Integrand size = 31, antiderivative size = 202 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(6 A-13 B) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
-1/2*(6*A-13*B)*arctanh(sin(d*x+c))/a^3/d+8/15*(9*A-19*B)*tan(d*x+c)/a^3/d -1/2*(6*A-13*B)*sec(d*x+c)*tan(d*x+c)/a^3/d+1/5*(A-B)*sec(d*x+c)^4*tan(d*x +c)/d/(a+a*sec(d*x+c))^3+1/15*(6*A-11*B)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a* sec(d*x+c))^2+4/15*(9*A-19*B)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c ))
Leaf count is larger than twice the leaf count of optimal. \(610\) vs. \(2(202)=404\).
Time = 5.69 (sec) , antiderivative size = 610, normalized size of antiderivative = 3.02 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {1920 (6 A-13 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 ^2(c+d x) \left ((-870 A+1235 B) \sin \left (\frac {d x}{2}\right )+5 (366 A-761 B) \sin \left (\frac {3 d x}{2}\right )-2094 A \sin \left (c-\frac {d x}{2}\right )+4329 B \sin \left (c-\frac {d x}{2}\right )+1314 A \sin \left (c+\frac {d x}{2}\right )-1989 B \sin \left (c+\frac {d x}{2}\right )-1650 A \sin \left (2 c+\frac {d x}{2}\right )+3575 B \sin \left (2 c+\frac {d x}{2}\right )-450 A \sin \left (c+\frac {3 d x}{2}\right )+475 B \sin \left (c+\frac {3 d x}{2}\right )+1230 A \sin \left (2 c+\frac {3 d x}{2}\right )-2005 B \sin \left (2 c+\frac {3 d x}{2}\right )-1050 A \sin \left (3 c+\frac {3 d x}{2}\right )+2275 B \sin \left (3 c+\frac {3 d x}{2}\right )+1278 A \sin \left (c+\frac {5 d x}{2}\right )-2673 B \sin \left (c+\frac {5 d x}{2}\right )-90 A \sin \left (2 c+\frac {5 d x}{2}\right )-105 B \sin \left (2 c+\frac {5 d x}{2}\right )+918 A \sin \left (3 c+\frac {5 d x}{2}\right )-1593 B \sin \left (3 c+\frac {5 d x}{2}\right )-450 A \sin \left (4 c+\frac {5 d x}{2}\right )+975 B \sin \left (4 c+\frac {5 d x}{2}\right )+630 A \sin \left (2 c+\frac {7 d x}{2}\right )-1325 B \sin \left (2 c+\frac {7 d x}{2}\right )+60 A \sin \left (3 c+\frac {7 d x}{2}\right )-255 B \sin \left (3 c+\frac {7 d x}{2}\right )+480 A \sin \left (4 c+\frac {7 d x}{2}\right )-875 B \sin \left (4 c+\frac {7 d x}{2}\right )-90 A \sin \left (5 c+\frac {7 d x}{2}\right )+195 B \sin \left (5 c+\frac {7 d x}{2}\right )+144 A \sin \left (3 c+\frac {9 d x}{2}\right )-304 B \sin \left (3 c+\frac {9 d x}{2}\right )+30 A \sin \left (4 c+\frac {9 d x}{2}\right )-90 B \sin \left (4 c+\frac {9 d x}{2}\right )+114 A \sin \left (5 c+\frac {9 d x}{2}\right )-214 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]
Output:
(1920*(6*A - 13*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]^2*((-870*A + 1235*B)*Sin[(d*x)/2] + 5*(366*A - 761 *B)*Sin[(3*d*x)/2] - 2094*A*Sin[c - (d*x)/2] + 4329*B*Sin[c - (d*x)/2] + 1 314*A*Sin[c + (d*x)/2] - 1989*B*Sin[c + (d*x)/2] - 1650*A*Sin[2*c + (d*x)/ 2] + 3575*B*Sin[2*c + (d*x)/2] - 450*A*Sin[c + (3*d*x)/2] + 475*B*Sin[c + (3*d*x)/2] + 1230*A*Sin[2*c + (3*d*x)/2] - 2005*B*Sin[2*c + (3*d*x)/2] - 1 050*A*Sin[3*c + (3*d*x)/2] + 2275*B*Sin[3*c + (3*d*x)/2] + 1278*A*Sin[c + (5*d*x)/2] - 2673*B*Sin[c + (5*d*x)/2] - 90*A*Sin[2*c + (5*d*x)/2] - 105*B *Sin[2*c + (5*d*x)/2] + 918*A*Sin[3*c + (5*d*x)/2] - 1593*B*Sin[3*c + (5*d *x)/2] - 450*A*Sin[4*c + (5*d*x)/2] + 975*B*Sin[4*c + (5*d*x)/2] + 630*A*S in[2*c + (7*d*x)/2] - 1325*B*Sin[2*c + (7*d*x)/2] + 60*A*Sin[3*c + (7*d*x) /2] - 255*B*Sin[3*c + (7*d*x)/2] + 480*A*Sin[4*c + (7*d*x)/2] - 875*B*Sin[ 4*c + (7*d*x)/2] - 90*A*Sin[5*c + (7*d*x)/2] + 195*B*Sin[5*c + (7*d*x)/2] + 144*A*Sin[3*c + (9*d*x)/2] - 304*B*Sin[3*c + (9*d*x)/2] + 30*A*Sin[4*c + (9*d*x)/2] - 90*B*Sin[4*c + (9*d*x)/2] + 114*A*Sin[5*c + (9*d*x)/2] - 214 *B*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.38 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4507, 3042, 4507, 3042, 4507, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (4 a (A-B)-a (2 A-7 B) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (4 a (A-B)-a (2 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\frac {\int \sec ^2(c+d x) \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \sec (c+d x)\right )dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {8 a^3 (9 A-19 B) \int \sec ^2(c+d x)dx-15 a^3 (6 A-13 B) \int \sec ^3(c+d x)dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {8 a^3 (9 A-19 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (6 A-13 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {-\frac {8 a^3 (9 A-19 B) \int 1d(-\tan (c+d x))}{d}-15 a^3 (6 A-13 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 A-19 B) \tan (c+d x)}{d}-15 a^3 (6 A-13 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 A-19 B) \tan (c+d x)}{d}-15 a^3 (6 A-13 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 A-19 B) \tan (c+d x)}{d}-15 a^3 (6 A-13 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {4 a^2 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {8 a^3 (9 A-19 B) \tan (c+d x)}{d}-15 a^3 (6 A-13 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}+\frac {a (6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]
Output:
((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((a*( 6*A - 11*B)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + (( 4*a^2*(9*A - 19*B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((8*a^3*(9*A - 19*B)*Tan[c + d*x])/d - 15*a^3*(6*A - 13*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*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_)]*(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[(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 = 0.54 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87
| method | result | size |
| parallelrisch | \(\frac {720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {13 B}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {13 B}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+72 \left (\left (\frac {21 A}{2}-\frac {87 B}{4}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {19 A}{4}-\frac {239 B}{24}\right ) \cos \left (3 d x +3 c \right )+\left (A -\frac {19 B}{9}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {191 A}{12}-\frac {259 B}{8}\right ) \cos \left (d x +c \right )+\frac {19 A}{2}-\frac {677 B}{36}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(175\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}+\frac {\left (3 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {25 \left (9 A -19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (25 A -51 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (27 A -59 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (345 A -721 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (549 A -1165 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}-\frac {\left (3123 A -6613 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a^{2}}+\frac {\left (6 A -13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}-\frac {\left (6 A -13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(280\) |
| risch | \(\frac {i \left (90 A \,{\mathrm e}^{8 i \left (d x +c \right )}-195 B \,{\mathrm e}^{8 i \left (d x +c \right )}+450 A \,{\mathrm e}^{7 i \left (d x +c \right )}-975 B \,{\mathrm e}^{7 i \left (d x +c \right )}+1050 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}+1650 A \,{\mathrm e}^{5 i \left (d x +c \right )}-3575 B \,{\mathrm e}^{5 i \left (d x +c \right )}+2094 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4329 B \,{\mathrm e}^{4 i \left (d x +c \right )}+1830 A \,{\mathrm e}^{3 i \left (d x +c \right )}-3805 B \,{\mathrm e}^{3 i \left (d x +c \right )}+1278 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2673 B \,{\mathrm e}^{2 i \left (d x +c \right )}+630 \,{\mathrm e}^{i \left (d x +c \right )} A -1325 B \,{\mathrm e}^{i \left (d x +c \right )}+144 A -304 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{3} d}\) | \(324\) |
Input:
int(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/240*(720*(1+cos(2*d*x+2*c))*(A-13/6*B)*ln(tan(1/2*d*x+1/2*c)-1)-720*(1+c os(2*d*x+2*c))*(A-13/6*B)*ln(tan(1/2*d*x+1/2*c)+1)+72*((21/2*A-87/4*B)*cos (2*d*x+2*c)+(19/4*A-239/24*B)*cos(3*d*x+3*c)+(A-19/9*B)*cos(4*d*x+4*c)+(19 1/12*A-259/8*B)*cos(d*x+c)+19/2*A-677/36*B)*sec(1/2*d*x+1/2*c)^4*tan(1/2*d *x+1/2*c))/d/a^3/(1+cos(2*d*x+2*c))
Time = 0.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="f ricas")
Output:
-1/60*(15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^4 + 3 *(6*A - 13*B)*cos(d*x + c)^3 + (6*A - 13*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A - 13*B)*cos(d*x + c)^2)*log(-sin(d* x + c) + 1) - 2*(16*(9*A - 19*B)*cos(d*x + c)^4 + 3*(114*A - 239*B)*cos(d* x + c)^3 + (234*A - 479*B)*cos(d*x + c)^2 + 15*(2*A - 3*B)*cos(d*x + c) + 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3 *d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)
Output:
(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**6/(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 = 377, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="m axima")
Output:
-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*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 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*A*(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))/d
Time = 0.20 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="g iac")
Output:
-1/60*(30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(2*A*tan(1/2*d*x + 1/2*c )^3 - 7*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 5*B*tan(1/2* d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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 + 30*A*a^12*tan(1/2*d*x + 1/ 2*c)^3 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1/2*c ) - 465*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 11.30 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (3\,A-5\,B\right )}{4\,a^3}+\frac {2\,A-10\,B}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-7\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-5\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {3\,A-5\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,A-13\,B\right )}{a^3\,d} \] Input:
int((A + B/cos(c + d*x))/(cos(c + d*x)^5*(a + a/cos(c + d*x))^3),x)
Output:
(tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^3) + (3*(3*A - 5*B))/(4*a^3) + (2*A - 10*B)/(4*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(2*A - 7*B) - tan(c/2 + (d*x)/ 2)*(2*A - 5*B))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) + (tan(c/2 + (d*x)/2)^3*((A - B)/(4*a^3) + (3*A - 5*B)/(12*a^3)))/ d + (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d) - (atanh(tan(c/2 + (d*x)/2)) *(6*A - 13*B))/(a^3*d)
Time = 0.21 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.18 \[ \int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x)
Output:
(180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a - 390*log(tan((c + d* x)/2) - 1)*tan((c + d*x)/2)**4*b - 360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 180*l og(tan((c + d*x)/2) - 1)*a - 390*log(tan((c + d*x)/2) - 1)*b - 180*log(tan ((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a + 390*log(tan((c + d*x)/2) + 1)*t an((c + d*x)/2)**4*b + 360*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b - 180*log(tan((c + d*x)/2) + 1)*a + 390*log(tan((c + d*x)/2) + 1)*b + 3*tan((c + d*x)/2)**9*a - 3*tan((c + d*x)/2)**9*b + 24*tan((c + d*x)/2)**7*a - 34*tan((c + d*x)/2 )**7*b + 198*tan((c + d*x)/2)**5*a - 388*tan((c + d*x)/2)**5*b - 600*tan(( c + d*x)/2)**3*a + 1310*tan((c + d*x)/2)**3*b + 375*tan((c + d*x)/2)*a - 7 65*tan((c + d*x)/2)*b)/(60*a**3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x)/2 )**2 + 1))