Integrand size = 31, antiderivative size = 69 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a A x+\frac {(2 A b+2 a B+b C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(b B+a C) \tan (c+d x)}{d}+\frac {b C \sec (c+d x) \tan (c+d x)}{2 d} \]
a*A*x+1/2*(2*A*b+2*B*a+C*b)*arctanh(sin(d*x+c))/d+(B*b+C*a)*tan(d*x+c)/d+1 /2*b*C*sec(d*x+c)*tan(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a A x+\frac {A b \text {arctanh}(\sin (c+d x))}{d}+\frac {a B \text {arctanh}(\sin (c+d x))}{d}+\frac {b C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b B \tan (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}+\frac {b C \sec (c+d x) \tan (c+d x)}{2 d} \]
a*A*x + (A*b*ArcTanh[Sin[c + d*x]])/d + (a*B*ArcTanh[Sin[c + d*x]])/d + (b *C*ArcTanh[Sin[c + d*x]])/(2*d) + (b*B*Tan[c + d*x])/d + (a*C*Tan[c + d*x] )/d + (b*C*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3042, 4536, 2009}
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.
| \(\displaystyle \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{2} \int \left (2 (b B+a C) \sec ^2(c+d x)+(2 A b+C b+2 a B) \sec (c+d x)+2 a A\right )dx+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 a B+2 A b+b C) \text {arctanh}(\sin (c+d x))}{d}+2 a A x+\frac {2 (a C+b B) \tan (c+d x)}{d}\right )+\frac {b C \tan (c+d x) \sec (c+d x)}{2 d}\) |
(b*C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a*A*x + ((2*A*b + 2*a*B + b*C)* ArcTanh[Sin[c + d*x]])/d + (2*(b*B + a*C)*Tan[c + d*x])/d)/2
3.9.64.3.1 Defintions of rubi rules used
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Time = 0.65 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23
| method | result | size |
| parts | \(a A x +\frac {\left (A b +a B \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (B b +C a \right ) \tan \left (d x +c \right )}{d}+\frac {C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(85\) |
| derivativedivides | \(\frac {a A \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \tan \left (d x +c \right )+A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b +C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(100\) |
| default | \(\frac {a A \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \tan \left (d x +c \right )+A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b +C b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(100\) |
| parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+a A x d \cos \left (2 d x +2 c \right )+\left (B b +C a \right ) \sin \left (2 d x +2 c \right )+a A x d +C \sin \left (d x +c \right ) b}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(135\) |
| norman | \(\frac {a A x +\frac {\left (2 B b +2 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {\left (2 B b +2 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-2 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 A b +2 a B +C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A b +2 a B +C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(166\) |
| risch | \(a A x -\frac {i \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -2 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a B}{d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a B}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(203\) |
a*A*x+(A*b+B*a)/d*ln(sec(d*x+c)+tan(d*x+c))+(B*b+C*a)*tan(d*x+c)/d+C*b/d*( 1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.71 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, A a d x \cos \left (d x + c\right )^{2} + {\left (2 \, B a + {\left (2 \, A + C\right )} b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, B a + {\left (2 \, A + C\right )} b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b + 2 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(4*A*a*d*x*cos(d*x + c)^2 + (2*B*a + (2*A + C)*b)*cos(d*x + c)^2*log(s in(d*x + c) + 1) - (2*B*a + (2*A + C)*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*b + 2*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 2)
\[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.68 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} A a - C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, A b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, C a \tan \left (d x + c\right ) + 4 \, B b \tan \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*A*a - C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin( d*x + c) + 1) + log(sin(d*x + c) - 1)) + 4*B*a*log(sec(d*x + c) + tan(d*x + c)) + 4*A*b*log(sec(d*x + c) + tan(d*x + c)) + 4*C*a*tan(d*x + c) + 4*B* b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (65) = 130\).
Time = 0.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.46 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (d x + c\right )} A a + {\left (2 \, B a + 2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, B a + 2 \, A b + C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(2*(d*x + c)*A*a + (2*B*a + 2*A*b + C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*B*a + 2*A*b + C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*C* a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 - C*b*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c) - 2*B*b*tan(1/2*d*x + 1/2*c) - C*b *tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 17.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38 \[ \int (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}\right )}{d}+\frac {\frac {C\,b\,\sin \left (c+d\,x\right )}{2}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
(2*(A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + A*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + B*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2)) + (C*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2))/d + ((C*b*s in(c + d*x))/2 + (B*b*sin(2*c + 2*d*x))/2 + (C*a*sin(2*c + 2*d*x))/2)/(d*( cos(2*c + 2*d*x)/2 + 1/2))