Integrand size = 15, antiderivative size = 75 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=\frac {B \text {arctanh}(\sin (x))}{a^3}+\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a+a \cos (x))^2}+\frac {2 (A-11 B) \sin (x)}{15 \left (a^3+a^3 \cos (x)\right )} \]
B*arctanh(sin(x))/a^3+1/5*(A-B)*sin(x)/(a+a*cos(x))^3+1/15*(2*A-7*B)*sin(x )/a/(a+a*cos(x))^2+2/15*(A-11*B)*sin(x)/(a^3+a^3*cos(x))
Time = 0.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=\frac {-120 B \cos ^6\left (\frac {x}{2}\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+(8 A-43 B+(6 A-51 B) \cos (x)+(A-11 B) \cos (2 x)) \sin (x)}{15 a^3 (1+\cos (x))^3} \]
(-120*B*Cos[x/2]^6*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (8*A - 43*B + (6*A - 51*B)*Cos[x] + (A - 11*B)*Cos[2*x])*Sin[x])/(15*a^3* (1 + Cos[x])^3)
Time = 0.69 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {3042, 3307, 3042, 3457, 3042, 3457, 3042, 3457, 27, 3042, 4257}
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.
| \(\displaystyle \int \frac {A+B \sec (x)}{(a \cos (x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (x+\frac {\pi }{2}\right )}{\left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3307 |
| \(\displaystyle \int \frac {\sec (x) (A \cos (x)+B)}{(a \cos (x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \sin \left (x+\frac {\pi }{2}\right )+B}{\sin \left (x+\frac {\pi }{2}\right ) \left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(5 a B+2 a (A-B) \cos (x)) \sec (x)}{(\cos (x) a+a)^2}dx}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a B+2 a (A-B) \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right ) \left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (15 B a^2+(2 A-7 B) \cos (x) a^2\right ) \sec (x)}{\cos (x) a+a}dx}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {15 B a^2+(2 A-7 B) \sin \left (x+\frac {\pi }{2}\right ) a^2}{\sin \left (x+\frac {\pi }{2}\right ) \left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int 15 a^3 B \sec (x)dx}{a^2}+\frac {2 a^2 (A-11 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {15 a B \int \sec (x)dx+\frac {2 a^2 (A-11 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a B \int \csc \left (x+\frac {\pi }{2}\right )dx+\frac {2 a^2 (A-11 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (A-11 B) \sin (x)}{a \cos (x)+a}+15 a B \text {arctanh}(\sin (x))}{3 a^2}+\frac {a (2 A-7 B) \sin (x)}{3 (a \cos (x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3}\) |
((A - B)*Sin[x])/(5*(a + a*Cos[x])^3) + ((a*(2*A - 7*B)*Sin[x])/(3*(a + a* Cos[x])^2) + (15*a*B*ArcTanh[Sin[x]] + (2*a^2*(A - 11*B)*Sin[x])/(a + a*Co s[x]))/(3*a^2))/(5*a^2)
3.2.91.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85
| method | result | size |
| parallelrisch | \(\frac {-20 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+20 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+\tan \left (\frac {x}{2}\right ) \left (\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{4}+\frac {10 \left (-2 B +A \right ) \tan \left (\frac {x}{2}\right )^{2}}{3}+5 A -35 B \right )}{20 a^{3}}\) | \(64\) |
| default | \(\frac {4 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+\frac {A \tan \left (\frac {x}{2}\right )^{5}}{5}-\frac {B \tan \left (\frac {x}{2}\right )^{5}}{5}-\frac {4 B \tan \left (\frac {x}{2}\right )^{3}}{3}+\frac {2 A \tan \left (\frac {x}{2}\right )^{3}}{3}-4 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+A \tan \left (\frac {x}{2}\right )-7 B \tan \left (\frac {x}{2}\right )}{4 a^{3}}\) | \(76\) |
| norman | \(\frac {\frac {\left (A -7 B \right ) \tan \left (\frac {x}{2}\right )}{4 a}+\frac {\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{5}}{20 a}+\frac {\left (-2 B +A \right ) \tan \left (\frac {x}{2}\right )^{3}}{6 a}}{a^{2}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a^{3}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{3}}\) | \(78\) |
| risch | \(-\frac {2 i \left (15 B \,{\mathrm e}^{4 i x}+75 B \,{\mathrm e}^{3 i x}-20 A \,{\mathrm e}^{2 i x}+145 B \,{\mathrm e}^{2 i x}-10 A \,{\mathrm e}^{i x}+95 B \,{\mathrm e}^{i x}-2 A +22 B \right )}{15 \left ({\mathrm e}^{i x}+1\right )^{5} a^{3}}+\frac {B \ln \left (i+{\mathrm e}^{i x}\right )}{a^{3}}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a^{3}}\) | \(101\) |
1/20*(-20*B*ln(tan(1/2*x)-1)+20*B*ln(tan(1/2*x)+1)+tan(1/2*x)*((A-B)*tan(1 /2*x)^4+10/3*(-2*B+A)*tan(1/2*x)^2+5*A-35*B))/a^3
Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.63 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=\frac {15 \, {\left (B \cos \left (x\right )^{3} + 3 \, B \cos \left (x\right )^{2} + 3 \, B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - 15 \, {\left (B \cos \left (x\right )^{3} + 3 \, B \cos \left (x\right )^{2} + 3 \, B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A - 11 \, B\right )} \cos \left (x\right )^{2} + 3 \, {\left (2 \, A - 17 \, B\right )} \cos \left (x\right ) + 7 \, A - 32 \, B\right )} \sin \left (x\right )}{30 \, {\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} + 3 \, a^{3} \cos \left (x\right ) + a^{3}\right )}} \]
1/30*(15*(B*cos(x)^3 + 3*B*cos(x)^2 + 3*B*cos(x) + B)*log(sin(x) + 1) - 15 *(B*cos(x)^3 + 3*B*cos(x)^2 + 3*B*cos(x) + B)*log(-sin(x) + 1) + 2*(2*(A - 11*B)*cos(x)^2 + 3*(2*A - 17*B)*cos(x) + 7*A - 32*B)*sin(x))/(a^3*cos(x)^ 3 + 3*a^3*cos(x)^2 + 3*a^3*cos(x) + a^3)
\[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=\frac {\int \frac {A}{\cos ^{3}{\left (x \right )} + 3 \cos ^{2}{\left (x \right )} + 3 \cos {\left (x \right )} + 1}\, dx + \int \frac {B \sec {\left (x \right )}}{\cos ^{3}{\left (x \right )} + 3 \cos ^{2}{\left (x \right )} + 3 \cos {\left (x \right )} + 1}\, dx}{a^{3}} \]
(Integral(A/(cos(x)**3 + 3*cos(x)**2 + 3*cos(x) + 1), x) + Integral(B*sec( x)/(cos(x)**3 + 3*cos(x)**2 + 3*cos(x) + 1), x))/a**3
Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.59 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=-\frac {1}{60} \, B {\left (\frac {\frac {105 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {20 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {10 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}}{60 \, a^{3}} \]
-1/60*B*((105*sin(x)/(cos(x) + 1) + 20*sin(x)^3/(cos(x) + 1)^3 + 3*sin(x)^ 5/(cos(x) + 1)^5)/a^3 - 60*log(sin(x)/(cos(x) + 1) + 1)/a^3 + 60*log(sin(x )/(cos(x) + 1) - 1)/a^3) + 1/60*A*(15*sin(x)/(cos(x) + 1) + 10*sin(x)^3/(c os(x) + 1)^3 + 3*sin(x)^5/(cos(x) + 1)^5)/a^3
Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx=\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a^{3}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, x\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, x\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, x\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )}{60 \, a^{15}} \]
B*log(abs(tan(1/2*x) + 1))/a^3 - B*log(abs(tan(1/2*x) - 1))/a^3 + 1/60*(3* A*a^12*tan(1/2*x)^5 - 3*B*a^12*tan(1/2*x)^5 + 10*A*a^12*tan(1/2*x)^3 - 20* B*a^12*tan(1/2*x)^3 + 15*A*a^12*tan(1/2*x) - 105*B*a^12*tan(1/2*x))/a^15
Time = 26.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx={\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {A-B}{12\,a^3}+\frac {A-3\,B}{12\,a^3}\right )+\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {A-B}{4\,a^3}+\frac {A-3\,B}{4\,a^3}-\frac {A+3\,B}{4\,a^3}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3} \]