Integrand size = 33, antiderivative size = 137 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d} \]
1/2*(2*A*b^2+a^2*(A+2*C))*arctanh(sin(d*x+c))/a^3/d-2*b*(A*b^2+C*a^2)*arct an((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/d/(a-b)^(1/2)/(a+b)^(1/ 2)-A*b*tan(d*x+c)/a^2/d+1/2*A*sec(d*x+c)*tan(d*x+c)/a/d
Result contains complex when optimal does not.
Time = 3.05 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (-2 \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 b \left (A b^2+a^2 C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))}{\sqrt {\left (-a^2+b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 a A b \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 a A b \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 a^3 d (2 A+C+C \cos (2 (c+d x)))} \]
(Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-2*(2*A*b^2 + a^2*(A + 2*C))*Log[C os[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[ (c + d*x)/2] + Sin[(c + d*x)/2]] + (8*b*(A*b^2 + a^2*C)*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(C os[c] - I*Sin[c])^2)]]*(I*Cos[c] + Sin[c]))/Sqrt[(-a^2 + b^2)*(Cos[c] - I* Sin[c])^2] + (a^2*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - (4*a*A*b*Si n[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (a^2*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (4*a*A*b*Sin[(d*x)/2]) /((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(2*a^3*d* (2*A + C + C*Cos[2*(c + d*x)]))
Time = 1.00 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3535, 25, 3042, 3534, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-A b \cos ^2(c+d x)-a (A+2 C) \cos (c+d x)+2 A b\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-A b \cos ^2(c+d x)-a (A+2 C) \cos (c+d x)+2 A b\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {\left ((A+2 C) a^2+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 A b \tan (c+d x)}{a d}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\int \frac {\left ((A+2 C) a^2+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\int \frac {(A+2 C) a^2+A b \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b \left (a^2 C+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 A b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
(A*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b*(A*b^2 + a^2*C)*ArcTan[( Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*A*b*Tan [c + d*x])/(a*d))/(2*a)
3.6.68.3.1 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[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[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, 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[((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 = 2.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.59
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-2 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(218\) |
| default | \(\frac {-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-2 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(218\) |
| risch | \(-\frac {i A \left ({\mathrm e}^{3 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}\) | \(498\) |
1/d*(-1/2*A/a/(tan(1/2*d*x+1/2*c)+1)^2+1/2*(A*a^2+2*A*b^2+2*C*a^2)/a^3*ln( tan(1/2*d*x+1/2*c)+1)+1/2*A*(a+2*b)/a^2/(tan(1/2*d*x+1/2*c)+1)+1/2*A/a/(ta n(1/2*d*x+1/2*c)-1)^2+1/2/a^3*(-A*a^2-2*A*b^2-2*C*a^2)*ln(tan(1/2*d*x+1/2* c)-1)+1/2*A*(a+2*b)/a^2/(tan(1/2*d*x+1/2*c)-1)-2*b*(A*b^2+C*a^2)/a^3/((a-b )*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 1.74 (sec) , antiderivative size = 535, normalized size of antiderivative = 3.91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, {\left (C a^{2} b + A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{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 ) - {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} - 2 \, {\left (A a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, {\left (C a^{2} b + A b^{3}\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 ) \cos \left (d x + c\right )^{2} - {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} - 2 \, {\left (A a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]
[-1/4*(2*(C*a^2*b + A*b^3)*sqrt(-a^2 + b^2)*cos(d*x + c)^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)) - ((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos(d*x + c)^2* log(sin(d*x + c) + 1) + ((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos( d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 - 2*(A*a^3*b - A* a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c)^2), -1/ 4*(4*(C*a^2*b + A*b^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt( a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - ((A + 2*C)*a^4 + (A - 2*C)*a^2* b^2 - 2*A*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + ((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 - 2*(A*a^3*b - A*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a ^3*b^2)*d*cos(d*x + c)^2)]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.33 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.77 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (C a^{2} b + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \]
1/2*((A*a^2 + 2*C*a^2 + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (A*a^2 + 2*C*a^2 + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 4*(C* a^2*b + A*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan( -(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt (a^2 - b^2)*a^3) + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 + 2*A*b*tan(1/2*d*x + 1/2 *c)^3 + A*a*tan(1/2*d*x + 1/2*c) - 2*A*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d *x + 1/2*c)^2 - 1)^2*a^2))/d
Time = 5.51 (sec) , antiderivative size = 3926, normalized size of antiderivative = 28.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
(A*a*sin(c + d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b*sin (2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*a*atan((s in(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1 i)/(d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b^2*atan((sin(c/2 + (d* x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(a^3* d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b^2*atan((sin(c/2 + (d*x)/2 )*1i)/cos(c/2 + (d*x)/2))*1i)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*b^3*sin(2*c + 2*d*x))/(2*a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2 )) - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x )*1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(d*(a^2 - b^2)*(cos( 2*c + 2*d*x)/2 + 1/2)) - (A*b^2*sin(c + d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1 /2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*A^2*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*C^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/ 2) + 4*A*C*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1 /2) + 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^...