Integrand size = 38, antiderivative size = 67 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {B x}{a}-\frac {2 (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d} \]
B*x/a-2*(B*b-C*a)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/d/ (a-b)^(1/2)/(a+b)^(1/2)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {B (c+d x)+\frac {2 (b B-a C) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{a d} \]
(B*(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])/(a*d)
Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3042, 4560, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {B+C \sec (c+d x)}{a+b \sec (c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {B x}{a}-\frac {(b B-a C) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B x}{a}-\frac {(b B-a C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {B x}{a}-\frac {(b B-a C) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B x}{a}-\frac {(b B-a C) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a b}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {B x}{a}-\frac {2 (b B-a C) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {B x}{a}-\frac {2 (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\) |
(B*x)/a - (2*(b*B - a*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b ]])/(a*Sqrt[a - 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\) | \(73\) |
| default | \(\frac {-\frac {2 \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\) | \(73\) |
| risch | \(\frac {B x}{a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d}\) | \(290\) |
1/d*(-2*(B*b-C*a)/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/( (a+b)*(a-b))^(1/2))+2*B/a*arctan(tan(1/2*d*x+1/2*c)))
Time = 0.29 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.73 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, {\left (B a^{2} - B b^{2}\right )} d x - {\left (C a - B b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {{\left (B a^{2} - B b^{2}\right )} d x + {\left (C a - B b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]
[1/2*(2*(B*a^2 - B*b^2)*d*x - (C*a - B*b)*sqrt(a^2 - b^2)*log((2*a*b*cos(d *x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)))/((a^3 - a*b^2)*d), ((B*a^2 - B*b^2)*d*x + (C*a - B*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))))/((a^3 - a*b^2)*d)]
\[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (58) = 116\).
Time = 0.32 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.09 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} B {\left (a - 2 \, b\right )} {\left | -a + b \right |} + \sqrt {-a^{2} + b^{2}} C a {\left | -a + b \right |} - \sqrt {-a^{2} + b^{2}} B {\left | a \right |} {\left | -a + b \right |} + \sqrt {-a^{2} + b^{2}} C {\left | a \right |} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} + \frac {{\left (B a + C a - 2 \, B b + B {\left | a \right |} - C {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{a^{2} - b {\left | a \right |}}}{d} \]
((sqrt(-a^2 + b^2)*B*(a - 2*b)*abs(-a + b) + sqrt(-a^2 + b^2)*C*a*abs(-a + b) - sqrt(-a^2 + b^2)*B*abs(a)*abs(-a + b) + sqrt(-a^2 + b^2)*C*abs(a)*ab s(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c) /sqrt(-(b + sqrt((a + b)*(a - b) + b^2))/(a - b))))/((a^2 - 2*a*b + b^2)*a ^2 + (a^2*b - 2*a*b^2 + b^3)*abs(a)) + (B*a + C*a - 2*B*b + B*abs(a) - C*a bs(a))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqr t(-(b - sqrt((a + b)*(a - b) + b^2))/(a - b))))/(a^2 - b*abs(a)))/d
Time = 19.00 (sec) , antiderivative size = 573, normalized size of antiderivative = 8.55 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {2\,B\,a\,\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 )}{d\,\left (a^2-b^2\right )}-\frac {C\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{d\,\left (a^2-b^2\right )}+\frac {C\,a^2\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {C\,b^2\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {2\,B\,b^2\,\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 )}{a\,d\,\left (a^2-b^2\right )}+\frac {B\,b^3\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {B\,a\,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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {B\,b\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{a\,d\,\left (a^2-b^2\right )} \]
(2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (C*l og((a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(d*(a^2 - b^2)) + (C*a^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d* x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) - (C*b ^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*( a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) - (2*B*b^2*at an(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d*(a^2 - b^2)) + (B*b^3*log( (a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b ^2)^(1/2))/cos(c/2 + (d*x)/2)))/(a*d*(a^2 - b^2)^(3/2)) - (B*a*b*log((a*si n(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^( 1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (B*b*log((a*cos(c/2 + ( d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos (c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(a*d*(a^2 - b^2))