Integrand size = 31, antiderivative size = 155 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {b \left (3 a^2-2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-b*(3*a^2-2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a -b)^(3/2)/(a+b)^(3/2)/d+arctanh(sin(d*x+c))/a^3/d-1/2*sin(d*x+c)/a/d/(a+b* cos(d*x+c))^2-1/2*(a^2-2*b^2)*sin(d*x+c)/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 1.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.16 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {2 b \left (-3 a^2+2 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {a \left (2 a^3-3 a b^2+b \left (a^2-2 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}}{2 a^3 d} \]
((2*b*(-3*a^2 + 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2] ])/(-a^2 + b^2)^(3/2) - 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*Log [Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (a*(2*a^3 - 3*a*b^2 + b*(a^2 - 2*b ^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2))/ (2*a^3*d)
Time = 1.00 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.28, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3535, 3042, 3535, 3042, 3480, 3042, 3138, 218, 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.
| \(\displaystyle \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int \frac {\left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (b^2-a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 \left (a^2-b^2\right )^2-a b \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (a^2-b^2\right )^2-a b \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{a}-\frac {b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}-\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a d (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {\sin (c+d x)}{2 a d (a+b \cos (c+d x))^2}\) |
-1/2*Sin[c + d*x]/(a*d*(a + b*Cos[c + d*x])^2) + (((-2*b*(3*a^2 - 2*b^2)*( a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(a*d))/(a*(a^ 2 - b^2)) - ((a^2 - 2*b^2)*Sin[c + d*x])/(a*d*(a + b*Cos[c + d*x])))/(2*a* (a^2 - b^2))
3.7.15.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}+\frac {\frac {2 \left (-\frac {\left (2 a^{2}+a b -2 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a +b \right )}-\frac {\left (2 a^{2}-a b -2 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}-\frac {b \left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{3}}}{d}\) | \(209\) |
| default | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}+\frac {\frac {2 \left (-\frac {\left (2 a^{2}+a b -2 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a +b \right )}-\frac {\left (2 a^{2}-a b -2 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}-\frac {b \left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{a^{3}}}{d}\) | \(209\) |
| risch | \(-\frac {i \left (-b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-7 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a +a^{2} b^{2}-2 b^{4}\right )}{b \,a^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\) | \(539\) |
1/d*(-1/a^3*ln(tan(1/2*d*x+1/2*c)-1)+1/a^3*ln(tan(1/2*d*x+1/2*c)+1)+2/a^3* ((-1/2*(2*a^2+a*b-2*b^2)*a/(a+b)*tan(1/2*d*x+1/2*c)^3-1/2*(2*a^2-a*b-2*b^2 )*a/(a-b)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c) ^2+a+b)^2-1/2*b*(3*a^2-2*b^2)/(a^2-b^2)/((a-b)*(a+b))^(1/2)*arctan((a-b)*t an(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (142) = 284\).
Time = 0.50 (sec) , antiderivative size = 915, normalized size of antiderivative = 5.90 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [-\frac {{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{7} b^{2} - 2 \, a^{5} b^{4} + a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d\right )}}, -\frac {{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} b^{2} - 2 \, a^{5} b^{4} + a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d\right )}}\right ] \]
[-1/4*((3*a^4*b - 2*a^2*b^3 + (3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + 2*(3*a^ 3*b^2 - 2*a*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin (d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^ 2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2* (a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2* a^6 - 5*a^4*b^2 + 3*a^2*b^4 + (a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c))* sin(d*x + c))/((a^7*b^2 - 2*a^5*b^4 + a^3*b^6)*d*cos(d*x + c)^2 + 2*(a^8*b - 2*a^6*b^3 + a^4*b^5)*d*cos(d*x + c) + (a^9 - 2*a^7*b^2 + a^5*b^4)*d), - 1/2*((3*a^4*b - 2*a^2*b^3 + (3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + 2*(3*a^3* b^2 - 2*a*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/ (sqrt(a^2 - b^2)*sin(d*x + c))) - (a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2* b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*lo g(-sin(d*x + c) + 1) + (2*a^6 - 5*a^4*b^2 + 3*a^2*b^4 + (a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b^2 - 2*a^5*b^4 + a^3*b^6)*d *cos(d*x + c)^2 + 2*(a^8*b - 2*a^6*b^3 + a^4*b^5)*d*cos(d*x + c) + (a^9...
\[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=- \int \left (- \frac {\sec {\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\, dx \]
-Integral(-sec(c + d*x)/(a**3 + 3*a**2*b*cos(c + d*x) + 3*a*b**2*cos(c + d *x)**2 + b**3*cos(c + d*x)**3), x) - Integral(cos(c + d*x)**2*sec(c + d*x) /(a**3 + 3*a**2*b*cos(c + d*x) + 3*a*b**2*cos(c + d*x)**2 + b**3*cos(c + d *x)**3), x)
Exception generated. \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (142) = 284\).
Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.01 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \]
-((3*a^2*b - 2*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arc tan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(( a^5 - a^3*b^2)*sqrt(a^2 - b^2)) + (2*a^3*tan(1/2*d*x + 1/2*c)^3 - a^2*b*ta n(1/2*d*x + 1/2*c)^3 - 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*a^3*tan(1/2*d*x + 1/2*c) + a^2*b*tan(1/2*d*x + 1/2*c) - 3*a *b^2*tan(1/2*d*x + 1/2*c) - 2*b^3*tan(1/2*d*x + 1/2*c))/((a^4 - a^2*b^2)*( a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) - log(abs( tan(1/2*d*x + 1/2*c) + 1))/a^3 + log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3)/d
Time = 7.96 (sec) , antiderivative size = 3083, normalized size of antiderivative = 19.89 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)*(a*b - 2*a^2 + 2*b^2))/(a^2*b - a^3) + (tan(c/2 + ( d*x)/2)^3*(a*b + 2*a^2 - 2*b^2))/(a^2*(a + b)))/(d*(2*a*b + tan(c/2 + (d*x )/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) - (atan((((((8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b ^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (8*tan(c/2 + (d*x)/2) *(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2) )/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4* a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a ^6*b + a^7 - a^4*b^3 - a^5*b^2))*1i)/a^3 - ((((8*(6*a^10*b - 4*a^11 + 4*a^ 6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7* b^2) + (8*tan(c/2 + (d*x)/2)*(8*a^11*b - 8*a^6*b^6 + 8*a^7*b^5 + 16*a^8*b^ 4 - 16*a^9*b^3 - 8*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))/a^3 + (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a*b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2))*1i)/a^3)/(((( 8*(6*a^10*b - 4*a^11 + 4*a^6*b^5 - 2*a^7*b^4 - 10*a^8*b^3 + 6*a^9*b^2))/(a ^8*b + a^9 - a^6*b^3 - a^7*b^2) - (8*tan(c/2 + (d*x)/2)*(8*a^11*b - 8*a^6* b^6 + 8*a^7*b^5 + 16*a^8*b^4 - 16*a^9*b^3 - 8*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*a^6 - 8*a^5*b - 8*a *b^5 + 8*b^6 - 16*a^2*b^4 + 16*a^3*b^3 + 5*a^4*b^2))/(a^6*b + a^7 - a^4*b^ 3 - a^5*b^2))/a^3 - (16*(6*a^4*b - 2*a*b^4 + 4*b^5 - 10*a^2*b^3 + 3*a^3...