Integrand size = 33, antiderivative size = 133 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}+\frac {(3 A+4 C) \tan (c+d x)}{a d}-\frac {(2 A+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d} \]
-1/2*(2*A+3*C)*arctanh(sin(d*x+c))/a/d+(3*A+4*C)*tan(d*x+c)/a/d-1/2*(2*A+3 *C)*sec(d*x+c)*tan(d*x+c)/a/d-(A+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x +c))+1/3*(3*A+4*C)*tan(d*x+c)^3/a/d
Leaf count is larger than twice the leaf count of optimal. \(1090\) vs. \(2(133)=266\).
Time = 8.43 (sec) , antiderivative size = 1090, normalized size of antiderivative = 8.20 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \]
(2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(d*(A + 2*C + A*Cos[2*c + 2*d* x])*(a + a*Sec[c + d*x])) - (2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d* x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(d *(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (4*Cos[c/2 + (d*x) /2]*Cos[c + d*x]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] + C*Sin[( d*x)/2]))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*C*C os[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(3*d *(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2]) *(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) - (2*Cos[c/2 + (d*x)/2]^2*Co s[c + d*x]*(A + C*Sec[c + d*x]^2)*(C*Cos[c/2] - 2*C*Sin[c/2]))/(3*d*(A + 2 *C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c /2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (4*Cos[c/2 + (d*x)/2]^2*Cos[c + d *x]*(A + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2])*(Co s[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(3*d*(A + 2*C + A*Cos[2*c + 2* d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin [c/2 + (d*x)/2])^3) + (2*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(C*Cos[c/2] + 2*C*Sin[c/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x]...
Time = 0.68 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4573, 3042, 4274, 3042, 4254, 2009, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4573 |
\(\displaystyle -\frac {\int \sec ^3(c+d x) (a (2 A+3 C)-a (3 A+4 C) \sec (c+d x))dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (2 A+3 C)-a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle -\frac {a (2 A+3 C) \int \sec ^3(c+d x)dx-a (3 A+4 C) \int \sec ^4(c+d x)dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-a (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {\frac {a (3 A+4 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}+a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\) |
-(((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))) - (a*(2* A + 3*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) + (a*(3*A + 4*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/a^2
3.2.22.3.1 Defintions of rubi rules used
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_)]^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 = 0.44 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\left (A +\frac {4 C}{3}\right ) \cos \left (3 d x +3 c \right )}{3}+\frac {\left (A +\frac {7 C}{6}\right ) \cos \left (2 d x +2 c \right )}{3}+\left (A +\frac {11 C}{9}\right ) \cos \left (d x +c \right )+\frac {A}{3}+\frac {11 C}{18}\right )}{a d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(165\) |
derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\) | \(172\) |
default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\) | \(172\) |
norman | \(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {\left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (2 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (30 A +37 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (36 A +49 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\) | \(195\) |
risch | \(\frac {i \left (6 A \,{\mathrm e}^{6 i \left (d x +c \right )}+9 C \,{\mathrm e}^{6 i \left (d x +c \right )}+6 A \,{\mathrm e}^{5 i \left (d x +c \right )}+9 C \,{\mathrm e}^{5 i \left (d x +c \right )}+24 A \,{\mathrm e}^{4 i \left (d x +c \right )}+24 C \,{\mathrm e}^{4 i \left (d x +c \right )}+12 A \,{\mathrm e}^{3 i \left (d x +c \right )}+24 C \,{\mathrm e}^{3 i \left (d x +c \right )}+30 A \,{\mathrm e}^{2 i \left (d x +c \right )}+39 C \,{\mathrm e}^{2 i \left (d x +c \right )}+6 A \,{\mathrm e}^{i \left (d x +c \right )}+7 C \,{\mathrm e}^{i \left (d x +c \right )}+12 A +16 C \right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}\) | \(275\) |
3*((A+3/2*C)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/2*c)-1)-(A+3 /2*C)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/2*c)+1)+2*tan(1/2*d *x+1/2*c)*(1/3*(A+4/3*C)*cos(3*d*x+3*c)+1/3*(A+7/6*C)*cos(2*d*x+2*c)+(A+11 /9*C)*cos(d*x+c)+1/3*A+11/18*C))/a/d/(cos(3*d*x+3*c)+3*cos(d*x+c))
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 \, {\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]
-1/12*(3*((2*A + 3*C)*cos(d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^3)*log(sin (d*x + c) + 1) - 3*((2*A + 3*C)*cos(d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^ 3)*log(-sin(d*x + c) + 1) - 2*(4*(3*A + 4*C)*cos(d*x + c)^3 + (6*A + 7*C)* cos(d*x + c)^2 - C*cos(d*x + c) + 2*C)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
(Integral(A*sec(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d* x)**5/(sec(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (127) = 254\).
Time = 0.23 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.44 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {C {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 6 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]
1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c )^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin (d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a* (cos(d*x + c) + 1))) - 6*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - l og(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d *x + c) + 1))))/d
Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {3 \, {\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]
-1/6*(3*(2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(2*A + 3*C)*l og(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/ 2*d*x + 1/2*c))/a + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2 *c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*ta n(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d
Time = 15.32 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (2\,A+5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{2}\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d} \]