Integrand size = 21, antiderivative size = 83 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {8 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {7 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]
1/5*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-8/15*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2 +7/15*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.55 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (2+6 \cos (c+d x)+7 \cos ^2(c+d x)\right ) \sin (c+d x)}{15 a^3 d (1+\cos (c+d x))^3} \]
Time = 0.42 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3237, 25, 3042, 3229, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3237 |
\(\displaystyle \frac {\int -\frac {3 a-5 a \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\int \frac {3 a-5 a \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\int \frac {3 a-5 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\frac {8 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {7}{3} \int \frac {1}{\cos (c+d x) a+a}dx}{5 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\frac {8 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {7}{3} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a^2}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\frac {8 a \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}-\frac {7 \sin (c+d x)}{3 d (a \cos (c+d x)+a)}}{5 a^2}\) |
Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) - ((8*a*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) - (7*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])))/(5*a^2)
3.1.66.3.1 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[(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 = 0.71 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54
method | result | size |
derivativedivides | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
default | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
parallelrisch | \(\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a^{3} d}\) | \(47\) |
risch | \(\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{3 i \left (d x +c \right )}+40 \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )}+7\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(69\) |
norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{30 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}}{a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(114\) |
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*(7*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Time = 1.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((tan(c/2 + d*x/2)**5/(20*a**3*d) - tan(c/2 + d*x/2)**3/(6*a**3*d ) + tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*cos(c)**2/(a*cos(c) + a)**3 , True))
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{60 \, a^{3} d} \]
1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^3*d)
Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.55 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
1/60*(3*tan(1/2*d*x + 1/2*c)^5 - 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d* x + 1/2*c))/(a^3*d)
Time = 14.54 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\right )}{60\,a^3\,d} \]