Integrand size = 23, antiderivative size = 102 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 A+3 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]
1/5*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(2*A+3*B)*sin(d*x+c)/a/d/(a +a*cos(d*x+c))^2+1/15*(2*A+3*B)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (7 A+3 B+(6 A+9 B) \cos (c+d x)+(2 A+3 B) \cos ^2(c+d x)\right ) \sin (c+d x)}{15 a^3 d (1+\cos (c+d x))^3} \]
((7*A + 3*B + (6*A + 9*B)*Cos[c + d*x] + (2*A + 3*B)*Cos[c + d*x]^2)*Sin[c + d*x])/(15*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.41 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {(2 A+3 B) \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{5 a}+\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(2 A+3 B) \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(2 A+3 B) \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(2 A+3 B) \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(2 A+3 B) \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}\) |
((A - B)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((2*A + 3*B)*(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*a)
3.1.61.3.1 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)]
Time = 1.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 A +5 B \right )}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 B \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+10 A \,{\mathrm e}^{i \left (d x +c \right )}+15 B \,{\mathrm e}^{i \left (d x +c \right )}+2 A +3 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(90\) |
norman | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (5 A +3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (13 A -3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}\) | \(117\) |
1/20*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^4+10/3*A*tan(1/2*d*x+1/2 *c)^2+5*A+5*B)/a^3/d
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left ({\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 3 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*((2*A + 3*B)*cos(d*x + c)^2 + 3*(2*A + 3*B)*cos(d*x + c) + 7*A + 3*B) *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 = 0.78 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((A*tan(c/2 + d*x/2)**5/(20*a**3*d) + A*tan(c/2 + d*x/2)**3/(6*a* *3*d) + A*tan(c/2 + d*x/2)/(4*a**3*d) - B*tan(c/2 + d*x/2)**5/(20*a**3*d) + B*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))/(a*cos(c) + a)**3, True))
Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
1/60*(A*(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 + 3*B*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 10*A*tan(1 /2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) + 15*B*tan(1/2*d*x + 1/2*c)) /(a^3*d)
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B+10\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]