Integrand size = 33, antiderivative size = 94 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b \sqrt {a+b} d} \]
A*x/a+C*arctanh(sin(d*x+c))/b/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/b/d/(a-b)^(1/2)/(a+b)^(1/2)
Result contains complex when optimal does not.
Time = 0.98 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.78 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (\sqrt {a^2-b^2} \left (A b d x-a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sqrt {(\cos (c)-i \sin (c))^2}+2 \left (A b^2+a (-b B+a C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (i \cos (c)+\sin (c))\right )}{a b \sqrt {a^2-b^2} d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sqrt {(\cos (c)-i \sin (c))^2}} \]
(2*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(Sqrt[a^2 - b^2]*(A*b*d*x - a*C *Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + a*C*Log[Cos[(c + d*x)/2] + Sin [(c + d*x)/2]])*Sqrt[(Cos[c] - I*Sin[c])^2] + 2*(A*b^2 + a*(-(b*B) + a*C)) *ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(S qrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(I*Cos[c] + Sin[c])))/(a*b*Sq rt[a^2 - b^2]*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt[(Co s[c] - I*Sin[c])^2])
Time = 0.59 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4538, 3042, 4257, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4538 |
| \(\displaystyle \frac {\int \frac {A b+(b B-a C) \sec (c+d x)}{a+b \sec (c+d x)}dx}{b}+\frac {C \int \sec (c+d x)dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A b+(b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\int \frac {A b+(b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a b}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a b}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {A b x}{a}-\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{b}+\frac {C \text {arctanh}(\sin (c+d x))}{b d}\) |
(C*ArcTanh[Sin[c + d*x]])/(b*d) + ((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))/b
3.10.3.3.1 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[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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_)]*(b_.) + (a_)), x_Symbol] :> Simp[C/b Int[Csc[ e + f*x], x], x] + Simp[1/b Int[(A*b + (b*B - a*C)*Csc[e + f*x])/(a + b*C sc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]
Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {2 \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(119\) |
| default | \(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {2 \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b a \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(119\) |
| risch | \(\frac {A x}{a}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d b}\) | \(478\) |
1/d*(2*A/a*arctan(tan(1/2*d*x+1/2*c))-C/b*ln(tan(1/2*d*x+1/2*c)-1)+C/b*ln( tan(1/2*d*x+1/2*c)+1)-2/b*(A*b^2-B*a*b+C*a^2)/a/((a+b)*(a-b))^(1/2)*arctan h((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
Time = 1.60 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.96 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x + {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}, \frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x - 2 \, {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}\right ] \]
[1/2*(2*(A*a^2*b - A*b^3)*d*x + (C*a^2 - B*a*b + A*b^2)*sqrt(a^2 - b^2)*lo g((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)) + (C*a^3 - C*a*b^2)*log(sin(d*x + c) + 1) - (C*a^3 - C*a*b^2)*log(-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d), 1/2*(2*(A*a^2*b - A*b^3)*d*x - 2*(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))) + (C*a^3 - C*a*b^2 )*log(sin(d*x + c) + 1) - (C*a^3 - C*a*b^2)*log(-sin(d*x + c) + 1))/((a^3* b - a*b^3)*d)]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.59 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b} - \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a b}}{d} \]
((d*x + c)*A/a + C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b - C*log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/b - 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*b))/d
Time = 28.53 (sec) , antiderivative size = 18184, normalized size of antiderivative = 193.45 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
(2*C*atanh((16384*C^5*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4 *a^5 + 32768*B*C^4*a^5 - 16384*A^4*C*b^5 - 16384*C^5*a^4*b + 16384*B^2*C^3 *a^5 - 32768*A^2*C^3*a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b ^3 + 32768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^4*C*a*b^4 - 16384 *B^2*C^3*a^4*b - (32768*B*C^4*a^6)/b + 32768*A*B*C^3*a^3*b^2 - 32768*A^2*B *C^2*a^4*b - 32768*A^3*B*C*a^2*b^3 + 32768*A^2*B*C^2*a^3*b^2 - 16384*A^2*B ^2*C*a^2*b^3 + 16384*A^2*B^2*C*a^3*b^2 - 32768*A*B*C^3*a^4*b + 32768*A^3*B *C*a*b^4) + (16384*C^5*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^ 4*a^4 + 16384*A^4*C*b^4 + 16384*B^2*C^3*a^4 - (16384*C^5*a^5)/b + 32768*A^ 2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - (16384*B^2*C^3*a^5)/b + 32768*A*B* C^3*a^4 - 16384*A^4*C*a*b^3 + 32768*A^2*B*C^2*a^4 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b - (32768*B*C^4*a^5)/b + (32768*B *C^4*a^6)/b^2 - 32768*A^2*B*C^2*a^3*b - 16384*A^2*B^2*C*a^3*b + 32768*A^3* B*C*a^2*b^2 + 16384*A^2*B^2*C*a^2*b^2 - 32768*A*B*C^3*a^3*b - 32768*A^3*B* C*a*b^3) + (16384*B^2*C^3*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A *C^4*a^4 + 16384*A^4*C*b^4 + 16384*B^2*C^3*a^4 - (16384*C^5*a^5)/b + 32768 *A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - (16384*B^2*C^3*a^5)/b + 32768*A *B*C^3*a^4 - 16384*A^4*C*a*b^3 + 32768*A^2*B*C^2*a^4 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b - (32768*B*C^4*a^5)/b + (3276 8*B*C^4*a^6)/b^2 - 32768*A^2*B*C^2*a^3*b - 16384*A^2*B^2*C*a^3*b + 3276...