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]

-1/7*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+4/35*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3+8/105*sin(d*x+c)/d/(a^2+a^2*cos(d*
x+c))^2+8/105*sin(d*x+c)/d/(a^4+a^4*cos(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 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]

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 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)]

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*}

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Mathematica [A]  time = 0.24, 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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fricas [A]  time = 0.92, size = 99, normalized size = 0.88 \[ \frac {{\left (8 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 52 \, \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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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giac [A]  time = 0.56, size = 59, normalized size = 0.53 \[ -\frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

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)

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maple [A]  time = 0.04, size = 58, normalized size = 0.52 \[ \frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a+a*cos(d*x+c))^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 = 1.20, size = 87, normalized size = 0.78 \[ \frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} d} \]

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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mupad [B]  time = 0.39, size = 58, normalized size = 0.52 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+105\right )}{840\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 - 15*tan(c/2 + (d*x)/2)^6 + 105))/(840*
a^4*d)

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sympy [A]  time = 4.46, size = 85, normalized size = 0.76 \[ \begin {cases} - \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

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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