5 d x}{2}\right )+12 \sin \left (3 c+\frac{7 d x}{2}\right )+210 \sin \left (\frac{d x}{2}\right )\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right )}{2240 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(210*Sin[(d*x)/2] - 210*Sin[c + (d*x)/2] + 147*Sin[c + (3*d*x)/2] - 105*Sin[2*c +
(3*d*x)/2] + 49*Sin[2*c + (5*d*x)/2] - 35*Sin[3*c + (5*d*x)/2] + 12*Sin[3*c + (7*d*x)/2]))/(2240*a^4*d)

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Maple [A]  time = 0.04, 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{3}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}- \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)^3/(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+3/5*tan(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.42387, size = 117, normalized size = 1.03 \begin{align*} \frac{\frac{35 \, \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{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/280*(35*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 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*d)

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Fricas [A]  time = 1.54253, size = 248, normalized size = 2.18 \begin{align*} \frac{{\left (12 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{35 \,{\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)^3/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/35*(12*cos(d*x + c)^3 + 13*cos(d*x + c)^2 + 8*cos(d*x + c) + 2)*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 = 22.5238, size = 88, normalized size = 0.77 \begin{align*} \begin{cases} - \frac{\tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} + \frac{3 \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 )}}{8 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 ^{3}{\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)**3/(a+a*cos(d*x+c))**4,x)

[Out]

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

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Giac [A]  time = 1.4228, size = 80, normalized size = 0.7 \begin{align*} -\frac{5 \, \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} - 35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{280 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/280*(5*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 - 35*tan(1/2*d*x + 1/
2*c))/(a^4*d)