\(\int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [54]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {7 x}{2 a^2}-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

7/2*x/a^2-16/3*sin(d*x+c)/a^2/d+7/2*cos(d*x+c)*sin(d*x+c)/a^2/d-8/3*cos(d*x+c)^2*sin(d*x+c)/a^2/d/(1+cos(d*x+c
))-1/3*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2844, 3056, 2813} \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {16 \sin (c+d x)}{3 a^2 d}-\frac {8 \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {7 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {7 x}{2 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^2,x]

[Out]

(7*x)/(2*a^2) - (16*Sin[c + d*x])/(3*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) - (8*Cos[c + d*x]^2*Sin[
c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2813

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

Rule 2844

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 3056

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {\cos ^2(c+d x) (3 a-5 a \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = -\frac {8 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \cos (c+d x) \left (16 a^2-21 a^2 \cos (c+d x)\right ) \, dx}{3 a^4} \\ & = \frac {7 x}{2 a^2}-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sin (c+d x) \left (-84 \arcsin (\cos (c+d x)) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\left (-32-43 \cos (c+d x)-6 \cos ^2(c+d x)+3 \cos ^3(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{6 a^2 d \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{5/2}} \]

[In]

Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^2,x]

[Out]

(Sin[c + d*x]*(-84*ArcSin[Cos[c + d*x]]*Cos[(c + d*x)/2]^4 + (-32 - 43*Cos[c + d*x] - 6*Cos[c + d*x]^2 + 3*Cos
[c + d*x]^3)*Sqrt[Sin[c + d*x]^2]))/(6*a^2*d*Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {-163 \left (\cos \left (d x +c \right )+\frac {12 \cos \left (2 d x +2 c \right )}{163}-\frac {3 \cos \left (3 d x +3 c \right )}{163}+\frac {140}{163}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 d x}{48 a^{2} d}\)	\(66\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(88\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(88\)
risch	\(\frac {7 x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 i \left (12 \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{i \left (d x +c \right )}+11\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(126\)
norman	\(\frac {\frac {7 x}{2 a}-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {149 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}-\frac {100 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {18 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}+\frac {14 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {21 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {14 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {7 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(207\)

[In]

int(cos(d*x+c)^4/(a+cos(d*x+c)*a)^2,x,method=_RETURNVERBOSE)

[Out]

1/48*(-163*(cos(d*x+c)+12/163*cos(2*d*x+2*c)-3/163*cos(3*d*x+3*c)+140/163)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*
c)^2+168*d*x)/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {21 \, d x \cos \left (d x + c\right )^{2} + 42 \, d x \cos \left (d x + c\right ) + 21 \, d x + {\left (3 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 43 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(21*d*x*cos(d*x + c)^2 + 42*d*x*cos(d*x + c) + 21*d*x + (3*cos(d*x + c)^3 - 6*cos(d*x + c)^2 - 43*cos(d*x
+ c) - 32)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (107) = 214\).

Time = 1.77 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.62 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} \frac {21 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {42 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {21 d x}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {19 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {71 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {39 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((21*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d
) + 42*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*
d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + tan(c/2 + d*x/2)**7/(6*a**2*d*
tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d
*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 71*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 +
12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(
c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*cos(c)**4/(a*cos(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(
d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} - \frac {6 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(21*(d*x + c)/a^2 - 6*(5*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*
a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 21*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 14.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-22\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-30\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+21\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

[In]

int(cos(c + d*x)^4/(a + a*cos(c + d*x))^2,x)

[Out]

(sin(c/2 + (d*x)/2) - 22*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 30*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 12*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) + 21*cos(c/2 + (d*x)/2)^3*(c + d*x))/(6*a^2*d*cos(c/2 + (d*x)/2)^
3)