Integrand size = 25, antiderivative size = 98 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {2 (A-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 A+7 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/5*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+2/15*(A-4*C)*sin(d*x+c)/a/d/(a+a *cos(d*x+c))^2+1/15*(2*A+7*C)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Time = 0.98 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.32 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (20 (A+2 C) \sin \left (\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )+10 A \sin \left (c+\frac {3 d x}{2}\right )+20 C \sin \left (c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {3 d x}{2}\right )+2 A \sin \left (2 c+\frac {5 d x}{2}\right )+7 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^3,x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(20*(A + 2*C)*Sin[(d*x)/2] - 30*C*Sin[c + (d*x) /2] + 10*A*Sin[c + (3*d*x)/2] + 20*C*Sin[c + (3*d*x)/2] - 15*C*Sin[2*c + ( 3*d*x)/2] + 2*A*Sin[2*c + (5*d*x)/2] + 7*C*Sin[2*c + (5*d*x)/2]))/(30*a^3* d*(1 + Cos[c + d*x])^3)
Time = 0.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3499, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3499 |
| \(\displaystyle \frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {\int -\frac {a (2 A-3 C)+5 a C \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (2 A-3 C)+5 a C \cos (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (2 A-3 C)+5 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {\frac {1}{3} (2 A+7 C) \int \frac {1}{\cos (c+d x) a+a}dx+\frac {2 a (A-4 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} (2 A+7 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a (A-4 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {(2 A+7 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)}+\frac {2 a (A-4 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^3,x]
Output:
((A + C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((2*a*(A - 4*C)*Sin[ c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((2*A + 7*C)*Sin[c + d*x])/(3*d*( a + a*Cos[c + d*x])))/(5*a^2)
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b*(A + 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) - a*C*m + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63
| method | result | size |
| parallelrisch | \(\frac {\left (\frac {\left (A +\frac {7 C}{2}\right ) \cos \left (2 d x +2 c \right )}{6}+\cos \left (d x +c \right ) \left (A +C \right )+\frac {4 A}{3}+\frac {11 C}{12}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{10 a^{3} d}\) | \(62\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(88\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(88\) |
| risch | \(\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}+30 C \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{2 i \left (d x +c \right )}+40 C \,{\mathrm e}^{2 i \left (d x +c \right )}+10 A \,{\mathrm e}^{i \left (d x +c \right )}+20 C \,{\mathrm e}^{i \left (d x +c \right )}+2 A +7 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(102\) |
| norman | \(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}+\frac {\left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (4 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 a d}+\frac {\left (19 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{2}}\) | \(139\) |
Input:
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/10*(1/6*(A+7/2*C)*cos(2*d*x+2*c)+cos(d*x+c)*(A+C)+4/3*A+11/12*C)*tan(1/2 *d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4/a^3/d
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left ({\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A + C\right )} \cos \left (d x + c\right ) + 7 \, A + 2 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
1/15*((2*A + 7*C)*cos(d*x + c)^2 + 6*(A + C)*cos(d*x + c) + 7*A + 2*C)*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 = 1.00 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.39 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
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) + C*tan(c/2 + d*x/2)**5/(20*a**3*d) - C*tan(c/2 + d*x/2)**3/(6*a**3*d) + C*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + C*cos(c)**2)/(a*cos(c) + a)**3, True))
Time = 0.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
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 + C*(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 = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 + 10*A*tan(1 /2*d*x + 1/2*c)^3 - 10*C*tan(1/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c ) + 15*C*tan(1/2*d*x + 1/2*c))/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{4\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-2\,C\right )}{12\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \] Input:
int((A + C*cos(c + d*x)^2)/(a + a*cos(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)*(A + C))/(4*a^3*d) + (tan(c/2 + (d*x)/2)^3*(2*A - 2*C) )/(12*a^3*d) + (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} c +10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +15 a +15 c \right )}{60 a^{3} d} \] Input:
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
(tan((c + d*x)/2)*(3*tan((c + d*x)/2)**4*a + 3*tan((c + d*x)/2)**4*c + 10* tan((c + d*x)/2)**2*a - 10*tan((c + d*x)/2)**2*c + 15*a + 15*c))/(60*a**3* d)