Integrand size = 21, antiderivative size = 133 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {3 a \left (a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a \left (a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^2 b \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
3/8*a*(a^2+4*b^2)*arctanh(sin(d*x+c))/d+b*(2*a^2+b^2)*tan(d*x+c)/d+3/8*a*( a^2+4*b^2)*sec(d*x+c)*tan(d*x+c)/d+3/4*a^2*b*sec(d*x+c)^2*tan(d*x+c)/d+1/4 *a^2*(a+b*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.68 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {3 a \left (a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 a \left (a^2+4 b^2\right ) \sec (c+d x)+2 a^3 \sec ^3(c+d x)+8 b \left (3 a^2+b^2+a^2 \tan ^2(c+d x)\right )\right )}{8 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^5,x]
Output:
(3*a*(a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*a*(a^2 + 4*b^2) *Sec[c + d*x] + 2*a^3*Sec[c + d*x]^3 + 8*b*(3*a^2 + b^2 + a^2*Tan[c + d*x] ^2)))/(8*d)
Time = 0.85 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3271, 3042, 3500, 27, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {1}{4} \int \left (9 b a^2+3 \left (a^2+4 b^2\right ) \cos (c+d x) a+2 b \left (a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {9 b a^2+3 \left (a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 b \left (a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int 3 \left (3 a \left (a^2+4 b^2\right )+4 b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\int \left (3 a \left (a^2+4 b^2\right )+4 b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {3 a \left (a^2+4 b^2\right )+4 b \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \int \sec ^3(c+d x)dx+4 b \left (2 a^2+b^2\right ) \int \sec ^2(c+d x)dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (4 b \left (2 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 a \left (a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (-\frac {4 b \left (2 a^2+b^2\right ) \int 1d(-\tan (c+d x))}{d}+3 a \left (a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^5,x]
Output:
(a^2*(a + b*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*b*(2*a^ 2 + b^2)*Tan[c + d*x])/d + (3*a^2*b*Sec[c + d*x]^2*Tan[c + d*x])/d + 3*a*( a^2 + 4*b^2)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2 *d)))/4
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[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 = 7.84 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{3}}{d}\) | \(125\) |
| default | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{3}}{d}\) | \(125\) |
| parts | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{3} \tan \left (d x +c \right )}{d}-\frac {3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 b^{2} a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(133\) |
| parallelrisch | \(\frac {-6 a \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 a \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \left (4 a^{2} b +b^{3}\right ) \sin \left (2 d x +2 c \right )+3 \left (a^{3}+4 b^{2} a \right ) \sin \left (3 d x +3 c \right )+4 \left (2 a^{2} b +b^{3}\right ) \sin \left (4 d x +4 c \right )+\left (11 a^{3}+12 b^{2} a \right ) \sin \left (d x +c \right )}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(205\) |
| risch | \(-\frac {i \left (3 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+12 b^{2} a \,{\mathrm e}^{7 i \left (d x +c \right )}-8 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+11 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+12 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-48 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-11 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{3}-12 b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}-64 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-24 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{3} {\mathrm e}^{i \left (d x +c \right )}-12 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}-16 a^{2} b -8 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\) | \(307\) |
| norman | \(\frac {\frac {a \left (7 a^{2}-12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (5 a^{3}-24 a^{2} b +12 b^{2} a -8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (5 a^{3}+24 a^{2} b +12 b^{2} a +8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (27 a^{3}-8 a^{2} b -12 b^{2} a -24 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (27 a^{3}+8 a^{2} b -12 b^{2} a +24 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {a \left (9 a^{2}-16 a b +12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d}+\frac {a \left (9 a^{2}+16 a b +12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {3 a \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(333\) |
Input:
int((a+cos(d*x+c)*b)^3*sec(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(-(-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)))-3*a^2*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3*b^2*a*(1/2*sec( d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+tan(d*x+c)*b^3)
Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a^{2} b \cos \left (d x + c\right ) + 8 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, a^{3} + 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="fricas")
Output:
1/16*(3*(a^3 + 4*a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^3 + 4* a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*a^2*b*cos(d*x + c) + 8 *(2*a^2*b + b^3)*cos(d*x + c)^3 + 2*a^3 + 3*(a^3 + 4*a*b^2)*cos(d*x + c)^2 )*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*sec(d*x+c)**5,x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.19 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b - a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, b^{3} \tan \left (d x + c\right )}{16 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="maxima")
Output:
1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b - a^3*(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)) - 12*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 16*b^3*tan (d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (125) = 250\).
Time = 0.55 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.48 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="giac")
Output:
1/8*(3*(a^3 + 4*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(a^3 + 4*a*b ^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 8*b^3 *tan(1/2*d*x + 1/2*c)^7 + 3*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^2*b*tan(1/2* d*x + 1/2*c)^5 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*b^3*tan(1/2*d*x + 1/ 2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*ta n(1/2*d*x + 1/2*c) + 24*a^2*b*tan(1/2*d*x + 1/2*c) + 12*a*b^2*tan(1/2*d*x + 1/2*c) + 8*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 44.95 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.68 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx=\frac {\left (\frac {5\,a^3}{4}-6\,a^2\,b+3\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^3}{4}+10\,a^2\,b-3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^3}{4}-10\,a^2\,b-3\,a\,b^2-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^3}{4}+6\,a^2\,b+3\,a\,b^2+2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+4\,b^2\right )}{4\,d} \] Input:
int((a + b*cos(c + d*x))^3/cos(c + d*x)^5,x)
Output:
(tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 6*a^2*b + (5*a^3)/4 - 2*b^3) - tan(c/2 + (d*x)/2)^3*(3*a*b^2 + 10*a^2*b - (3*a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)^5 *(10*a^2*b - 3*a*b^2 + (3*a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)*(3*a*b^2 + 6*a^2*b + (5*a^3)/4 + 2*b^3))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d* x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*a*atanh (tan(c/2 + (d*x)/2))*(a^2 + 4*b^2))/(4*d)
Time = 0.18 (sec) , antiderivative size = 533, normalized size of antiderivative = 4.01 \[ \int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 - 12*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**2 + 6*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 24*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - 3*cos(c + d*x)*log(tan((c + d* x)/2) - 1)*a**3 - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2 + 3*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 + 12*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2 - 6*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 24*cos(c + d*x)*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1 )*a**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 - 3*cos(c + d*x) *sin(c + d*x)**3*a**3 - 12*cos(c + d*x)*sin(c + d*x)**3*a*b**2 + 5*cos(c + d*x)*sin(c + d*x)*a**3 + 12*cos(c + d*x)*sin(c + d*x)*a*b**2 + 16*sin(c + d*x)**5*a**2*b + 8*sin(c + d*x)**5*b**3 - 40*sin(c + d*x)**3*a**2*b - 16* sin(c + d*x)**3*b**3 + 24*sin(c + d*x)*a**2*b + 8*sin(c + d*x)*b**3)/(8*co s(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))