Integrand size = 39, antiderivative size = 98 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {(A b-a B) x}{a^2}+\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}+\frac {A \sin (c+d x)}{a d} \]
-(A*b-B*a)*x/a^2+A*sin(d*x+c)/a/d+2*(A*b^2-a*(B*b-C*a))*arctanh((a-b)^(1/2 )*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {(-A b+a B) (c+d x)-\frac {2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a A \sin (c+d x)}{a^2 d} \]
((-(A*b) + a*B)*(c + d*x) - (2*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((-a + b )*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + a*A*Sin[c + d*x])/ (a^2*d)
Time = 0.60 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 4592, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\int \frac {A b-a B-a C \sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\int \frac {A b-a B-a C \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a b}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a b}}{a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}}{a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {A \sin (c+d x)}{a d}-\frac {\frac {x (A b-a B)}{a}-\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}\) |
-((((A*b - a*B)*x)/a - (2*(A*b^2 - a*(b*B - a*C))*ArcTanh[(Sqrt[a - b]*Tan [(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/a) + (A*Sin[c + d*x])/(a*d)
3.10.4.3.1 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_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-A \,b^{2}+B a b -C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (A b -a B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{2}}}{d}\) | \(119\) |
| default | \(\frac {-\frac {2 \left (-A \,b^{2}+B a b -C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (A b -a B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{2}}}{d}\) | \(119\) |
| risch | \(-\frac {A b x}{a^{2}}+\frac {B x}{a}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A \,b^{2}}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A \,b^{2}}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d}\) | \(491\) |
1/d*(-2*(-A*b^2+B*a*b-C*a^2)/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2 *d*x+1/2*c)/((a+b)*(a-b))^(1/2))-2/a^2*(-A*a*tan(1/2*d*x+1/2*c)/(1+tan(1/2 *d*x+1/2*c)^2)+(A*b-B*a)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.32 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.43 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x + {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d}, \frac {{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x + {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - a^{2} b^{2}\right )} d}\right ] \]
[1/2*(2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*d*x + (C*a^2 - B*a*b + A*b^2)* 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)) + 2*(A*a^3 - A*a*b^2)*sin(d*x + c ))/((a^4 - a^2*b^2)*d), ((B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*d*x + (C*a^2 - B*a*b + A*b^2)*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*a^3 - A*a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d)]
\[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.49 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2}} + \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a} + \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{2}}}{d} \]
((B*a - A*b)*(d*x + c)/a^2 + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2* c)^2 + 1)*a) + 2*(C*a^2 - B*a*b + A*b^2)*(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^2))/d
Time = 20.63 (sec) , antiderivative size = 4410, normalized size of antiderivative = 45.00 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
(2*A*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^4 - a^2*b^2)) + (2*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^4 - a^2*b^2) ) + (A*a^3*sin(c + d*x))/(d*(a^4 - a^2*b^2)) - (A*a*b^2*sin(c + d*x))/(d*( a^4 - a^2*b^2)) + (A*b^2*atan((A^2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2 )*2i + A^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*2i - B^2*a^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i - C^2*a^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1 /2)*1i + B^2*a^6*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i + C^2*a^4*b*sin (c/2 + (d*x)/2)*(a^2 - b^2)^(3/2)*2i - C^2*a^6*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i - A^2*a^2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*3i + A^2 *a^3*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i + A^2*a^4*b^3*sin(c/2 + ( d*x)/2)*(a^2 - b^2)^(1/2)*1i - A^2*a^5*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^ (1/2)*1i + B^2*a^2*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2)*2i + B^2*a^2*b ^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*2i - B^2*a^4*b^3*sin(c/2 + (d*x)/2 )*(a^2 - b^2)^(1/2)*3i + B^2*a^5*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)* 1i + C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i + C^2*a^5*b^2*sin (c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*1i - A*B*a*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2)*4i - A*B*a*b^6*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*4i + A*B*a ^6*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*2i + B*C*a^6*b*sin(c/2 + (d*x)/2 )*(a^2 - b^2)^(1/2)*2i + A*B*a^3*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)* 6i - A*B*a^4*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*2i - A*B*a^5*b^2*...