Integrand size = 41, antiderivative size = 161 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(B-3 C) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
(B-3*C)*arctanh(sin(d*x+c))/a^3/d+1/15*(2*A-7*B+27*C)*tan(d*x+c)/a^3/d-1/5 *(A-B+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(A+4*B-9*C)*sec (d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-(B-3*C)*tan(d*x+c)/d/(a^3+a^3* sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(644\) vs. \(2(161)=322\).
Time = 6.84 (sec) , antiderivative size = 644, normalized size of antiderivative = 4.00 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^3,x]
Output:
((C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[c + d*x]*(-960*(B - 3*C)*Cos[ (c + d*x)/2]^6*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Lo g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c] *(-5*(4*A - 32*B + 51*C)*Sin[(d*x)/2] + (22*A - 167*B + 567*C)*Sin[(3*d*x) /2] - 10*A*Sin[c - (d*x)/2] + 170*B*Sin[c - (d*x)/2] - 600*C*Sin[c - (d*x) /2] + 10*A*Sin[c + (d*x)/2] - 170*B*Sin[c + (d*x)/2] + 375*C*Sin[c + (d*x) /2] - 20*A*Sin[2*c + (d*x)/2] + 160*B*Sin[2*c + (d*x)/2] - 480*C*Sin[2*c + (d*x)/2] + 75*B*Sin[c + (3*d*x)/2] - 60*C*Sin[c + (3*d*x)/2] + 22*A*Sin[2 *c + (3*d*x)/2] - 167*B*Sin[2*c + (3*d*x)/2] + 402*C*Sin[2*c + (3*d*x)/2] + 75*B*Sin[3*c + (3*d*x)/2] - 225*C*Sin[3*c + (3*d*x)/2] + 10*A*Sin[c + (5 *d*x)/2] - 95*B*Sin[c + (5*d*x)/2] + 315*C*Sin[c + (5*d*x)/2] + 15*B*Sin[2 *c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] + 10*A*Sin[3*c + (5*d*x)/2] - 95*B*Sin[3*c + (5*d*x)/2] + 240*C*Sin[3*c + (5*d*x)/2] + 15*B*Sin[4*c + (5 *d*x)/2] - 45*C*Sin[4*c + (5*d*x)/2] + 2*A*Sin[2*c + (7*d*x)/2] - 22*B*Sin [2*c + (7*d*x)/2] + 72*C*Sin[2*c + (7*d*x)/2] + 15*C*Sin[3*c + (7*d*x)/2] + 2*A*Sin[4*c + (7*d*x)/2] - 22*B*Sin[4*c + (7*d*x)/2] + 57*C*Sin[4*c + (7 *d*x)/2])))/(60*a^3*d*(1 + Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A *Cos[2*(c + d*x)]))
Time = 1.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4572, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (a (2 A+3 B-3 C)+a (A-B+6 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (2 A+3 B-3 C)+a (A-B+6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 (A+4 B-9 C) a^2+(2 A-7 B+27 C) \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 (A+4 B-9 C) a^2+(2 A-7 B+27 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {-\frac {\int -\sec (c+d x) \left (15 (B-3 C) a^3+(2 A-7 B+27 C) \sec (c+d x) a^3\right )dx}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \sec (c+d x) \left (15 (B-3 C) a^3+(2 A-7 B+27 C) \sec (c+d x) a^3\right )dx}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 (B-3 C) a^3+(2 A-7 B+27 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {a^3 (2 A-7 B+27 C) \int \sec ^2(c+d x)dx+15 a^3 (B-3 C) \int \sec (c+d x)dx}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^3 (2 A-7 B+27 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+15 a^3 (B-3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (B-3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (2 A-7 B+27 C) \int 1d(-\tan (c+d x))}{d}}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (B-3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^3 (2 A-7 B+27 C) \tan (c+d x)}{d}}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {a^3 (2 A-7 B+27 C) \tan (c+d x)}{d}+\frac {15 a^3 (B-3 C) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {15 a^2 (B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
Output:
-1/5*((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((a*(A + 4*B - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x ])^2) + ((-15*a^2*(B - 3*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((15* a^3*(B - 3*C)*ArcTanh[Sin[c + d*x]])/d + (a^3*(2*A - 7*B + 27*C)*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]
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 = 0.47 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\frac {-120 \cos \left (d x +c \right ) \left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \cos \left (d x +c \right ) \left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (6 A -51 B +171 C \right ) \cos \left (2 d x +2 c \right )+\left (A -11 B +36 C \right ) \cos \left (3 d x +3 c \right )+\left (17 A -97 B +342 C \right ) \cos \left (d x +c \right )+6 A -51 B +201 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{120 d \,a^{3} \cos \left (d x +c \right )}\) | \(154\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (12 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 C +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(201\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (12 C -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 C +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(201\) |
| norman | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (A +4 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}-\frac {\left (7 A +43 B -153 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (-553 B +1773 C +53 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}-\frac {\left (-26 B +81 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}-\frac {5 \left (-8 B +27 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (-7 B +25 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} a^{2}}+\frac {\left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\left (B -3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(260\) |
| risch | \(-\frac {2 i \left (15 B \,{\mathrm e}^{6 i \left (d x +c \right )}-45 C \,{\mathrm e}^{6 i \left (d x +c \right )}+75 B \,{\mathrm e}^{5 i \left (d x +c \right )}-225 C \,{\mathrm e}^{5 i \left (d x +c \right )}-20 A \,{\mathrm e}^{4 i \left (d x +c \right )}+160 B \,{\mathrm e}^{4 i \left (d x +c \right )}-480 C \,{\mathrm e}^{4 i \left (d x +c \right )}-10 A \,{\mathrm e}^{3 i \left (d x +c \right )}+170 B \,{\mathrm e}^{3 i \left (d x +c \right )}-600 C \,{\mathrm e}^{3 i \left (d x +c \right )}-22 A \,{\mathrm e}^{2 i \left (d x +c \right )}+167 B \,{\mathrm e}^{2 i \left (d x +c \right )}-567 C \,{\mathrm e}^{2 i \left (d x +c \right )}-10 A \,{\mathrm e}^{i \left (d x +c \right )}+95 B \,{\mathrm e}^{i \left (d x +c \right )}-315 C \,{\mathrm e}^{i \left (d x +c \right )}-2 A +22 B -72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\) | \(326\) |
Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,meth od=_RETURNVERBOSE)
Output:
1/120*(-120*cos(d*x+c)*(B-3*C)*ln(tan(1/2*d*x+1/2*c)-1)+120*cos(d*x+c)*(B- 3*C)*ln(tan(1/2*d*x+1/2*c)+1)+((6*A-51*B+171*C)*cos(2*d*x+2*c)+(A-11*B+36* C)*cos(3*d*x+3*c)+(17*A-97*B+342*C)*cos(d*x+c)+6*A-51*B+201*C)*tan(1/2*d*x +1/2*c)*sec(1/2*d*x+1/2*c)^4)/d/a^3/cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.63 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A - 11 \, B + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A - 17 \, B + 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 32 \, B + 117 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="fricas")
Output:
1/30*(15*((B - 3*C)*cos(d*x + c)^4 + 3*(B - 3*C)*cos(d*x + c)^3 + 3*(B - 3 *C)*cos(d*x + c)^2 + (B - 3*C)*cos(d*x + c))*log(sin(d*x + c) + 1) - 15*(( B - 3*C)*cos(d*x + c)^4 + 3*(B - 3*C)*cos(d*x + c)^3 + 3*(B - 3*C)*cos(d*x + c)^2 + (B - 3*C)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2*(A - 11*B + 36*C)*cos(d*x + c)^3 + 3*(2*A - 17*B + 57*C)*cos(d*x + c)^2 + (7*A - 32* B + 117*C)*cos(d*x + c) + 15*C)*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))
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *3,x)
Output:
(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*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
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (155) = 310\).
Time = 0.05 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="maxima")
Output:
1/60*(3*C*(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) - B*((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) + A*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10 *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)/d
Time = 0.30 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="giac")
Output:
1/60*(60*(B - 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(B - 3*C)*l og(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 120*C*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 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 10*A*a^12*tan(1 /2*d*x + 1/2*c)^3 - 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d *x + 1/2*c)^3 + 15*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 13.28 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{6\,a^3}-\frac {2\,B-4\,C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,A-6\,C}{4\,a^3}-\frac {3\,\left (A-B+C\right )}{4\,a^3}+\frac {2\,B-4\,C}{2\,a^3}\right )}{d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-3\,C\right )}{a^3\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*((A - B + C)/(6*a^3) - (2*B - 4*C)/(12*a^3)))/d - (t an(c/2 + (d*x)/2)*((2*A - 6*C)/(4*a^3) - (3*(A - B + C))/(4*a^3) + (2*B - 4*C)/(2*a^3)))/d + (2*atanh(tan(c/2 + (d*x)/2))*(B - 3*C))/(a^3*d) - (2*C* tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x)/2)^2 - a^3)) + (tan(c/2 + (d*x )/2)^5*(A - B + C))/(20*a^3*d)
Time = 0.15 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -85 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b +225 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -375 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{60 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)
Output:
( - 60*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c + 60*log(tan((c + d*x)/2) - 1)*b - 180* log(tan((c + d*x)/2) - 1)*c + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2 )**2*b - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c - 60*log(tan( (c + d*x)/2) + 1)*b + 180*log(tan((c + d*x)/2) + 1)*c + 3*tan((c + d*x)/2) **7*a - 3*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2)**7*c + 7*tan((c + d*x )/2)**5*a - 17*tan((c + d*x)/2)**5*b + 27*tan((c + d*x)/2)**5*c + 5*tan((c + d*x)/2)**3*a - 85*tan((c + d*x)/2)**3*b + 225*tan((c + d*x)/2)**3*c - 1 5*tan((c + d*x)/2)*a + 105*tan((c + d*x)/2)*b - 375*tan((c + d*x)/2)*c)/(6 0*a**3*d*(tan((c + d*x)/2)**2 - 1))