Integrand size = 33, antiderivative size = 180 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \left (3 a^2 A b^2-2 A b^4+a^4 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {2 A b \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
2*(3*A*a^2*b^2-2*A*b^4+C*a^4)*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-2*A*b*arctanh(sin(d*x+c))/a^3/d-(2*A* b^2-a^2*(A-C))*tan(d*x+c)/a^2/(a^2-b^2)/d+(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2- b^2)/d/(a+b*cos(d*x+c))
Time = 3.43 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.22 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (\frac {2 \left (3 a^2 A b^2-2 A b^4+a^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+2 A b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\frac {a b \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+a A \tan (c+d x)\right )}{a^3 d (2 A+C+C \cos (2 (c+d x)))} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x ]
Output:
(2*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*((2*(3*a^2*A*b^2 - 2*A*b^4 + a^4* C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2 ) + 2*A*b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - (a*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b )*(a + b*Cos[c + d*x])) + a*A*Tan[c + d*x]))/(a^3*d*(2*A + C + C*Cos[2*(c + d*x)]))
Time = 1.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3535, 25, 3042, 3534, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-\left ((A-C) a^2\right )+b (A+C) \cos (c+d x) a+2 A b^2-\left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {\left (-\left ((A-C) a^2\right )+b (A+C) \cos (c+d x) a+2 A b^2-\left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((A-C) a^2\right )+b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2+\left (-C a^2-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int \frac {\left (2 A b \left (a^2-b^2\right )-a \left (C a^2+A b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int \frac {2 A b \left (a^2-b^2\right )-a \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 A b \left (a^2-b^2\right ) \int \sec (c+d x)dx}{a}-\frac {\left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 A b \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {\left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 A b \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 \left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 A b \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 \left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}+\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a d}+\frac {\frac {2 A b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 \left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]
Output:
((A*b^2 + a^2*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) - (( (-2*(3*a^2*A*b^2 - 2*A*b^4 + a^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/ Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*A*b*(a^2 - b^2)*ArcTanh[S in[c + d*x]])/(a*d))/a + ((2*A*b^2 - a^2*(A - C))*Tan[c + d*x])/(a*d))/(a* (a^2 - b^2))
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 = 0.70 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {-\frac {2 a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (3 A \,a^{2} b^{2}-2 A \,b^{4}+a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\) | \(227\) |
| default | \(\frac {\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {-\frac {2 a b \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (3 A \,a^{2} b^{2}-2 A \,b^{4}+a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{3}}-\frac {A}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\) | \(227\) |
| risch | \(\frac {2 i \left (A a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+C \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-A \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +2 A \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+C \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 A \,{\mathrm e}^{i \left (d x +c \right )} a^{3}+3 A \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}+C \,{\mathrm e}^{i \left (d x +c \right )} a^{3}-A \,a^{2} b +2 A \,b^{3}+a^{2} b C \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {2 A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {2 A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}\) | \(765\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x,method=_RETURNVER BOSE)
Output:
1/d*(2*A*b/a^3*ln(tan(1/2*d*x+1/2*c)-1)-A/a^2/(tan(1/2*d*x+1/2*c)-1)+2/a^3 *(-a*b*(A*b^2+C*a^2)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a- tan(1/2*d*x+1/2*c)^2*b+a+b)+(3*A*a^2*b^2-2*A*b^4+C*a^4)/(a+b)/(a-b)/((a+b) *(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-A/a^2/ (tan(1/2*d*x+1/2*c)+1)-2*A*b/a^3*ln(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (169) = 338\).
Time = 1.89 (sec) , antiderivative size = 842, normalized size of antiderivative = 4.68 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm= "fricas")
Output:
[-1/2*(((C*a^4*b + 3*A*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^2 + (C*a^5 + 3*A*a^ 3*b^2 - 2*A*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)*s in(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*a^4*b^2 - 2*A*a^2*b^4 + A*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a ^3*b^3 + A*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*((A*a^4*b^2 - 2* A*a^2*b^4 + A*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos( d*x + c))*log(-sin(d*x + c) + 1) - 2*(A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4 + (( A - C)*a^5*b - (3*A - C)*a^3*b^3 + 2*A*a*b^5)*cos(d*x + c))*sin(d*x + c))/ ((a^7*b - 2*a^5*b^3 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 + a^4*b ^4)*d*cos(d*x + c)), (((C*a^4*b + 3*A*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^2 + (C*a^5 + 3*A*a^3*b^2 - 2*A*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*a^4*b^2 - 2*A*a^2 *b^4 + A*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((A*a^4*b^2 - 2*A*a^2*b^4 + A*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + (A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4 + ((A - C)*a^5*b - (3*A - C)*a^3*b^ 3 + 2*A*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^3*b^5)* d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4)*d*cos(d*x + c))]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**2,x)
Output:
Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a + b*cos(c + d*x))**2, x)
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm= "maxima")
Output:
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. 382 vs. \(2 (169) = 338\).
Time = 0.19 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.12 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (C a^{4} + 3 \, A a^{2} b^{2} - 2 \, A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}}\right )}}{d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm= "giac")
Output:
-2*((C*a^4 + 3*A*a^2*b^2 - 2*A*b^4)*(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))/sq rt(a^2 - b^2)))/((a^5 - a^3*b^2)*sqrt(a^2 - b^2)) + A*b*log(abs(tan(1/2*d* x + 1/2*c) + 1))/a^3 - A*b*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + (A*a^3 *tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + C*a^2*b*tan(1/2 *d*x + 1/2*c)^3 - A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^3*tan(1/2*d*x + 1 /2*c)^3 + A*a^3*tan(1/2*d*x + 1/2*c) + A*a^2*b*tan(1/2*d*x + 1/2*c) - C*a^ 2*b*tan(1/2*d*x + 1/2*c) - A*a*b^2*tan(1/2*d*x + 1/2*c) - 2*A*b^3*tan(1/2* d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*b* tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)))/d
Time = 4.99 (sec) , antiderivative size = 4118, normalized size of antiderivative = 22.88 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^2),x)
Output:
((2*tan(c/2 + (d*x)/2)^3*(A*a^3 + 2*A*b^3 - A*a*b^2 - A*a^2*b + C*a^2*b))/ (a^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(2*A*b^3 - A*a^3 + A*a*b^2 - A*a^2*b + C*a^2*b))/(a^2*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2) ^4*(a - b) - 2*b*tan(c/2 + (d*x)/2)^2)) + (A*b*atan(((A*b*((32*tan(c/2 + ( d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b ^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C *a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (2*A*b*((32*(C*a^12 + 2*A*a ^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (64*A*b*tan(c/2 + (d*x)/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2))))/a^3)* 2i)/a^3 + (A*b*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2* a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2 ) + (2*A*b*((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9* b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8 *b + a^9 - a^6*b^3 - a^7*b^2) + (64*A*b*tan(c/2 + (d*x)/2)*(2*a^11*b - 2*a ^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^ 7 - a^4*b^3 - a^5*b^2))))/a^3)*2i)/a^3)/((64*(8*A^3*b^8 - 4*A^3*a*b^7 - 20 *A^3*a^2*b^6 + 6*A^3*a^3*b^5 + 12*A^3*a^4*b^4 + 2*A*C^2*a^7*b - 4*A^2*C...
Time = 0.18 (sec) , antiderivative size = 1194, normalized size of antiderivative = 6.63 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a **2 - b**2))*cos(c + d*x)*a**4*c + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x) /2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b**2 - 4* sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**4 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) *a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**3*b*c - 6*s qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**3 + 4*sqrt(a**2 - b**2)*atan((tan((c + d* x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*b**5 + 2* sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**3*b*c + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**2*b**3 - 4*sqrt(a**2 - b**2)*atan((tan ((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*b**5 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**5*b - 4*cos(c + d*x)*log(tan((c + d*x)/2 ) - 1)*a**3*b**3 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**5 - 2*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*a**5*b + 4*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*a**3*b**3 - 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**5 + cos(c + d*x)*sin(c + d*x)*a**5*b - cos(c + d*x)*sin(c + d*x)*a**4*b*c - 3* cos(c + d*x)*sin(c + d*x)*a**3*b**3 + cos(c + d*x)*sin(c + d*x)*a**2*b**3* c + 2*cos(c + d*x)*sin(c + d*x)*a*b**5 - 2*log(tan((c + d*x)/2) - 1)*si...