Integrand size = 29, antiderivative size = 102 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 A+7 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]
-1/5*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(3*A-8*B)*sin(d*x+c)/a/d/( a+a*cos(d*x+c))^2+1/15*(3*A+7*B)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 0.82 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (3 A+8 B) \sin \left (\frac {d x}{2}\right )-15 (A+2 B) \sin \left (c+\frac {d x}{2}\right )+15 A \sin \left (c+\frac {3 d x}{2}\right )+20 B \sin \left (c+\frac {3 d x}{2}\right )-15 B \sin \left (2 c+\frac {3 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+7 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \]
(Cos[(c + d*x)/2]*Sec[c/2]*(5*(3*A + 8*B)*Sin[(d*x)/2] - 15*(A + 2*B)*Sin[ c + (d*x)/2] + 15*A*Sin[c + (3*d*x)/2] + 20*B*Sin[c + (3*d*x)/2] - 15*B*Si n[2*c + (3*d*x)/2] + 3*A*Sin[2*c + (5*d*x)/2] + 7*B*Sin[2*c + (5*d*x)/2])) /(30*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.58 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3447, 3042, 3498, 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.
\(\displaystyle \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3498 |
\(\displaystyle -\frac {\int -\frac {3 a (A-B)+5 a B \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {3 a (A-B)+5 a B \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a (A-B)+5 a B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {\frac {1}{3} (3 A+7 B) \int \frac {1}{\cos (c+d x) a+a}dx+\frac {a (3 A-8 B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} (3 A+7 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a (3 A-8 B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\frac {(3 A+7 B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)}+\frac {a (3 A-8 B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
-1/5*((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + ((a*(3*A - 8*B)*S in[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((3*A + 7*B)*Sin[c + d*x])/(3* d*(a + a*Cos[c + d*x])))/(5*a^2)
3.1.60.3.1 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[((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]
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] + Simp[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]
Time = 0.91 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-5 A -5 B \right )}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (-A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (-A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+30 B \,{\mathrm e}^{3 i \left (d x +c \right )}+15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+40 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A \,{\mathrm e}^{i \left (d x +c \right )}+20 B \,{\mathrm e}^{i \left (d x +c \right )}+3 A +7 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(114\) |
norman | \(\frac {-\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (3 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (3 A +2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (6 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2}}\) | \(143\) |
-1/20*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^4+10/3*B*tan(1/2*d*x+1/ 2*c)^2-5*A-5*B)/a^3/d
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left ({\left (3 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 2 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*((3*A + 7*B)*cos(d*x + c)^2 + 3*(3*A + 2*B)*cos(d*x + c) + 3*A + 2*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 = 1.01 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((-A*tan(c/2 + d*x/2)**5/(20*a**3*d) + A*tan(c/2 + d*x/2)/(4*a**3 *d) + B*tan(c/2 + d*x/2)**5/(20*a**3*d) - B*tan(c/2 + d*x/2)**3/(6*a**3*d) + B*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)/(a*c os(c) + a)**3, True))
Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
1/60*(B*(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*A*(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 = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
-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*B*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 = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B-3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]