Integrand size = 31, antiderivative size = 134 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b \left (2 a^2 A-A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
-2*b*(2*A*a^2-A*b^2+C*a^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/ 2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+A*arctanh(sin(d*x+c))/a^2/d+(A*b^2+C*a^2 )*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))
Result contains complex when optimal does not.
Time = 3.01 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.28 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 \cos (c+d x) (C \cos (c+d x)+A \sec (c+d x)) \left (-A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 b \left (-A b^2+a^2 (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right ) \sqrt {\left (-a^2+b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac {a \left (A b^2+a^2 C\right ) (-a \sin (c)+b \sin (d x))}{(a-b) b (a+b) (a+b \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{a^2 d (2 A+C+C \cos (2 (c+d x)))} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^2,x]
Output:
(2*Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*(-(A*Log[Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2]]) + A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*b *(-(A*b^2) + a^2*(2*A + C))*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(I*Co s[c] + Sin[c]))/((a^2 - b^2)*Sqrt[(-a^2 + b^2)*(Cos[c] - I*Sin[c])^2]) + ( a*(A*b^2 + a^2*C)*(-(a*Sin[c]) + b*Sin[d*x]))/((a - b)*b*(a + b)*(a + b*Co s[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(a^2*d*(2*A + C + C*Cos[2*(c + d*x)]))
Time = 0.76 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int \frac {\left (A \left (a^2-b^2\right )-a b (A+C) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A \left (a^2-b^2\right )-a b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {b \left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}+\frac {A \left (a^2-b^2\right ) \int \sec (c+d x)dx}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b \left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 b \left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}+\frac {A \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {A \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 b \left (A b^2-a^2 (2 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {2 b \left (A b^2-a^2 (2 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}+\frac {A \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^2,x]
Output:
((2*b*(A*b^2 - a^2*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (A*(a^2 - b^2)*ArcTanh[Sin[c + d*x ]])/(a*d))/(a*(a^2 - b^2)) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2) *d*(a + b*Cos[c + d*x]))
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 = 0.57 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.37
| method | result | size |
| derivativedivides | \(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {2 \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2}-A \,b^{2}+a^{2} C \right ) b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(183\) |
| default | \(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {2 \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 A \,a^{2}-A \,b^{2}+a^{2} C \right ) b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(183\) |
| risch | \(-\frac {2 i \left (A \,b^{2}+a^{2} C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{\left (-a^{2}+b^{2}\right ) a d b \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,a^{2}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}\) | \(625\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
1/d*(-A/a^2*ln(tan(1/2*d*x+1/2*c)-1)-2/a^2*(-a*(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)+(2*A*a ^2-A*b^2+C*a^2)*b/(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*ln(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (125) = 250\).
Time = 1.22 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.97 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="f ricas")
Output:
[-1/2*(((2*A + C)*a^3*b - A*a*b^3 + ((2*A + C)*a^2*b^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)) - (A*a^5 - 2*A*a^3*b^2 + A* a*b^4 + (A*a^4*b - 2*A*a^2*b^3 + A*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1 ) + (A*a^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b - 2*A*a^2*b^3 + A*b^5)*cos(d *x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^5 + (A - C)*a^3*b^2 - A*a*b^4)*si n(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b ^2 + a^3*b^4)*d), -1/2*(2*((2*A + C)*a^3*b - A*a*b^3 + ((2*A + C)*a^2*b^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*a^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b - 2* A*a^2*b^3 + A*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + (A*a^5 - 2*A*a^3* b^2 + A*a*b^4 + (A*a^4*b - 2*A*a^2*b^3 + A*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^5 + (A - C)*a^3*b^2 - A*a*b^4)*sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d)]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c))**2,x)
Output:
Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x))**2, x)
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="m axima")
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
Time = 0.16 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.69 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (2 \, A a^{2} b + C a^{2} b - A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="g iac")
Output:
-(2*(2*A*a^2*b + 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)))/((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) - A*log(abs(tan(1/2*d*x + 1/ 2*c) + 1))/a^2 + A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*(C*a^2*tan(1 /2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2*d* x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)))/d
Time = 4.77 (sec) , antiderivative size = 3850, normalized size of antiderivative = 28.73 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + b*cos(c + d*x))^2),x)
Output:
- (A*atan(((A*((A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 + C*a^ 5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b ^3 - a^4*b^2) - (32*A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^ 2))))/a^2 - (32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 - 2*A^2*a*b^5 - 2* A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2 - (A*((A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4 *b^2) + (32*A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6* b^4 - 4*a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2))))/a^ 2 + (32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 - 2*A^2*a*b^5 - 2*A^2*a^5* b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 2*A*C*a^ 2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2)/((A*(( A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 + C*a^5*b^4 - C*a^6*b^ 3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4* a^7*b^3 - 2*a^8*b^2))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2))))/a^2 - (32* tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^ 2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 2*A*C*a^2*b^4...
Time = 0.17 (sec) , antiderivative size = 710, normalized size of antiderivative = 5.30 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x)
Output:
( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t(a**2 - b**2))*cos(c + d*x)*a**2*b**2 - 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*b**2*c + 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)*b**4 - 4*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 - 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**2*b *c + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*a*b**3 - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**3 - cos(c + d*x)*log(tan( (c + d*x)/2) - 1)*b**5 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**4*b - 2 *cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**3 + cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*b**5 - log(tan((c + d*x)/2) - 1)*a**5 + 2*log(tan((c + d* x)/2) - 1)*a**3*b**2 - log(tan((c + d*x)/2) - 1)*a*b**4 + log(tan((c + d*x )/2) + 1)*a**5 - 2*log(tan((c + d*x)/2) + 1)*a**3*b**2 + log(tan((c + d*x) /2) + 1)*a*b**4 + sin(c + d*x)*a**4*c + sin(c + d*x)*a**3*b**2 - sin(c + d *x)*a**2*b**2*c - sin(c + d*x)*a*b**4)/(a*d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))