Integrand size = 41, antiderivative size = 119 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(2 A-2 B+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(A-2 B+2 C) \tan (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))} \] Output:
1/2*(2*A-2*B+3*C)*arctanh(sin(d*x+c))/a/d-(A-2*B+2*C)*tan(d*x+c)/a/d+1/2*( 2*A-2*B+3*C)*sec(d*x+c)*tan(d*x+c)/a/d-(A-B+C)*sec(d*x+c)^2*tan(d*x+c)/d/( a+a*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(392\) vs. \(2(119)=238\).
Time = 4.06 (sec) , antiderivative size = 392, normalized size of antiderivative = 3.29 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (2 A-2 B+3 C) \cos \left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 (A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {C \cos \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{a d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (1+\sec (c+d x))} \] Input:
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*Cos[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-2 *(2*A - 2*B + 3*C)*Cos[(c + d*x)/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ]] + 2*(2*A - 2*B + 3*C)*Cos[(c + d*x)/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 4*(A - B + C)*Sec[c/2]*Sin[(d*x)/2] + (C*Cos[(c + d*x)/2])/(Cos [(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*(B - C)*Cos[(c + d*x)/2]*Sin[(d*x )/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (C*C os[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*(B - C)*Cos[ (c + d*x)/2]*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[ (c + d*x)/2]))))/(a*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x]))
Time = 0.71 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 4572, 25, 3042, 4274, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int -\sec ^2(c+d x) (a (A-2 B+2 C)-a (2 A-2 B+3 C) \sec (c+d x))dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \sec ^2(c+d x) (a (A-2 B+2 C)-a (2 A-2 B+3 C) \sec (c+d x))dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (A-2 B+2 C)-a (2 A-2 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {a (A-2 B+2 C) \int \sec ^2(c+d x)dx-a (2 A-2 B+3 C) \int \sec ^3(c+d x)dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a (A-2 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-a (2 A-2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {-\frac {a (A-2 B+2 C) \int 1d(-\tan (c+d x))}{d}-a (2 A-2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {a (A-2 B+2 C) \tan (c+d x)}{d}-a (2 A-2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {\frac {a (A-2 B+2 C) \tan (c+d x)}{d}-a (2 A-2 B+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a (A-2 B+2 C) \tan (c+d x)}{d}-a (2 A-2 B+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {a (A-2 B+2 C) \tan (c+d x)}{d}-a (2 A-2 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
Output:
-(((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))) - (( a*(A - 2*B + 2*C)*Tan[c + d*x])/d - a*(2*A - 2*B + 3*C)*(ArcTanh[Sin[c + d *x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
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*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.44 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13
| method | result | size |
| parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\left (\left (A -2 B +2 C \right ) \cos \left (2 d x +2 c \right )+\left (-2 B +C \right ) \cos \left (d x +c \right )+A -2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(134\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 C}{2}+B -A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\) | \(158\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 C}{2}-B +A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {C}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 C}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 C}{2}+B -A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\) | \(158\) |
| norman | \(\frac {\frac {\left (A -3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (3 A -5 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (3 A -7 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 A -2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (2 A -2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\) | \(185\) |
| risch | \(-\frac {i \left (2 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{3 i \left (d x +c \right )}+3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}+5 C \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{i \left (d x +c \right )}+C \,{\mathrm e}^{i \left (d x +c \right )}+2 A -4 B +4 C \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}\) | \(296\) |
Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method =_RETURNVERBOSE)
Output:
(-(1+cos(2*d*x+2*c))*(3/2*C-B+A)*ln(tan(1/2*d*x+1/2*c)-1)+(1+cos(2*d*x+2*c ))*(3/2*C-B+A)*ln(tan(1/2*d*x+1/2*c)+1)-((A-2*B+2*C)*cos(2*d*x+2*c)+(-2*B+ C)*cos(d*x+c)+A-2*B+C)*tan(1/2*d*x+1/2*c))/d/a/(1+cos(2*d*x+2*c))
Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left ({\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 2 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, B - C\right )} \cos \left (d x + c\right ) - C\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/4*(((2*A - 2*B + 3*C)*cos(d*x + c)^3 + (2*A - 2*B + 3*C)*cos(d*x + c)^2) *log(sin(d*x + c) + 1) - ((2*A - 2*B + 3*C)*cos(d*x + c)^3 + (2*A - 2*B + 3*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(2*(A - 2*B + 2*C)*cos(d*x + c)^2 - (2*B - C)*cos(d*x + c) - C)*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)), x)
Output:
(Integral(A*sec(c + d*x)**2/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d* x)**3/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (115) = 230\).
Time = 0.04 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.99 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {C {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 2 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(C*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4 /(cos(d*x + c) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3* log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 2*B*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(co s(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1 ))) - 2*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/( cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (2 \, A - 2 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (2 \, A - 2 \, B + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/2*((2*A - 2*B + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (2*A - 2*B + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 2*(A*tan(1/2*d*x + 1/2*c) - B *tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a - 2*(2*B*tan(1/2*d*x + 1 /2*c)^3 - 3*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + C*tan(1/ 2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d
Time = 12.18 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B+\frac {3\,C}{2}\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-3\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-C\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))),x)
Output:
(2*atanh(tan(c/2 + (d*x)/2))*(A - B + (3*C)/2))/(a*d) - (tan(c/2 + (d*x)/2 )*(A - B + C))/(a*d) - (tan(c/2 + (d*x)/2)^3*(2*B - 3*C) - tan(c/2 + (d*x) /2)*(2*B - C))/(d*(a - 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4))
Time = 0.18 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.47 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -2 \cos \left (d x +c \right ) a +2 \cos \left (d x +c \right ) b -2 \cos \left (d x +c \right ) c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) c -2 \sin \left (d x +c \right )^{2} a +2 \sin \left (d x +c \right )^{2} b -3 \sin \left (d x +c \right )^{2} c +2 a -2 b +2 c}{2 \sin \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)
Output:
(2*cos(c + d*x)*sin(c + d*x)**2*a - 4*cos(c + d*x)*sin(c + d*x)**2*b + 4*c os(c + d*x)*sin(c + d*x)**2*c - 2*cos(c + d*x)*a + 2*cos(c + d*x)*b - 2*co s(c + d*x)*c - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a + 2*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**3*b - 3*log(tan((c + d*x)/2) - 1)*sin(c + d *x)**3*c + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a - 2*log(tan((c + d*x )/2) - 1)*sin(c + d*x)*b + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*c + 2* log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a - 2*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**3*b + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*c - 2*log( tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*c - 2*sin(c + d*x)**2*a + 2*sin(c + d*x)**2*b - 3*sin(c + d*x)**2*c + 2*a - 2*b + 2*c)/(2*sin(c + d*x)*a*d*(sin(c + d*x)**2 - 1))