Integrand size = 31, antiderivative size = 156 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-3 B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
(A-3*B)*arctanh(sin(d*x+c))/a^3/d-1/15*(7*A-27*B)*tan(d*x+c)/a^3/d+1/5*(A- B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(4*A-9*B)*sec(d*x+c)^ 2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-(A-3*B)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+ c))
Leaf count is larger than twice the leaf count of optimal. \(480\) vs. \(2(156)=312\).
Time = 4.11 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.08 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {-960 (A-3 B) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (5 (32 A-51 B) \sin \left (\frac {d x}{2}\right )+(-167 A+567 B) \sin \left (\frac {3 d x}{2}\right )+170 A \sin \left (c-\frac {d x}{2}\right )-600 B \sin \left (c-\frac {d x}{2}\right )-170 A \sin \left (c+\frac {d x}{2}\right )+375 B \sin \left (c+\frac {d x}{2}\right )+160 A \sin \left (2 c+\frac {d x}{2}\right )-480 B \sin \left (2 c+\frac {d x}{2}\right )+75 A \sin \left (c+\frac {3 d x}{2}\right )-60 B \sin \left (c+\frac {3 d x}{2}\right )-167 A \sin \left (2 c+\frac {3 d x}{2}\right )+402 B \sin \left (2 c+\frac {3 d x}{2}\right )+75 A \sin \left (3 c+\frac {3 d x}{2}\right )-225 B \sin \left (3 c+\frac {3 d x}{2}\right )-95 A \sin \left (c+\frac {5 d x}{2}\right )+315 B \sin \left (c+\frac {5 d x}{2}\right )+15 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )-95 A \sin \left (3 c+\frac {5 d x}{2}\right )+240 B \sin \left (3 c+\frac {5 d x}{2}\right )+15 A \sin \left (4 c+\frac {5 d x}{2}\right )-45 B \sin \left (4 c+\frac {5 d x}{2}\right )-22 A \sin \left (2 c+\frac {7 d x}{2}\right )+72 B \sin \left (2 c+\frac {7 d x}{2}\right )+15 B \sin \left (3 c+\frac {7 d x}{2}\right )-22 A \sin \left (4 c+\frac {7 d x}{2}\right )+57 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]
Output:
(-960*(A - 3*B)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2] *Sec[c]*Sec[c + d*x]*(5*(32*A - 51*B)*Sin[(d*x)/2] + (-167*A + 567*B)*Sin[ (3*d*x)/2] + 170*A*Sin[c - (d*x)/2] - 600*B*Sin[c - (d*x)/2] - 170*A*Sin[c + (d*x)/2] + 375*B*Sin[c + (d*x)/2] + 160*A*Sin[2*c + (d*x)/2] - 480*B*Si n[2*c + (d*x)/2] + 75*A*Sin[c + (3*d*x)/2] - 60*B*Sin[c + (3*d*x)/2] - 167 *A*Sin[2*c + (3*d*x)/2] + 402*B*Sin[2*c + (3*d*x)/2] + 75*A*Sin[3*c + (3*d *x)/2] - 225*B*Sin[3*c + (3*d*x)/2] - 95*A*Sin[c + (5*d*x)/2] + 315*B*Sin[ c + (5*d*x)/2] + 15*A*Sin[2*c + (5*d*x)/2] + 30*B*Sin[2*c + (5*d*x)/2] - 9 5*A*Sin[3*c + (5*d*x)/2] + 240*B*Sin[3*c + (5*d*x)/2] + 15*A*Sin[4*c + (5* d*x)/2] - 45*B*Sin[4*c + (5*d*x)/2] - 22*A*Sin[2*c + (7*d*x)/2] + 72*B*Sin [2*c + (7*d*x)/2] + 15*B*Sin[3*c + (7*d*x)/2] - 22*A*Sin[4*c + (7*d*x)/2] + 57*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4507, 3042, 4507, 3042, 4496, 25, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)-a (A-6 B) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a (A-B)-a (A-6 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {-\frac {\int -\sec (c+d x) \left (15 a^3 (A-3 B)-a^3 (7 A-27 B) \sec (c+d x)\right )dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \sec (c+d x) \left (15 a^3 (A-3 B)-a^3 (7 A-27 B) \sec (c+d x)\right )dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 a^3 (A-3 B)-a^3 (7 A-27 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (A-3 B) \int \sec (c+d x)dx-a^3 (7 A-27 B) \int \sec ^2(c+d x)dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (A-3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a^3 (7 A-27 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^3 (7 A-27 B) \int 1d(-\tan (c+d x))}{d}+15 a^3 (A-3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (A-3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (7 A-27 B) \tan (c+d x)}{d}}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {15 a^3 (A-3 B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (7 A-27 B) \tan (c+d x)}{d}}{a^2}-\frac {15 a^2 (A-3 B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]
Output:
((A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((a*( 4*A - 9*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((- 15*a^2*(A - 3*B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((15*a^3*(A - 3* B)*ArcTanh[Sin[c + d*x]])/d - (a^3*(7*A - 27*B)*Tan[c + d*x])/d)/a^2)/(3*a ^2))/(5*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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(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 - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Time = 0.43 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \(\frac {-120 \cos \left (d x +c \right ) \left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \cos \left (d x +c \right ) \left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-97 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {3 \left (17 A -57 B \right ) \cos \left (2 d x +2 c \right )}{97}+\frac {\left (11 A -36 B \right ) \cos \left (3 d x +3 c \right )}{97}+\left (A -\frac {342 B}{97}\right ) \cos \left (d x +c \right )+\frac {51 A}{97}-\frac {201 B}{97}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{120 \cos \left (d x +c \right ) a^{3} d}\) | \(145\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\left (12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(162\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\left (12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(162\) |
| norman | \(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (4 A -9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}-\frac {\left (7 A -25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {5 \left (8 A -27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (26 A -81 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}-\frac {\left (43 A -153 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\left (553 A -1773 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} a^{2}}+\frac {\left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\left (A -3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(249\) |
| risch | \(-\frac {2 i \left (15 A \,{\mathrm e}^{6 i \left (d x +c \right )}-45 B \,{\mathrm e}^{6 i \left (d x +c \right )}+75 A \,{\mathrm e}^{5 i \left (d x +c \right )}-225 B \,{\mathrm e}^{5 i \left (d x +c \right )}+160 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 B \,{\mathrm e}^{4 i \left (d x +c \right )}+170 A \,{\mathrm e}^{3 i \left (d x +c \right )}-600 B \,{\mathrm e}^{3 i \left (d x +c \right )}+167 A \,{\mathrm e}^{2 i \left (d x +c \right )}-567 B \,{\mathrm e}^{2 i \left (d x +c \right )}+95 \,{\mathrm e}^{i \left (d x +c \right )} A -315 B \,{\mathrm e}^{i \left (d x +c \right )}+22 A -72 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}\) | \(275\) |
Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/120*(-120*cos(d*x+c)*(A-3*B)*ln(tan(1/2*d*x+1/2*c)-1)+120*cos(d*x+c)*(A- 3*B)*ln(tan(1/2*d*x+1/2*c)+1)-97*tan(1/2*d*x+1/2*c)*(3/97*(17*A-57*B)*cos( 2*d*x+2*c)+1/97*(11*A-36*B)*cos(3*d*x+3*c)+(A-342/97*B)*cos(d*x+c)+51/97*A -201/97*B)*sec(1/2*d*x+1/2*c)^4)/cos(d*x+c)/a^3/d
Time = 0.09 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (11 \, A - 36 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="f ricas")
Output:
1/30*(15*((A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3 *B)*cos(d*x + c)^2 + (A - 3*B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 15*(( A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3*B)*cos(d*x + c)^2 + (A - 3*B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(11*A - 36 *B)*cos(d*x + c)^3 + 3*(17*A - 57*B)*cos(d*x + c)^2 + (32*A - 117*B)*cos(d *x + c) - 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c) ^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)
Output:
(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.04 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.83 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="m axima")
Output:
1/60*(3*B*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 )*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60 *log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d *x + c) + 1) - 1)/a^3) - A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin( d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a ^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/a^3))/d
Time = 0.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="g iac")
Output:
1/60*(60*(A - 3*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(A - 3*B)*l og(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 120*B*tan(1/2*d*x + 1/2*c)/((tan(1 /2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12* tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan( 1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2* d*x + 1/2*c))/a^15)/d
Time = 11.77 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{4\,a^3}-\frac {3\,B}{2\,a^3}+\frac {2\,A-4\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {2\,A-4\,B}{12\,a^3}\right )}{d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \] Input:
int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^3),x)
Output:
(2*atanh(tan(c/2 + (d*x)/2))*(A - 3*B))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3* (A - B))/(4*a^3) - (3*B)/(2*a^3) + (2*A - 4*B)/(2*a^3)))/d - (tan(c/2 + (d *x)/2)^5*(A - B))/(20*a^3*d) - (tan(c/2 + (d*x)/2)^3*((A - B)/(6*a^3) + (2 *A - 4*B)/(12*a^3)))/d - (2*B*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/ 2)^2 - a^3))
Time = 0.18 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.90 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b -17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -85 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -375 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x)
Output:
( - 60*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 60*log(tan((c + d*x)/2) - 1)*a - 180* log(tan((c + d*x)/2) - 1)*b + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2 )**2*a - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b - 60*log(tan( (c + d*x)/2) + 1)*a + 180*log(tan((c + d*x)/2) + 1)*b - 3*tan((c + d*x)/2) **7*a + 3*tan((c + d*x)/2)**7*b - 17*tan((c + d*x)/2)**5*a + 27*tan((c + d *x)/2)**5*b - 85*tan((c + d*x)/2)**3*a + 225*tan((c + d*x)/2)**3*b + 105*t an((c + d*x)/2)*a - 375*tan((c + d*x)/2)*b)/(60*a**3*d*(tan((c + d*x)/2)** 2 - 1))