Integrand size = 38, antiderivative size = 67 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {C x}{b}+\frac {2 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \]
C*x/b+2*(B*b-C*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b/d/( a-b)^(1/2)/(a+b)^(1/2)
Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {C (c+d x)+\frac {2 (-b B+a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}}{b d} \]
(C*(c + d*x) + (2*(-(b*B) + a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[- a^2 + b^2]])/Sqrt[-a^2 + b^2])/(b*d)
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 3508, 3042, 3214, 3042, 3138, 218}
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.
\(\displaystyle \int \frac {\sec (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {B+C \cos (c+d x)}{a+b \cos (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {(b B-a C) \int \frac {1}{a+b \cos (c+d x)}dx}{b}+\frac {C x}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(b B-a C) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C x}{b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {2 (b B-a C) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}+\frac {C x}{b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {C x}{b}\) |
(C*x)/b + (2*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b] ])/(Sqrt[a - b]*b*Sqrt[a + b]*d)
3.8.97.3.1 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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 1.00 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (B b -a C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(73\) |
default | \(\frac {\frac {2 \left (B b -a C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(73\) |
risch | \(\frac {C x}{b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a C}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a C}{\sqrt {-a^{2}+b^{2}}\, d b}\) | \(294\) |
1/d*(2*(B*b-C*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))+2*C/b*arctan(tan(1/2*d*x+1/2*c)))
Time = 0.30 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.61 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (C a^{2} - C b^{2}\right )} d x + {\left (C a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, \frac {{\left (C a^{2} - C b^{2}\right )} d x - {\left (C a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \]
[1/2*(2*(C*a^2 - C*b^2)*d*x + (C*a - B*b)*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^2*b - b^3)*d), ((C*a^2 - C*b^2)*d*x - (C*a - B*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))))/((a^ 2*b - b^3)*d)]
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (58) = 116\).
Time = 0.33 (sec) , antiderivative size = 296, normalized size of antiderivative = 4.42 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} C {\left (2 \, a - b\right )} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} B b {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} C {\left | a - b \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (2 \, C a - B b - C b + B {\left | b \right |} - C {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}}}{d} \]
-((sqrt(a^2 - b^2)*C*(2*a - b)*abs(a - b) - sqrt(a^2 - b^2)*B*b*abs(a - b) - sqrt(a^2 - b^2)*B*abs(a - b)*abs(b) + sqrt(a^2 - b^2)*C*abs(a - b)*abs( b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1 /2*c)/sqrt((2*a + sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/((a^2 - 2*a *b + b^2)*b^2 + (a^3 - 2*a^2*b + a*b^2)*abs(b)) + (2*C*a - B*b - C*b + B*a bs(b) - C*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*t an(1/2*d*x + 1/2*c)/sqrt((2*a - sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b)) ))/(b^2 - a*abs(b)))/d
Time = 3.42 (sec) , antiderivative size = 344, normalized size of antiderivative = 5.13 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {a\,\left (C\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}-C\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}\right )-B\,b\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}+B\,b\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b\,d\,\left (a^2-b^2\right )}+\frac {2\,C\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d} \]
(a*(C*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2 )*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2))*(-(a + b)*(a - b))^(1/2) - C*log( (a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a ^2)^(1/2))/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2)) - B*b*log((b*sin(c/2 + ( d*x)/2) - a*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos (c/2 + (d*x)/2))*(-(a + b)*(a - b))^(1/2) + B*b*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + ( d*x)/2))*(b^2 - a^2)^(1/2))/(b*d*(a^2 - b^2)) + (2*C*atan(sin(c/2 + (d*x)/ 2)/cos(c/2 + (d*x)/2)))/(b*d)