3.78 \(\int \frac{1}{(a+a \cos (c+d x))^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0697271, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ \frac{2 \sin (c+d x)}{35 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac{2 \sin (c+d x)}{35 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac{3 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}+\frac{\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(-4),x]

[Out]

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

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

Mathematica [A]  time = 0.160391, size = 77, normalized size = 0.69 \[ \frac{\left (35 \sin \left (\frac{1}{2} (c+d x)\right )+21 \sin \left (\frac{3}{2} (c+d x)\right )+7 \sin \left (\frac{5}{2} (c+d x)\right )+\sin \left (\frac{7}{2} (c+d x)\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right )}{70 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(35*Sin[(c + d*x)/2] + 21*Sin[(3*(c + d*x))/2] + 7*Sin[(5*(c + d*x))/2] + Sin[(7*(c + d*x))/
2]))/(70*a^4*d*(1 + Cos[c + d*x])^4)

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Maple [A]  time = 0.035, size = 56, 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(1/(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 = 1.14996, size = 117, normalized size = 1.04 \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(1/(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 = 1.54941, size = 248, normalized size = 2.21 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 12\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(1/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/35*(2*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 13*cos(d*x + c) + 12)*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 = 5.56325, size = 83, normalized size = 0.74 \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}{\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(1/(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/(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.30866, size = 80, normalized size = 0.71 \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(1/(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)