Integrand size = 23, antiderivative size = 138 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A+4 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {2 (3 A+4 B) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {2 (3 A+4 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \]
1/7*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/35*(3*A+4*B)*sin(d*x+c)/a/d/(a +a*cos(d*x+c))^3+2/105*(3*A+4*B)*sin(d*x+c)/d/(a^2+a^2*cos(d*x+c))^2+2/105 *(3*A+4*B)*sin(d*x+c)/d/(a^4+a^4*cos(d*x+c))
Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\left (36 A+13 B+13 (3 A+4 B) \cos (c+d x)+8 (3 A+4 B) \cos ^2(c+d x)+(6 A+8 B) \cos ^3(c+d x)\right ) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^4} \]
((36*A + 13*B + 13*(3*A + 4*B)*Cos[c + d*x] + 8*(3*A + 4*B)*Cos[c + d*x]^2 + (6*A + 8*B)*Cos[c + d*x]^3)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x]) ^4)
Time = 0.52 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3229, 3042, 3129, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {(3 A+4 B) \int \frac {1}{(\cos (c+d x) a+a)^3}dx}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+4 B) \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {(3 A+4 B) \left (\frac {2 \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+4 B) \left (\frac {2 \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {(3 A+4 B) \left (\frac {2 \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+4 B) \left (\frac {2 \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a}+\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac {(3 A+4 B) \left (\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {2 \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}\right )}{7 a}\) |
((A - B)*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((3*A + 4*B)*(Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) + (2*(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)))/(7*a)
3.1.70.3.1 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 = 1.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.58
| method | result | size |
| parallelrisch | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (3 A -B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+7 \left (A +\frac {B}{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 A +7 B \right )}{56 a^{4} d}\) | \(80\) |
| derivativedivides | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(88\) |
| default | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (3 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(88\) |
| risch | \(\frac {4 i \left (70 B \,{\mathrm e}^{4 i \left (d x +c \right )}+105 A \,{\mathrm e}^{3 i \left (d x +c \right )}+70 B \,{\mathrm e}^{3 i \left (d x +c \right )}+63 A \,{\mathrm e}^{2 i \left (d x +c \right )}+84 B \,{\mathrm e}^{2 i \left (d x +c \right )}+21 A \,{\mathrm e}^{i \left (d x +c \right )}+28 B \,{\mathrm e}^{i \left (d x +c \right )}+3 A +4 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(114\) |
| norman | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (3 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (12 A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (13 A -6 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{140 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}\) | \(141\) |
1/56*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^6+7/5*(3*A-B)*tan(1/2*d* x+1/2*c)^4+7*(A+1/3*B)*tan(1/2*d*x+1/2*c)^2+7*A+7*B)/a^4/d
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left (2 \, {\left (3 \, A + 4 \, B\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, A + 4 \, B\right )} \cos \left (d x + c\right )^{2} + 13 \, {\left (3 \, A + 4 \, B\right )} \cos \left (d x + c\right ) + 36 \, A + 13 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/105*(2*(3*A + 4*B)*cos(d*x + c)^3 + 8*(3*A + 4*B)*cos(d*x + c)^2 + 13*(3 *A + 4*B)*cos(d*x + c) + 36*A + 13*B)*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 = 1.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.28 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} - \frac {B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]
Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*A*tan(c/2 + d*x/2)**5/(40 *a**4*d) + A*tan(c/2 + d*x/2)**3/(8*a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d ) - B*tan(c/2 + d*x/2)**7/(56*a**4*d) - B*tan(c/2 + d*x/2)**5/(40*a**4*d) + B*tan(c/2 + d*x/2)**3/(24*a**4*d) + B*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + B*cos(c))/(a*cos(c) + a)**4, True))
Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.27 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]
1/840*(B*(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/ (cos(d*x + c) + 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*s in(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^ 5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]
1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 63*A*ta n(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1 /2*c)^3 + 35*B*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105*B *tan(1/2*d*x + 1/2*c))/(a^4*d)
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A+B\right )}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,A-B\right )}{40\,a^4}}{d} \]