Integrand size = 19, antiderivative size = 107 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=2 a b \left (2 a^2+b^2\right ) x+\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{3 d}+\frac {4 a b^3 \cos (c+d x) \sin (c+d x)}{3 d}+\frac {b^2 (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \] Output:
2*a*b*(2*a^2+b^2)*x+a^4*arctanh(sin(d*x+c))/d+1/3*b^2*(17*a^2+2*b^2)*sin(d *x+c)/d+4/3*a*b^3*cos(d*x+c)*sin(d*x+c)/d+1/3*b^2*(a+b*cos(d*x+c))^2*sin(d *x+c)/d
Time = 0.89 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.20 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {24 a b \left (2 a^2+b^2\right ) (c+d x)-12 a^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 b^2 \left (8 a^2+b^2\right ) \sin (c+d x)+12 a b^3 \sin (2 (c+d x))+b^4 \sin (3 (c+d x))}{12 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x],x]
Output:
(24*a*b*(2*a^2 + b^2)*(c + d*x) - 12*a^4*Log[Cos[(c + d*x)/2] - Sin[(c + d *x)/2]] + 12*a^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*b^2*(8*a^2 + b^2)*Sin[c + d*x] + 12*a*b^3*Sin[2*(c + d*x)] + b^4*Sin[3*(c + d*x)])/(12 *d)
Time = 0.83 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 3272, 3042, 3512, 27, 3042, 3502, 27, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \cos (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (3 a^3+8 b^2 \cos ^2(c+d x) a+b \left (9 a^2+2 b^2\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 a^3+8 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (9 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int 2 \left (3 a^4+6 b \left (2 a^2+b^2\right ) \cos (c+d x) a+b^2 \left (17 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \left (3 a^4+6 b \left (2 a^2+b^2\right ) \cos (c+d x) a+b^2 \left (17 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 a^4+6 b \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2 \left (17 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\int 3 \left (a^4+2 b \left (2 a^2+b^2\right ) \cos (c+d x) a\right ) \sec (c+d x)dx+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (3 \int \left (a^4+2 b \left (2 a^2+b^2\right ) \cos (c+d x) a\right ) \sec (c+d x)dx+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 \int \frac {a^4+2 b \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (3 \left (a^4 \int \sec (c+d x)dx+2 a b x \left (2 a^2+b^2\right )\right )+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 \left (a^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a b x \left (2 a^2+b^2\right )\right )+\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {b^2 \left (17 a^2+2 b^2\right ) \sin (c+d x)}{d}+3 \left (\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+2 a b x \left (2 a^2+b^2\right )\right )+\frac {4 a b^3 \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {b^2 \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x],x]
Output:
(b^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (3*(2*a*b*(2*a^2 + b^2)* x + (a^4*ArcTanh[Sin[c + d*x]])/d) + (b^2*(17*a^2 + 2*b^2)*Sin[c + d*x])/d + (4*a*b^3*Cos[c + d*x]*Sin[c + d*x])/d)/3
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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[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[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 12.71 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{3} b \left (d x +c \right )+6 a^{2} b^{2} \sin \left (d x +c \right )+4 a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(98\) |
| default | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{3} b \left (d x +c \right )+6 a^{2} b^{2} \sin \left (d x +c \right )+4 a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(98\) |
| parallelrisch | \(\frac {-12 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \sin \left (2 d x +2 c \right ) a \,b^{3}+\sin \left (3 d x +3 c \right ) b^{4}+9 \left (8 a^{2} b^{2}+b^{4}\right ) \sin \left (d x +c \right )+48 a d b x \left (a^{2}+\frac {b^{2}}{2}\right )}{12 d}\) | \(104\) |
| parts | \(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {4 a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {6 \sin \left (d x +c \right ) a^{2} b^{2}}{d}+\frac {4 a^{3} b \left (d x +c \right )}{d}\) | \(109\) |
| risch | \(4 a^{3} b x +2 a \,b^{3} x -\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b^{2}}{d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} b^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{2}}{d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} b^{4}}{8 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) b^{4}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a \,b^{3}}{d}\) | \(169\) |
| norman | \(\frac {\left (4 a^{3} b +2 a \,b^{3}\right ) x +\left (4 a^{3} b +2 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (16 a^{3} b +8 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (16 a^{3} b +8 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (24 a^{3} b +12 a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {2 b^{2} \left (6 a^{2}-2 a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 b^{2} \left (6 a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 b^{2} \left (54 a^{2}-6 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {2 b^{2} \left (54 a^{2}+6 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(307\) |
Input:
int((a+cos(d*x+c)*b)^4*sec(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*(a^4*ln(sec(d*x+c)+tan(d*x+c))+4*a^3*b*(d*x+c)+6*a^2*b^2*sin(d*x+c)+4* a*b^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*b^4*(cos(d*x+c)^2+2)*s in(d*x+c))
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {3 \, a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, {\left (2 \, a^{3} b + a b^{3}\right )} d x + 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} + 6 \, a b^{3} \cos \left (d x + c\right ) + 18 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c),x, algorithm="fricas")
Output:
1/6*(3*a^4*log(sin(d*x + c) + 1) - 3*a^4*log(-sin(d*x + c) + 1) + 12*(2*a^ 3*b + a*b^3)*d*x + 2*(b^4*cos(d*x + c)^2 + 6*a*b^3*cos(d*x + c) + 18*a^2*b ^2 + 2*b^4)*sin(d*x + c))/d
\[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{4} \sec {\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))**4*sec(d*x+c),x)
Output:
Integral((a + b*cos(c + d*x))**4*sec(c + d*x), x)
Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} a^{3} b + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} - {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{4} + 3 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 18 \, a^{2} b^{2} \sin \left (d x + c\right )}{3 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c),x, algorithm="maxima")
Output:
1/3*(12*(d*x + c)*a^3*b + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^3 - (sin( d*x + c)^3 - 3*sin(d*x + c))*b^4 + 3*a^4*log(sec(d*x + c) + tan(d*x + c)) + 18*a^2*b^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (101) = 202\).
Time = 0.56 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.98 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {3 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 6 \, {\left (2 \, a^{3} b + a b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c),x, algorithm="giac")
Output:
1/3*(3*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(2*a^3*b + a*b^3)*(d*x + c) + 2*(18*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*b^4*tan(1/2*d*x + 1/2*c)^ 5 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*b^4*tan(1/2*d*x + 1/2*c)^3 + 18* a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*a*b^3*tan(1/2*d*x + 1/2*c) + 3*b^4*tan(1/ 2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 40.64 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.48 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {3\,b^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \] Input:
int((a + b*cos(c + d*x))^4/cos(c + d*x),x)
Output:
(3*b^4*sin(c + d*x))/(4*d) + (2*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d* x)/2)))/d + (b^4*sin(3*c + 3*d*x))/(12*d) + (a*b^3*sin(2*c + 2*d*x))/d + ( 6*a^2*b^2*sin(c + d*x))/d + (4*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d* x)/2)))/d + (8*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4}-\sin \left (d x +c \right )^{3} b^{4}+18 \sin \left (d x +c \right ) a^{2} b^{2}+3 \sin \left (d x +c \right ) b^{4}+12 a^{3} b d x +6 a \,b^{3} d x}{3 d} \] Input:
int((a+b*cos(d*x+c))^4*sec(d*x+c),x)
Output:
(6*cos(c + d*x)*sin(c + d*x)*a*b**3 - 3*log(tan((c + d*x)/2) - 1)*a**4 + 3 *log(tan((c + d*x)/2) + 1)*a**4 - sin(c + d*x)**3*b**4 + 18*sin(c + d*x)*a **2*b**2 + 3*sin(c + d*x)*b**4 + 12*a**3*b*d*x + 6*a*b**3*d*x)/(3*d)