\(\int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx\) [85]

Optimal result

Integrand size = 21, antiderivative size = 155 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {67 \sin (c+d x)}{315 a^2 d (a+a \cos (c+d x))^3}-\frac {142 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}+\frac {83 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )} \]

[Out]

-1/9*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^5-11/63*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^4+67/315*
sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^3-142/315*sin(d*x+c)/a^3/d/(a+a*cos(d*x+c))^2+83/315*sin(d*x+c)/d/(a^5+a^5*c
os(d*x+c))

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2844, 3056, 3047, 3098, 2829, 2727} \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {83 \sin (c+d x)}{315 d \left (a^5 \cos (c+d x)+a^5\right )}-\frac {142 \sin (c+d x)}{315 a^3 d (a \cos (c+d x)+a)^2}+\frac {67 \sin (c+d x)}{315 a^2 d (a \cos (c+d x)+a)^3}-\frac {\sin (c+d x) \cos ^3(c+d x)}{9 d (a \cos (c+d x)+a)^5}-\frac {11 \sin (c+d x) \cos ^2(c+d x)}{63 a d (a \cos (c+d x)+a)^4} \]

[In]

Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^5,x]

[Out]

-1/9*(Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^5) - (11*Cos[c + d*x]^2*Sin[c + d*x])/(63*a*d*(a +
a*Cos[c + d*x])^4) + (67*Sin[c + d*x])/(315*a^2*d*(a + a*Cos[c + d*x])^3) - (142*Sin[c + d*x])/(315*a^3*d*(a +
a*Cos[c + d*x])^2) + (83*Sin[c + d*x])/(315*d*(a^5 + a^5*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 2829

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 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 3047

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

Rule 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {\int \frac {\cos ^2(c+d x) (3 a-8 a \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx}{9 a^2} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac {\int \frac {\cos (c+d x) \left (22 a^2-45 a^2 \cos (c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx}{63 a^4} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac {\int \frac {22 a^2 \cos (c+d x)-45 a^2 \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{63 a^4} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {67 \sin (c+d x)}{315 a^2 d (a+a \cos (c+d x))^3}+\frac {\int \frac {-201 a^3+225 a^3 \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{315 a^6} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {67 \sin (c+d x)}{315 a^2 d (a+a \cos (c+d x))^3}-\frac {142 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}+\frac {83 \int \frac {1}{a+a \cos (c+d x)} \, dx}{315 a^4} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {11 \cos ^2(c+d x) \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {67 \sin (c+d x)}{315 a^2 d (a+a \cos (c+d x))^3}-\frac {142 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}+\frac {83 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.74 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\left (8+40 \cos (c+d x)+84 \cos ^2(c+d x)+100 \cos ^3(c+d x)+83 \cos ^4(c+d x)\right ) \sin (c+d x)}{315 a^5 d (1+\cos (c+d x))^5} \]

[In]

Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^5,x]

[Out]

((8 + 40*Cos[c + d*x] + 84*Cos[c + d*x]^2 + 100*Cos[c + d*x]^3 + 83*Cos[c + d*x]^4)*Sin[c + d*x])/(315*a^5*d*(
1 + Cos[c + d*x])^5)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(71\)
default	\(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(71\)
parallelrisch	\(\frac {35 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-180 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+378 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-420 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5040 a^{5} d}\)	\(73\)
risch	\(\frac {2 i \left (315 \,{\mathrm e}^{8 i \left (d x +c \right )}+1260 \,{\mathrm e}^{7 i \left (d x +c \right )}+3360 \,{\mathrm e}^{6 i \left (d x +c \right )}+5040 \,{\mathrm e}^{5 i \left (d x +c \right )}+5418 \,{\mathrm e}^{4 i \left (d x +c \right )}+3612 \,{\mathrm e}^{3 i \left (d x +c \right )}+1728 \,{\mathrm e}^{2 i \left (d x +c \right )}+432 \,{\mathrm e}^{i \left (d x +c \right )}+83\right )}{315 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\)	\(113\)
norman	\(\frac {\frac {\tan ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{70 d a}+\frac {109 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2520 d a}+\frac {19 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{630 d a}-\frac {11 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 d a}-\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{126 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{4}}\)	\(190\)

[In]

int(cos(d*x+c)^4/(a+cos(d*x+c)*a)^5,x,method=_RETURNVERBOSE)

[Out]

1/16/d/a^5*(1/9*tan(1/2*d*x+1/2*c)^9-4/7*tan(1/2*d*x+1/2*c)^7+6/5*tan(1/2*d*x+1/2*c)^5-4/3*tan(1/2*d*x+1/2*c)^
3+tan(1/2*d*x+1/2*c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {{\left (83 \, \cos \left (d x + c\right )^{4} + 100 \, \cos \left (d x + c\right )^{3} + 84 \, \cos \left (d x + c\right )^{2} + 40 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{315 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^5,x, algorithm="fricas")

[Out]

1/315*(83*cos(d*x + c)^4 + 100*cos(d*x + c)^3 + 84*cos(d*x + c)^2 + 40*cos(d*x + c) + 8)*sin(d*x + c)/(a^5*d*c
os(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x +
c) + a^5*d)

Sympy [A] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\begin {cases} \frac {\tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{144 a^{5} d} - \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{28 a^{5} d} + \frac {3 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{5} d} - \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{12 a^{5} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{16 a^{5} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4/(a+a*cos(d*x+c))**5,x)

[Out]

Piecewise((tan(c/2 + d*x/2)**9/(144*a**5*d) - tan(c/2 + d*x/2)**7/(28*a**5*d) + 3*tan(c/2 + d*x/2)**5/(40*a**5
*d) - tan(c/2 + d*x/2)**3/(12*a**5*d) + tan(c/2 + d*x/2)/(16*a**5*d), Ne(d, 0)), (x*cos(c)**4/(a*cos(c) + a)**
5, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {420 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {180 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{5040 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^5,x, algorithm="maxima")

[Out]

1/5040*(315*sin(d*x + c)/(cos(d*x + c) + 1) - 420*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 180*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{5040 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^5,x, algorithm="giac")

[Out]

1/5040*(35*tan(1/2*d*x + 1/2*c)^9 - 180*tan(1/2*d*x + 1/2*c)^7 + 378*tan(1/2*d*x + 1/2*c)^5 - 420*tan(1/2*d*x
+ 1/2*c)^3 + 315*tan(1/2*d*x + 1/2*c))/(a^5*d)

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+378\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-180\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{5040\,a^5\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]

[In]

int(cos(c + d*x)^4/(a + a*cos(c + d*x))^5,x)

[Out]

(sin(c/2 + (d*x)/2)*(315*cos(c/2 + (d*x)/2)^8 + 35*sin(c/2 + (d*x)/2)^8 - 180*cos(c/2 + (d*x)/2)^2*sin(c/2 + (
d*x)/2)^6 + 378*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 - 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(5
040*a^5*d*cos(c/2 + (d*x)/2)^9)