Integrand size = 33, antiderivative size = 99 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 C \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {(A+4 C) \tan (c+d x)}{3 a^2 d}+\frac {2 C \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:
-2*C*arctanh(sin(d*x+c))/a^2/d+1/3*(A+4*C)*tan(d*x+c)/a^2/d+2*C*tan(d*x+c) /a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c)) ^2
Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(99)=198\).
Time = 2.04 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.83 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left ((A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (A+7 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+6 C \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+(A+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \] Input:
Integrate[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x ]
Output:
(4*Cos[(c + d*x)/2]*(A + C*Sec[c + d*x]^2)*((A + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(A + 7*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 6*C*Cos[(c + d*x) /2]^3*(2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2 ])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x) /2]))) + (A + C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(A + 2*C + A*Cos[2*( c + d*x)])*(1 + Sec[c + d*x])^2)
Time = 0.79 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4573, 25, 3042, 4496, 3042, 4274, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^2(c+d x) (a (A-2 C)+a (A+4 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (a (A-2 C)+a (A+4 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (A-2 C)+a (A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {\int \sec (c+d x) \left (6 a^2 C-a^2 (A+4 C) \sec (c+d x)\right )dx}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (6 a^2 C-a^2 (A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {6 a^2 C \int \sec (c+d x)dx-a^2 (A+4 C) \int \sec ^2(c+d x)dx}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {6 a^2 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a^2 (A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {\frac {a^2 (A+4 C) \int 1d(-\tan (c+d x))}{d}+6 a^2 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {6 a^2 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^2 (A+4 C) \tan (c+d x)}{d}}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {6 C \tan (c+d x)}{d (\sec (c+d x)+1)}-\frac {\frac {6 a^2 C \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (A+4 C) \tan (c+d x)}{d}}{a^2}}{3 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
Output:
-1/3*((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (( 6*C*Tan[c + d*x])/(d*(1 + Sec[c + d*x])) - ((6*a^2*C*ArcTanh[Sin[c + d*x]] )/d - (a^2*(A + 4*C)*Tan[c + d*x])/d)/a^2)/(3*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_)], 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[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.19
| method | result | size |
| parallelrisch | \(\frac {6 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-6 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\frac {\left (\frac {A}{2}+5 C \right ) \cos \left (2 d x +2 c \right )}{2}+\left (A +7 C \right ) \cos \left (d x +c \right )+\frac {A}{4}+4 C \right )}{3 d \,a^{2} \cos \left (d x +c \right )}\) | \(118\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(123\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(123\) |
| risch | \(\frac {2 i \left (6 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3 A \,{\mathrm e}^{3 i \left (d x +c \right )}+18 C \,{\mathrm e}^{3 i \left (d x +c \right )}+A \,{\mathrm e}^{2 i \left (d x +c \right )}+22 C \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A \,{\mathrm e}^{i \left (d x +c \right )}+24 C \,{\mathrm e}^{i \left (d x +c \right )}+A +10 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}\) | \(169\) |
| norman | \(\frac {\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\left (A +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (A +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {2 \left (17 C +2 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a}+\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}\) | \(178\) |
Input:
int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVER BOSE)
Output:
1/3*(6*C*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)-6*C*ln(tan(1/2*d*x+1/2*c)+1)* cos(d*x+c)+tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2*(1/2*(1/2*A+5*C)*cos(2* d*x+2*c)+(A+7*C)*cos(d*x+c)+1/4*A+4*C))/d/a^2/cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left (C \cos \left (d x + c\right )^{3} + 2 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C \cos \left (d x + c\right )^{3} + 2 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 7 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm= "fricas")
Output:
-1/3*(3*(C*cos(d*x + c)^3 + 2*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(sin(d *x + c) + 1) - 3*(C*cos(d*x + c)^3 + 2*C*cos(d*x + c)^2 + C*cos(d*x + c))* log(-sin(d*x + c) + 1) - ((A + 10*C)*cos(d*x + c)^2 + 2*(A + 7*C)*cos(d*x + c) + 3*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)
Output:
(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + I ntegral(C*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (95) = 190\).
Time = 0.04 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.93 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm= "maxima")
Output:
1/6*(C*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(si n(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*sin(d *x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + A*(3*sin(d*x + c)/( cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d
Time = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {12 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {12 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm= "giac")
Output:
-1/6*(12*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 12*C*log(abs(tan(1/2*d *x + 1/2*c) - 1))/a^2 + 12*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 13.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}-\frac {A-3\,C}{2\,a^2}\right )}{d}-\frac {4\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \] Input:
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^2),x)
Output:
(tan(c/2 + (d*x)/2)*((A + C)/a^2 - (A - 3*C)/(2*a^2)))/d - (4*C*atanh(tan( c/2 + (d*x)/2)))/(a^2*d) - (2*C*tan(c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x )/2)^2 - a^2)) + (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)
Time = 0.16 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.86 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
Output:
(12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c - 12*log(tan((c + d*x) /2) - 1)*c - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 12*log(t an((c + d*x)/2) + 1)*c + tan((c + d*x)/2)**5*a + tan((c + d*x)/2)**5*c + 2 *tan((c + d*x)/2)**3*a + 14*tan((c + d*x)/2)**3*c - 3*tan((c + d*x)/2)*a - 27*tan((c + d*x)/2)*c)/(6*a**2*d*(tan((c + d*x)/2)**2 - 1))