Integrand size = 31, antiderivative size = 173 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=a^4 B x+\frac {a^4 (35 A+48 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
a^4*B*x+1/8*a^4*(35*A+48*B)*arctanh(sin(d*x+c))/d+5/8*a^4*(7*A+8*B)*tan(d* x+c)/d+1/24*(35*A+32*B)*(a^4+a^4*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/12* (7*A+4*B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*A*(a+a*co s(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d
Time = 3.68 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=a^4 B x+\frac {35 a^4 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {6 a^4 B \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {7 a^4 B \tan (c+d x)}{d}+\frac {27 a^4 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {2 a^4 B \sec (c+d x) \tan (c+d x)}{d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {4 a^4 A \tan ^3(c+d x)}{3 d}+\frac {a^4 B \tan ^3(c+d x)}{3 d} \]
a^4*B*x + (35*a^4*A*ArcTanh[Sin[c + d*x]])/(8*d) + (6*a^4*B*ArcTanh[Sin[c + d*x]])/d + (8*a^4*A*Tan[c + d*x])/d + (7*a^4*B*Tan[c + d*x])/d + (27*a^4 *A*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (2*a^4*B*Sec[c + d*x]*Tan[c + d*x])/ d + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*A*Tan[c + d*x]^3)/( 3*d) + (a^4*B*Tan[c + d*x]^3)/(3*d)
Time = 1.38 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3454, 3042, 3454, 3042, 3454, 27, 3042, 3447, 3042, 3500, 3042, 3214, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^4 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{4} \int (\cos (c+d x) a+a)^3 (a (7 A+4 B)+4 a B \cos (c+d x)) \sec ^4(c+d x)dx+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (7 A+4 B)+4 a B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left ((35 A+32 B) a^2+12 B \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((35 A+32 B) a^2+12 B \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (5 (7 A+8 B) a^3+8 B \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int (\cos (c+d x) a+a) \left (5 (7 A+8 B) a^3+8 B \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (7 A+8 B) a^3+8 B \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \left (8 B \cos ^2(c+d x) a^4+5 (7 A+8 B) a^4+\left (8 B a^4+5 (7 A+8 B) a^4\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \frac {8 B \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+5 (7 A+8 B) a^4+\left (8 B a^4+5 (7 A+8 B) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\int \left ((35 A+48 B) a^4+8 B \cos (c+d x) a^4\right ) \sec (c+d x)dx+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{d}\right )+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\int \frac {(35 A+48 B) a^4+8 B \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{d}\right )+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (a^4 (35 A+48 B) \int \sec (c+d x)dx+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{d}+8 a^4 B x\right )+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (a^4 (35 A+48 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{d}+8 a^4 B x\right )+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\frac {a^4 (35 A+48 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{d}+8 a^4 B x\right )+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{2 d}\right )+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d}\) |
(a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (((7*A + 4*B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((35 *A + 32*B)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3* (8*a^4*B*x + (a^4*(35*A + 48*B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 8 *B)*Tan[c + d*x])/d))/2)/3)/4
3.1.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 4.78 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.17
| method | result | size |
| parts | \(\frac {a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (a^{4} A +4 B \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {\left (4 a^{4} A +B \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +6 B \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \,a^{4} \left (d x +c \right )}{d}\) | \(203\) |
| parallelrisch | \(\frac {56 a^{4} \left (-\frac {15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {48 B}{35}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16}+\frac {15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {48 B}{35}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16}+\frac {3 d x B \cos \left (2 d x +2 c \right )}{14}+\frac {3 d x B \cos \left (4 d x +4 c \right )}{56}+\left (A +\frac {11 B}{14}\right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (\frac {27 A}{16}+B \right ) \sin \left (3 d x +3 c \right )}{14}+\frac {5 \left (A +B \right ) \sin \left (4 d x +4 c \right )}{14}+\frac {3 \left (\frac {5 A}{16}+\frac {B}{7}\right ) \sin \left (d x +c \right )}{2}+\frac {9 d x B}{56}\right )}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(209\) |
| derivativedivides | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 B \,a^{4} \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(250\) |
| default | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 B \,a^{4} \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(250\) |
| risch | \(a^{4} B x -\frac {i a^{4} \left (81 A \,{\mathrm e}^{7 i \left (d x +c \right )}+48 B \,{\mathrm e}^{7 i \left (d x +c \right )}-96 A \,{\mathrm e}^{6 i \left (d x +c \right )}-144 B \,{\mathrm e}^{6 i \left (d x +c \right )}+105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+48 B \,{\mathrm e}^{5 i \left (d x +c \right )}-480 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 B \,{\mathrm e}^{4 i \left (d x +c \right )}-105 A \,{\mathrm e}^{3 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-544 A \,{\mathrm e}^{2 i \left (d x +c \right )}-496 B \,{\mathrm e}^{2 i \left (d x +c \right )}-81 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}-160 A -160 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {35 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {35 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(293\) |
a^4*A/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+ tan(d*x+c)))+(A*a^4+4*B*a^4)/d*ln(sec(d*x+c)+tan(d*x+c))-(4*A*a^4+B*a^4)/d *(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(4*A*a^4+6*B*a^4)/d*tan(d*x+c)+(6*A*a^ 4+4*B*a^4)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+B*a ^4/d*(d*x+c)
Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {48 \, B a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, {\left (A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(48*B*a^4*d*x*cos(d*x + c)^4 + 3*(35*A + 48*B)*a^4*cos(d*x + c)^4*log (sin(d*x + c) + 1) - 3*(35*A + 48*B)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(160*(A + B)*a^4*cos(d*x + c)^3 + 3*(27*A + 16*B)*a^4*cos(d*x + c )^2 + 8*(4*A + B)*a^4*cos(d*x + c) + 6*A*a^4)*sin(d*x + c))/(d*cos(d*x + c )^4)
Timed out. \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{4} - 3 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right )}{48 \, d} \]
1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 16*(tan(d*x + c)^3 + 3* tan(d*x + c))*B*a^4 + 48*(d*x + c)*B*a^4 - 3*A*a^4*(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)) - 72*A*a^4*(2*sin(d*x + c)/(sin(d*x + c )^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 48*B*a^4*(2*si n(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*B*a ^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 192*A*a^4*tan(d*x + c ) + 288*B*a^4*tan(d*x + c))/d
Time = 0.37 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {24 \, {\left (d x + c\right )} B a^{4} + 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(24*(d*x + c)*B*a^4 + 3*(35*A*a^4 + 48*B*a^4)*log(abs(tan(1/2*d*x + 1 /2*c) + 1)) - 3*(35*A*a^4 + 48*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 511* A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 279*A*a^ 4*tan(1/2*d*x + 1/2*c) - 216*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/ 2*c)^2 - 1)^4)/d
Time = 0.47 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {35\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,A\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]
(35*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (2*B*a^4*a tan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d*x )) + (27*A*a^4*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (4*A*a^4*sin(c + d*x)) /(3*d*cos(c + d*x)^3) + (A*a^4*sin(c + d*x))/(4*d*cos(c + d*x)^4) + (20*B* a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*B*a^4*sin(c + d*x))/(d*cos(c + d *x)^2) + (B*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3)