Integrand size = 40, antiderivative size = 169 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (7 B+6 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d} \] Output:
1/8*a^2*(7*B+6*C)*arctanh(sin(d*x+c))/d+1/5*a^2*(10*B+9*C)*tan(d*x+c)/d+1/ 8*a^2*(7*B+6*C)*sec(d*x+c)*tan(d*x+c)/d+1/20*a^2*(5*B+6*C)*sec(d*x+c)^3*ta n(d*x+c)/d+1/5*C*sec(d*x+c)^3*(a^2+a^2*sec(d*x+c))*tan(d*x+c)/d+1/15*a^2*( 10*B+9*C)*tan(d*x+c)^3/d
Leaf count is larger than twice the leaf count of optimal. \(391\) vs. \(2(169)=338\).
Time = 1.26 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.31 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^2 \sec ^5(c+d x) \left (105 B \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+90 C \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+150 (7 B+6 C) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+75 (7 B+6 C) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-105 B \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 C \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-640 B \sin (c+d x)-960 C \sin (c+d x)-660 B \sin (2 (c+d x))-840 C \sin (2 (c+d x))-800 B \sin (3 (c+d x))-720 C \sin (3 (c+d x))-210 B \sin (4 (c+d x))-180 C \sin (4 (c+d x))-160 B \sin (5 (c+d x))-144 C \sin (5 (c+d x))\right )}{1920 d} \] Input:
Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
-1/1920*(a^2*Sec[c + d*x]^5*(105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 90*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*(7*B + 6*C)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 75*(7*B + 6*C)*Cos[ 3*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x) /2] + Sin[(c + d*x)/2]]) - 105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + S in[(c + d*x)/2]] - 90*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d *x)/2]] - 640*B*Sin[c + d*x] - 960*C*Sin[c + d*x] - 660*B*Sin[2*(c + d*x)] - 840*C*Sin[2*(c + d*x)] - 800*B*Sin[3*(c + d*x)] - 720*C*Sin[3*(c + d*x) ] - 210*B*Sin[4*(c + d*x)] - 180*C*Sin[4*(c + d*x)] - 160*B*Sin[5*(c + d*x )] - 144*C*Sin[5*(c + d*x)]))/d
Time = 1.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4560, 3042, 4506, 3042, 4485, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a)^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^2 (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{5} \int \sec ^3(c+d x) (\sec (c+d x) a+a) (a (5 B+3 C)+a (5 B+6 C) \sec (c+d x))dx+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 B+3 C)+a (5 B+6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \sec ^3(c+d x) \left (5 (7 B+6 C) a^2+4 (10 B+9 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 (7 B+6 C) a^2+4 (10 B+9 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (4 a^2 (10 B+9 C) \int \sec ^4(c+d x)dx+5 a^2 (7 B+6 C) \int \sec ^3(c+d x)dx\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 a^2 (10 B+9 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^2 (10 B+9 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^2 (10 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (10 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (10 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (10 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
Input:
Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x] ^2),x]
Output:
(C*Sec[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Tan[c + d*x])/(5*d) + ((a^2*(5* B + 6*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^2*(7*B + 6*C)*(ArcTanh[ Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*a^2*(10*B + 9*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4)/5
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_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.84 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02
| method | result | size |
| parts | \(\frac {\left (B \,a^{2}+2 C \,a^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (2 B \,a^{2}+C \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(173\) |
| norman | \(\frac {\frac {7 a^{2} \left (7 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{2} \left (7 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{2} \left (25 B +26 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {8 a^{2} \left (25 B +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a^{2} \left (79 B +54 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{2} \left (7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(201\) |
| parallelrisch | \(\frac {16 \left (-\frac {105 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (B +\frac {6 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}+\frac {105 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (B +\frac {6 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}+\left (\frac {33 B}{32}+\frac {21 C}{16}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 B}{4}+\frac {9 C}{8}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {21 B}{64}+\frac {9 C}{32}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {B}{4}+\frac {9 C}{40}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (B +\frac {3 C}{2}\right )\right ) a^{2}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(217\) |
| derivativedivides | \(\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-2 B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(223\) |
| default | \(\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-2 B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(223\) |
| risch | \(-\frac {i a^{2} \left (105 B \,{\mathrm e}^{9 i \left (d x +c \right )}+90 C \,{\mathrm e}^{9 i \left (d x +c \right )}+330 B \,{\mathrm e}^{7 i \left (d x +c \right )}+420 C \,{\mathrm e}^{7 i \left (d x +c \right )}-480 B \,{\mathrm e}^{6 i \left (d x +c \right )}-240 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-1200 C \,{\mathrm e}^{4 i \left (d x +c \right )}-330 B \,{\mathrm e}^{3 i \left (d x +c \right )}-420 C \,{\mathrm e}^{3 i \left (d x +c \right )}-800 B \,{\mathrm e}^{2 i \left (d x +c \right )}-720 C \,{\mathrm e}^{2 i \left (d x +c \right )}-105 B \,{\mathrm e}^{i \left (d x +c \right )}-90 C \,{\mathrm e}^{i \left (d x +c \right )}-160 B -144 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}\) | \(287\) |
Input:
int(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method =_RETURNVERBOSE)
Output:
(B*a^2+2*C*a^2)/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(s ec(d*x+c)+tan(d*x+c)))-(2*B*a^2+C*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c )+B*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-C*a^2/ d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)
Time = 0.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, C a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/240*(15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(10*B + 9*C)*a^2*c os(d*x + c)^4 + 15*(7*B + 6*C)*a^2*cos(d*x + c)^3 + 8*(10*B + 9*C)*a^2*cos (d*x + c)^2 + 30*(B + 2*C)*a^2*cos(d*x + c) + 24*C*a^2)*sin(d*x + c))/(d*c os(d*x + c)^5)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))**2*(B*sec(d*x+c)+C*sec(d*x+c)**2) ,x)
Output:
a**2*(Integral(B*sec(c + d*x)**3, x) + Integral(2*B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**4, x) + Integral (2*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))
Time = 0.04 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.64 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/240*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^2 + 80*(tan(d*x + c)^3 + 3*tan( d*x + c))*C*a^2 - 15*B*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*C*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c) ^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Time = 0.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.46 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 90 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 420 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 800 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 864 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 390 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/120*(15*(7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(7*B *a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(105*B*a^2*tan(1/2* d*x + 1/2*c)^9 + 90*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 490*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 420*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 800*B*a^2*tan(1/2*d*x + 1/2 *c)^5 + 864*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 790*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 540*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 375*B*a^2*tan(1/2*d*x + 1/2*c) + 39 0*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 13.96 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,B+6\,C\right )}{4\,d}-\frac {\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {49\,B\,a^2}{6}-7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,B\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {79\,B\,a^2}{6}-9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,B\,a^2}{4}+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/cos(c + d *x)^2,x)
Output:
(a^2*atanh(tan(c/2 + (d*x)/2))*(7*B + 6*C))/(4*d) - (tan(c/2 + (d*x)/2)*(( 25*B*a^2)/4 + (13*C*a^2)/2) + tan(c/2 + (d*x)/2)^9*((7*B*a^2)/4 + (3*C*a^2 )/2) - tan(c/2 + (d*x)/2)^7*((49*B*a^2)/6 + 7*C*a^2) - tan(c/2 + (d*x)/2)^ 3*((79*B*a^2)/6 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*((40*B*a^2)/3 + (72*C*a^ 2)/5))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.15 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.85 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**2*( - 105*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 9 0*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 210*cos(c + d *x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 180*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 105*cos(c + d*x)*log(tan((c + d*x)/ 2) - 1)*b - 90*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 105*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 210*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 180*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c + 105*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 90*cos(c + d*x )*log(tan((c + d*x)/2) + 1)*c - 105*cos(c + d*x)*sin(c + d*x)**3*b - 90*co s(c + d*x)*sin(c + d*x)**3*c + 135*cos(c + d*x)*sin(c + d*x)*b + 150*cos(c + d*x)*sin(c + d*x)*c + 160*sin(c + d*x)**5*b + 144*sin(c + d*x)**5*c - 4 00*sin(c + d*x)**3*b - 360*sin(c + d*x)**3*c + 240*sin(c + d*x)*b + 240*si n(c + d*x)*c))/(120*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))