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

Optimal. Leaf size=112 \[ \frac{8 \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac{8 \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac{4 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

-Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) + (4*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3) + (8*Sin[c + d*x
])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + (8*Sin[c + d*x])/(105*d*(a^4 + a^4*Cos[c + d*x]))

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Rubi [A]  time = 0.0783698, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2750, 2650, 2648} \[ \frac{8 \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac{8 \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac{4 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) + (4*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3) + (8*Sin[c + d*x
])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + (8*Sin[c + d*x])/(105*d*(a^4 + a^4*Cos[c + d*x]))

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 (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{4 \int \frac{1}{(a+a \cos (c+d x))^3} \, dx}{7 a}\\ &=-\frac{\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{4 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{8 \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{4 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{8 \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac{8 \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{4 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{8 \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac{8 \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.220062, size = 87, normalized size = 0.78 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-35 \sin \left (c+\frac{d x}{2}\right )+2 \left (21 \sin \left (c+\frac{3 d x}{2}\right )+7 \sin \left (2 c+\frac{5 d x}{2}\right )+\sin \left (3 c+\frac{7 d x}{2}\right )\right )+35 \sin \left (\frac{d x}{2}\right )\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right )}{1680 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(35*Sin[(d*x)/2] - 35*Sin[c + (d*x)/2] + 2*(21*Sin[c + (3*d*x)/2] + 7*Sin[2*c + (
5*d*x)/2] + Sin[3*c + (7*d*x)/2])))/(1680*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)/(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 = 1.16435, 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)/(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 = 1.53257, size = 251, normalized size = 2.24 \begin{align*} \frac{{\left (8 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 52 \, \cos \left (d x + c\right ) + 13\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)/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(8*cos(d*x + c)^3 + 32*cos(d*x + c)^2 + 52*cos(d*x + c) + 13)*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 = 7.78498, size = 85, normalized size = 0.76 \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{\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)/(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)/(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.35611, 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)/(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)