\(\int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx\) [87]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{45 a^3 d (a+a \cos (c+d x))^2}+\frac {2 \sin (c+d x)}{45 d \left (a^5+a^5 \cos (c+d x)\right )} \]

[Out]

1/9*sin(d*x+c)/d/(a+a*cos(d*x+c))^5-2/9*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^4+1/15*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c
))^3+2/45*sin(d*x+c)/a^3/d/(a+a*cos(d*x+c))^2+2/45*sin(d*x+c)/d/(a^5+a^5*cos(d*x+c))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2837, 2829, 2729, 2727} \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {2 \sin (c+d x)}{45 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac {2 \sin (c+d x)}{45 a^3 d (a \cos (c+d x)+a)^2}+\frac {\sin (c+d x)}{15 a^2 d (a \cos (c+d x)+a)^3}-\frac {2 \sin (c+d x)}{9 a d (a \cos (c+d x)+a)^4}+\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

[In]

Int[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^5,x]

[Out]

Sin[c + d*x]/(9*d*(a + a*Cos[c + d*x])^5) - (2*Sin[c + d*x])/(9*a*d*(a + a*Cos[c + d*x])^4) + Sin[c + d*x]/(15
*a^2*d*(a + a*Cos[c + d*x])^3) + (2*Sin[c + d*x])/(45*a^3*d*(a + a*Cos[c + d*x])^2) + (2*Sin[c + d*x])/(45*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 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]

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 2837

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {\int \frac {-5 a+9 a \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx}{9 a^2} \\ & = \frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac {\int \frac {1}{(a+a \cos (c+d x))^3} \, dx}{3 a^2} \\ & = \frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{15 a^3} \\ & = \frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{45 a^3 d (a+a \cos (c+d x))^2}+\frac {2 \int \frac {1}{a+a \cos (c+d x)} \, dx}{45 a^4} \\ & = \frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {2 \sin (c+d x)}{9 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{15 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{45 a^3 d (a+a \cos (c+d x))^2}+\frac {2 \sin (c+d x)}{45 d \left (a^5+a^5 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.47 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\left (2+10 \cos (c+d x)+21 \cos ^2(c+d x)+10 \cos ^3(c+d x)+2 \cos ^4(c+d x)\right ) \sin (c+d x)}{45 a^5 d (1+\cos (c+d x))^5} \]

[In]

Integrate[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^5,x]

[Out]

((2 + 10*Cos[c + d*x] + 21*Cos[c + d*x]^2 + 10*Cos[c + d*x]^3 + 2*Cos[c + d*x]^4)*Sin[c + d*x])/(45*a^5*d*(1 +
Cos[c + d*x])^5)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.32

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(45\)
default	\(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(45\)
parallelrisch	\(\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+45 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{720 a^{5} d}\)	\(47\)
risch	\(\frac {4 i \left (30 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{5 i \left (d x +c \right )}+81 \,{\mathrm e}^{4 i \left (d x +c \right )}+54 \,{\mathrm e}^{3 i \left (d x +c \right )}+36 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{45 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\)	\(91\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}-\frac {13 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{720 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{72 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{4}}\)	\(152\)

[In]

int(cos(d*x+c)^2/(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)^9-2/5*tan(1/2*d*x+1/2*c)^5+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 = 0.88 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{45 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^5,x, algorithm="fricas")

[Out]

1/45*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d*x + c) + 2)*sin(d*x + c)/(a^5*d*cos(
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 = 3.42 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.49 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\begin {cases} \frac {\tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{144 a^{5} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{5} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{16 a^{5} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2/(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)**5/(40*a**5*d) + tan(c/2 + d*x/2)/(16*a**5*d),
Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.48 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {18 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{720 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^5,x, algorithm="maxima")

[Out]

1/720*(45*sin(d*x + c)/(cos(d*x + c) + 1) - 18*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^9/(cos(d*x
+ c) + 1)^9)/(a^5*d)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.33 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{720 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^5,x, algorithm="giac")

[Out]

1/720*(5*tan(1/2*d*x + 1/2*c)^9 - 18*tan(1/2*d*x + 1/2*c)^5 + 45*tan(1/2*d*x + 1/2*c))/(a^5*d)

Mupad [B] (verification not implemented)

Time = 14.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.32 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+45\right )}{720\,a^5\,d} \]

[In]

int(cos(c + d*x)^2/(a + a*cos(c + d*x))^5,x)

[Out]

(tan(c/2 + (d*x)/2)*(5*tan(c/2 + (d*x)/2)^8 - 18*tan(c/2 + (d*x)/2)^4 + 45))/(720*a^5*d)