Integrand size = 21, antiderivative size = 112 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {11 \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {13 \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {13 \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \]
1/7*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-11/35*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^ 3+13/105*sin(d*x+c)/d/(a^2+a^2*cos(d*x+c))^2+13/105*sin(d*x+c)/d/(a^4+a^4* cos(d*x+c))
Time = 0.88 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\left (8+32 \cos (c+d x)+52 \cos ^2(c+d x)+13 \cos ^3(c+d x)\right ) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^4} \]
((8 + 32*Cos[c + d*x] + 52*Cos[c + d*x]^2 + 13*Cos[c + d*x]^3)*Sin[c + d*x ])/(105*a^4*d*(1 + Cos[c + d*x])^4)
Time = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3237, 25, 3042, 3229, 3042, 3129, 3042, 3127}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
\(\Big \downarrow \) 3237 |
\(\displaystyle \frac {\int -\frac {4 a-7 a \cos (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}+\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\int \frac {4 a-7 a \cos (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\int \frac {4 a-7 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\frac {11 a \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {13}{5} \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{7 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\frac {11 a \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {13}{5} \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{7 a^2}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\frac {11 a \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{7 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\frac {11 a \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{7 a^2}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\frac {11 a \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{7 a^2}\) |
Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) - ((11*a*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (13*(Sin[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2) + Sin [c + d*x]/(3*a*d*(a + a*Cos[c + d*x]))))/5)/(7*a^2)
3.1.76.3.1 Defintions of rubi rules used
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]
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] + Simp[(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]
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] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(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)]
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] - Simp[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)]
Time = 0.78 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(58\) |
default | \(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(58\) |
parallelrisch | \(\frac {15 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-21 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{840 a^{4} d}\) | \(60\) |
risch | \(\frac {2 i \left (105 \,{\mathrm e}^{5 i \left (d x +c \right )}+175 \,{\mathrm e}^{4 i \left (d x +c \right )}+280 \,{\mathrm e}^{3 i \left (d x +c \right )}+168 \,{\mathrm e}^{2 i \left (d x +c \right )}+91 \,{\mathrm e}^{i \left (d x +c \right )}+13\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(80\) |
norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{60 d a}-\frac {31 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 d a}+\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d a}}{a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(133\) |
1/8/d/a^4*(1/7*tan(1/2*d*x+1/2*c)^7-1/5*tan(1/2*d*x+1/2*c)^5-1/3*tan(1/2*d *x+1/2*c)^3+tan(1/2*d*x+1/2*c))
Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left (13 \, \cos \left (d x + c\right )^{3} + 52 \, \cos \left (d x + c\right )^{2} + 32 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/105*(13*cos(d*x + c)^3 + 52*cos(d*x + c)^2 + 32*cos(d*x + c) + 8)*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)
Time = 1.89 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]
Piecewise((tan(c/2 + d*x/2)**7/(56*a**4*d) - tan(c/2 + d*x/2)**5/(40*a**4* d) - tan(c/2 + d*x/2)**3/(24*a**4*d) + tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**4, True))
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} d} \]
1/840*(105*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 + 15*sin(d*x + c)^7/(co s(d*x + c) + 1)^7)/(a^4*d)
Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]
1/840*(15*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 + 105*tan(1/2*d*x + 1/2*c))/(a^4*d)
Time = 14.94 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-105\right )}{840\,a^4\,d} \]