\(\int \frac {1}{(a+a \cos (c+d x))^4} \, dx\) [78]

Optimal result

Integrand size = 12, antiderivative size = 112 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {3 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{35 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {2 \sin (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2729, 2727} \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {2 \sin (c+d x)}{35 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac {2 \sin (c+d x)}{35 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac {3 \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}+\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[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 2727

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 2729

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (35 \sin \left (\frac {1}{2} (c+d x)\right )+21 \sin \left (\frac {3}{2} (c+d x)\right )+7 \sin \left (\frac {5}{2} (c+d x)\right )+\sin \left (\frac {7}{2} (c+d x)\right )\right )}{70 a^4 d (1+\cos (c+d x))^4} \]

[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)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(56\)
default	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(56\)
risch	\(\frac {4 i \left (35 \,{\mathrm e}^{3 i \left (d x +c \right )}+21 \,{\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{35 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(58\)
parallelrisch	\(\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{280 a^{4} d}\)	\(60\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d a}}{a^{3}}\)	\(80\)

[In]

int(1/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBOSE)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 12\right )} \sin \left (d x + c\right )}{35 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{280 \, a^{4} d} \]

[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)

Mupad [B] (verification not implemented)

Time = 15.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+a \cos (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+35\right )}{280\,a^4\,d} \]

[In]

int(1/(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 + 5*tan(c/2 + (d*x)/2)^6 + 35))/(280*a^
4*d)