Integrand size = 21, antiderivative size = 108 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=4 a b^3 x+\frac {a^2 \left (a^2+12 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a^3 b \tan (c+d x)}{d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
4*a*b^3*x+1/2*a^2*(a^2+12*b^2)*arctanh(sin(d*x+c))/d-1/2*b^2*(a^2-2*b^2)*s in(d*x+c)/d+3*a^3*b*tan(d*x+c)/d+1/2*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)*tan (d*x+c)/d
Time = 3.27 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.61 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {a \left (16 b^3 c+16 b^3 d x-2 a \left (a^2+12 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (a^2+12 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )+4 b^4 \sin (c+d x)+16 a^3 b \tan (c+d x)}{4 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3,x]
Output:
(a*(16*b^3*c + 16*b^3*d*x - 2*a*(a^2 + 12*b^2)*Log[Cos[(c + d*x)/2] - Sin[ (c + d*x)/2]] + 2*a*(a^2 + 12*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2] ] + a^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - a^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + 4*b^4*Sin[c + d*x] + 16*a^3*b*Tan[c + d*x])/(4*d)
Time = 0.84 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3271, 3042, 3510, 25, 3042, 3502, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {1}{2} \int (a+b \cos (c+d x)) \left (6 b a^2+\left (a^2+6 b^2\right ) \cos (c+d x) a-b \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)dx+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (6 b a^2+\left (a^2+6 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-b \left (a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{2} \left (\frac {6 a^3 b \tan (c+d x)}{d}-\int -\left (\left (8 a \cos (c+d x) b^3-\left (a^2-2 b^2\right ) \cos ^2(c+d x) b^2+a^2 \left (a^2+12 b^2\right )\right ) \sec (c+d x)\right )dx\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\int \left (8 a \cos (c+d x) b^3-\left (a^2-2 b^2\right ) \cos ^2(c+d x) b^2+a^2 \left (a^2+12 b^2\right )\right ) \sec (c+d x)dx+\frac {6 a^3 b \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {8 a \sin \left (c+d x+\frac {\pi }{2}\right ) b^3-\left (a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2+a^2 \left (a^2+12 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 a^3 b \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left (8 a \cos (c+d x) b^3+a^2 \left (a^2+12 b^2\right )\right ) \sec (c+d x)dx+\frac {6 a^3 b \tan (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {8 a \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+a^2 \left (a^2+12 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 a^3 b \tan (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a^2 \left (a^2+12 b^2\right ) \int \sec (c+d x)dx+\frac {6 a^3 b \tan (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{d}+8 a b^3 x\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a^2 \left (a^2+12 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {6 a^3 b \tan (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{d}+8 a b^3 x\right )+\frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {1}{2} \left (\frac {6 a^3 b \tan (c+d x)}{d}+\frac {a^2 \left (a^2+12 b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{d}+8 a b^3 x\right )\) |
Input:
Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^3,x]
Output:
(a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (8*a*b^3*x + (a^2*(a^2 + 12*b^2)*ArcTanh[Sin[c + d*x]])/d - (b^2*(a^2 - 2*b^2)*Sin[c + d*x])/d + (6*a^3*b*Tan[c + d*x])/d)/2
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*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[(-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[(-(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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 14.42 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{3} b \tan \left (d x +c \right )+6 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a \,b^{3} \left (d x +c \right )+\sin \left (d x +c \right ) b^{4}}{d}\) | \(96\) |
| default | \(\frac {a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{3} b \tan \left (d x +c \right )+6 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a \,b^{3} \left (d x +c \right )+\sin \left (d x +c \right ) b^{4}}{d}\) | \(96\) |
| parts | \(\frac {a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\sin \left (d x +c \right ) b^{4}}{d}+\frac {4 a^{3} b \tan \left (d x +c \right )}{d}+\frac {6 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a \,b^{3} \left (d x +c \right )}{d}\) | \(107\) |
| parallelrisch | \(\frac {-a^{2} \left (a^{2}+12 b^{2}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+a^{2} \left (a^{2}+12 b^{2}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 a \,b^{3} d x \cos \left (2 d x +2 c \right )+8 \sin \left (2 d x +2 c \right ) a^{3} b +\sin \left (3 d x +3 c \right ) b^{4}+\left (2 a^{4}+b^{4}\right ) \sin \left (d x +c \right )+8 a \,b^{3} d x}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(162\) |
| risch | \(4 a \,b^{3} x -\frac {i b^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{4}}{2 d}-\frac {i a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )} a -8 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-8 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}\) | \(196\) |
| norman | \(\frac {\frac {\left (a^{4}-8 a^{3} b +2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {\left (a^{4}+8 a^{3} b +2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (5 a^{4}-24 a^{3} b +2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (5 a^{4}+24 a^{3} b +2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+4 a \,b^{3} x +\frac {2 \left (5 a^{4}-8 a^{3} b -2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 \left (5 a^{4}+8 a^{3} b -2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+8 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-16 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+8 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+4 a \,b^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a^{2} \left (a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(395\) |
Input:
int((a+cos(d*x+c)*b)^4*sec(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^4*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4*a^3*b *tan(d*x+c)+6*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4*a*b^3*(d*x+c)+sin(d*x+c) *b^4)
Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.20 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {16 \, a b^{3} d x \cos \left (d x + c\right )^{2} + {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{3} b \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="fricas")
Output:
1/4*(16*a*b^3*d*x*cos(d*x + c)^2 + (a^4 + 12*a^2*b^2)*cos(d*x + c)^2*log(s in(d*x + c) + 1) - (a^4 + 12*a^2*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1 ) + 2*(2*b^4*cos(d*x + c)^2 + 8*a^3*b*cos(d*x + c) + a^4)*sin(d*x + c))/(d *cos(d*x + c)^2)
\[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{4} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**3,x)
Output:
Integral((a + b*cos(c + d*x))**4*sec(c + d*x)**3, x)
Time = 0.05 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {16 \, {\left (d x + c\right )} a b^{3} - a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, b^{4} \sin \left (d x + c\right ) + 16 \, a^{3} b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="maxima")
Output:
1/4*(16*(d*x + c)*a*b^3 - a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s in(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*b^4*sin(d*x + c) + 16*a^3*b*tan(d*x + c)) /d
Time = 0.53 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.64 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} a b^{3} + \frac {4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x, algorithm="giac")
Output:
1/2*(8*(d*x + c)*a*b^3 + 4*b^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^ 2 + 1) + (a^4 + 12*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (a^4 + 12 *a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(a^4*tan(1/2*d*x + 1/2*c) ^3 - 8*a^3*b*tan(1/2*d*x + 1/2*c)^3 + a^4*tan(1/2*d*x + 1/2*c) + 8*a^3*b*t an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 41.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.41 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {b^4\,\sin \left (c+d\,x\right )}{d}+\frac {a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {12\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a^3\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \] Input:
int((a + b*cos(c + d*x))^4/cos(c + d*x)^3,x)
Output:
(b^4*sin(c + d*x))/d + (a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/ d + (a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (12*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*a^3*b*sin(c + d*x))/(d*cos(c + d*x))
Time = 0.17 (sec) , antiderivative size = 367, normalized size of antiderivative = 3.40 \[ \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{4}-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}+12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{4}+12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4}-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{4}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{3} d x -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-8 \cos \left (d x +c \right ) a \,b^{3} d x +8 \sin \left (d x +c \right )^{3} a^{3} b -8 \sin \left (d x +c \right ) a^{3} b}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))^4*sec(d*x+c)^3,x)
Output:
( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 + cos(c + d*x) *log(tan((c + d*x)/2) - 1)*a**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1 )*a**2*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - co s(c + d*x)*log(tan((c + d*x)/2) + 1)*a**4 - 12*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*a**2*b**2 + 2*cos(c + d*x)*sin(c + d*x)**3*b**4 + 8*cos(c + d* x)*sin(c + d*x)**2*a*b**3*d*x - cos(c + d*x)*sin(c + d*x)*a**4 - 2*cos(c + d*x)*sin(c + d*x)*b**4 - 8*cos(c + d*x)*a*b**3*d*x + 8*sin(c + d*x)**3*a* *3*b - 8*sin(c + d*x)*a**3*b)/(2*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))