Integrand size = 21, antiderivative size = 188 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {a b \left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 d}+\frac {a b \left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {3 a^3 b \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:
1/2*a*b*(3*a^2+4*b^2)*arctanh(sin(d*x+c))/d+1/15*(8*a^4+60*a^2*b^2+15*b^4) *tan(d*x+c)/d+1/2*a*b*(3*a^2+4*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/15*a^2*(4*a^ 2+27*b^2)*sec(d*x+c)^2*tan(d*x+c)/d+3/5*a^3*b*sec(d*x+c)^3*tan(d*x+c)/d+1/ 5*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d
Time = 0.80 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.66 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {15 a b \left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (30 \left (a^4+6 a^2 b^2+b^4\right )+15 a b \left (3 a^2+4 b^2\right ) \sec (c+d x)+30 a^3 b \sec ^3(c+d x)+20 a^2 \left (a^2+3 b^2\right ) \tan ^2(c+d x)+6 a^4 \tan ^4(c+d x)\right )}{30 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^6,x]
Output:
(15*a*b*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(30*(a^4 + 6* a^2*b^2 + b^4) + 15*a*b*(3*a^2 + 4*b^2)*Sec[c + d*x] + 30*a^3*b*Sec[c + d* x]^3 + 20*a^2*(a^2 + 3*b^2)*Tan[c + d*x]^2 + 6*a^4*Tan[c + d*x]^4))/(30*d)
Time = 1.19 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3271, 3042, 3510, 27, 3042, 3500, 3042, 3227, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x)) \left (12 b a^2+\left (4 a^2+15 b^2\right ) \cos (c+d x) a+b \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x)dx+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (12 b a^2+\left (4 a^2+15 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (2 a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}-\frac {1}{4} \int -4 \left (\left (4 a^2+27 b^2\right ) a^2+5 b \left (3 a^2+4 b^2\right ) \cos (c+d x) a+b^2 \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)dx\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int \left (\left (4 a^2+27 b^2\right ) a^2+5 b \left (3 a^2+4 b^2\right ) \cos (c+d x) a+b^2 \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)dx+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {\left (4 a^2+27 b^2\right ) a^2+5 b \left (3 a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2 \left (2 a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \left (15 a b \left (3 a^2+4 b^2\right )+\left (8 a^4+60 b^2 a^2+15 b^4\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 a b \left (3 a^2+4 b^2\right )+\left (8 a^4+60 b^2 a^2+15 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \int \sec ^3(c+d x)dx+\left (8 a^4+60 a^2 b^2+15 b^4\right ) \int \sec ^2(c+d x)dx\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\left (8 a^4+60 a^2 b^2+15 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{d}\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{d}\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{d}\right )+\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}+\frac {1}{5} \left (\frac {3 a^3 b \tan (c+d x) \sec ^3(c+d x)}{d}+\frac {a^2 \left (4 a^2+27 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {1}{3} \left (15 a b \left (3 a^2+4 b^2\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (8 a^4+60 a^2 b^2+15 b^4\right ) \tan (c+d x)}{d}\right )\right )\) |
Input:
Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^6,x]
Output:
(a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a^2*(4* a^2 + 27*b^2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (3*a^3*b*Sec[c + d*x]^3 *Tan[c + d*x])/d + (((8*a^4 + 60*a^2*b^2 + 15*b^4)*Tan[c + d*x])/d + 15*a* b*(3*a^2 + 4*b^2)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x ])/(2*d)))/3)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* (n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin [e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] , x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
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]
Time = 18.38 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-6 a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{4}}{d}\) | \(162\) |
| default | \(\frac {-a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-6 a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{4}}{d}\) | \(162\) |
| parts | \(-\frac {a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{4} \tan \left (d x +c \right )}{d}+\frac {4 a^{3} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {6 a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(177\) |
| parallelrisch | \(\frac {-450 a b \left (a^{2}+\frac {4 b^{2}}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 a b \left (a^{2}+\frac {4 b^{2}}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (80 a^{4}+600 a^{2} b^{2}+90 b^{4}\right ) \sin \left (3 d x +3 c \right )+\left (16 a^{4}+120 a^{2} b^{2}+30 b^{4}\right ) \sin \left (5 d x +5 c \right )+\left (420 a^{3} b +240 a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (90 a^{3} b +120 a \,b^{3}\right ) \sin \left (4 d x +4 c \right )+160 \sin \left (d x +c \right ) \left (a^{4}+3 a^{2} b^{2}+\frac {3}{8} b^{4}\right )}{30 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(275\) |
| risch | \(\frac {i \left (-45 a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+30 b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-210 a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+360 a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+120 b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+160 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+840 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+180 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+210 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+120 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+80 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+600 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+120 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+60 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+16 a^{4}+120 a^{2} b^{2}+30 b^{4}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(383\) |
Input:
int((a+cos(d*x+c)*b)^4*sec(d*x+c)^6,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^4*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+4*a^3*b*(- (-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) ))-6*a^2*b^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+4*a*b^3*(1/2*sec(d*x+c)*ta n(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+tan(d*x+c)*b^4)
Time = 0.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.97 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (30 \, a^{3} b \cos \left (d x + c\right ) + 2 \, {\left (8 \, a^{4} + 60 \, a^{2} b^{2} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{4} + 15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, a^{4} + 15 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="fricas")
Output:
1/60*(15*(3*a^3*b + 4*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3* a^3*b + 4*a*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(30*a^3*b*cos(d *x + c) + 2*(8*a^4 + 60*a^2*b^2 + 15*b^4)*cos(d*x + c)^4 + 6*a^4 + 15*(3*a ^3*b + 4*a*b^3)*cos(d*x + c)^3 + 4*(2*a^4 + 15*a^2*b^2)*cos(d*x + c)^2)*si n(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**6,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.04 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2} - 15 \, a^{3} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, b^{4} \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="maxima")
Output:
1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 120 *(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b^2 - 15*a^3*b*(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*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 60*b^4*tan( d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (176) = 352\).
Time = 0.50 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.45 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x, algorithm="giac")
Output:
1/30*(15*(3*a^3*b + 4*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*a^ 3*b + 4*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(30*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*a^2*b^2*tan(1/2*d*x + 1 /2*c)^9 - 60*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*b^4*tan(1/2*d*x + 1/2*c)^9 - 40*a^4*tan(1/2*d*x + 1/2*c)^7 + 30*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 480*a^ 2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 120*b^4* tan(1/2*d*x + 1/2*c)^7 + 116*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*a^2*b^2*tan( 1/2*d*x + 1/2*c)^5 + 180*b^4*tan(1/2*d*x + 1/2*c)^5 - 40*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 480*a^2*b^2*tan(1/2*d*x + 1/ 2*c)^3 - 120*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 120*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*a^4*tan(1/2*d*x + 1/2*c) + 75*a^3*b*tan(1/2*d*x + 1/2*c) + 180*a^2*b ^2*tan(1/2*d*x + 1/2*c) + 60*a*b^3*tan(1/2*d*x + 1/2*c) + 30*b^4*tan(1/2*d *x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 44.94 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.62 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a^3\,b+4\,a\,b^3\right )}{d}-\frac {\left (2\,a^4-5\,a^3\,b+12\,a^2\,b^2-4\,a\,b^3+2\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^4}{3}+2\,a^3\,b-32\,a^2\,b^2+8\,a\,b^3-8\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^4}{15}+40\,a^2\,b^2+12\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^4}{3}-2\,a^3\,b-32\,a^2\,b^2-8\,a\,b^3-8\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^4+5\,a^3\,b+12\,a^2\,b^2+4\,a\,b^3+2\,b^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*cos(c + d*x))^4/cos(c + d*x)^6,x)
Output:
(atanh(tan(c/2 + (d*x)/2))*(4*a*b^3 + 3*a^3*b))/d - (tan(c/2 + (d*x)/2)^5* ((116*a^4)/15 + 12*b^4 + 40*a^2*b^2) + tan(c/2 + (d*x)/2)^9*(2*a^4 - 5*a^3 *b - 4*a*b^3 + 2*b^4 + 12*a^2*b^2) - tan(c/2 + (d*x)/2)^3*(8*a*b^3 + 2*a^3 *b + (8*a^4)/3 + 8*b^4 + 32*a^2*b^2) - tan(c/2 + (d*x)/2)^7*((8*a^4)/3 - 2 *a^3*b - 8*a*b^3 + 8*b^4 + 32*a^2*b^2) + tan(c/2 + (d*x)/2)*(4*a*b^3 + 5*a ^3*b + 2*a^4 + 2*b^4 + 12*a^2*b^2))/(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 = 0.18 (sec) , antiderivative size = 584, normalized size of antiderivative = 3.11 \[ \int (a+b \cos (c+d x))^4 \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^4*sec(d*x+c)^6,x)
Output:
( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b - 60* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**3 + 90*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b + 120*cos(c + d*x)* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 - 45*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*a**3*b - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b* *3 + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3*b + 60 *cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**3 - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b - 120*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3 + 45*cos(c + d*x)*log(ta n((c + d*x)/2) + 1)*a**3*b + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b **3 - 45*cos(c + d*x)*sin(c + d*x)**3*a**3*b - 60*cos(c + d*x)*sin(c + d*x )**3*a*b**3 + 75*cos(c + d*x)*sin(c + d*x)*a**3*b + 60*cos(c + d*x)*sin(c + d*x)*a*b**3 + 16*sin(c + d*x)**5*a**4 + 120*sin(c + d*x)**5*a**2*b**2 + 30*sin(c + d*x)**5*b**4 - 40*sin(c + d*x)**3*a**4 - 300*sin(c + d*x)**3*a* *2*b**2 - 60*sin(c + d*x)**3*b**4 + 30*sin(c + d*x)*a**4 + 180*sin(c + d*x )*a**2*b**2 + 30*sin(c + d*x)*b**4)/(30*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))