Integrand size = 21, antiderivative size = 109 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {b \left (3 a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{3 d}+\frac {7 a^2 b \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*b*(3*a^2+2*b^2)*arctanh(sin(d*x+c))/d+1/3*a*(2*a^2+9*b^2)*tan(d*x+c)/d +7/6*a^2*b*sec(d*x+c)*tan(d*x+c)/d+1/3*a^2*(a+b*cos(d*x+c))*sec(d*x+c)^2*t an(d*x+c)/d
Time = 0.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {6 b^3 \coth ^{-1}(\sin (c+d x))+a \left (9 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 a^2+18 b^2+9 a b \sec (c+d x)+2 a^2 \tan ^2(c+d x)\right )\right )}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4,x]
Output:
(6*b^3*ArcCoth[Sin[c + d*x]] + a*(9*a*b*ArcTanh[Sin[c + d*x]] + Tan[c + d* x]*(6*a^2 + 18*b^2 + 9*a*b*Sec[c + d*x] + 2*a^2*Tan[c + d*x]^2)))/(6*d)
Time = 0.73 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3271, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {1}{3} \int \left (7 b a^2+\left (2 a^2+9 b^2\right ) \cos (c+d x) a+b \left (a^2+3 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {7 b a^2+\left (2 a^2+9 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (a^2+3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (2 a \left (2 a^2+9 b^2\right )+3 b \left (3 a^2+2 b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 a \left (2 a^2+9 b^2\right )+3 b \left (3 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 a \left (2 a^2+9 b^2\right ) \int \sec ^2(c+d x)dx+3 b \left (3 a^2+2 b^2\right ) \int \sec (c+d x)dx\right )+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 b \left (3 a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a \left (2 a^2+9 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 b \left (3 a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 a \left (2 a^2+9 b^2\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 b \left (3 a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{d}\right )+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 b \left (3 a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{d}\right )+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4,x]
Output:
(a^2*(a + b*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((7*a^2*b*S ec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*b*(3*a^2 + 2*b^2)*ArcTanh[Sin[c + d* x]])/d + (2*a*(2*a^2 + 9*b^2)*Tan[c + d*x])/d)/2)/3
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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.80 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 \tan \left (d x +c \right ) a \,b^{2}+b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(96\) |
| default | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 \tan \left (d x +c \right ) a \,b^{2}+b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(96\) |
| parts | \(-\frac {a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 b^{2} a \tan \left (d x +c \right )}{d}\) | \(104\) |
| parallelrisch | \(\frac {-27 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \left (a^{2}+\frac {2 b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+27 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \left (a^{2}+\frac {2 b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (4 a^{3}+18 b^{2} a \right ) \sin \left (3 d x +3 c \right )+18 \sin \left (2 d x +2 c \right ) a^{2} b +\left (12 a^{3}+18 b^{2} a \right ) \sin \left (d x +c \right )}{6 d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\) | \(167\) |
| risch | \(-\frac {i a \left (9 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-18 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-36 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a b \,{\mathrm e}^{i \left (d x +c \right )}-4 a^{2}-18 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(186\) |
| norman | \(\frac {-\frac {2 a \left (2 a^{2}-3 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a \left (2 a^{2}-3 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 a \left (2 a^{2}+3 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a \left (2 a^{2}+3 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (14 a^{2}-27 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a \left (14 a^{2}+27 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(282\) |
Input:
int((a+cos(d*x+c)*b)^3*sec(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3*a^2*b*(1/2*sec(d*x+c)*tan(d *x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+3*tan(d*x+c)*a*b^2+b^3*ln(sec(d*x+c)+ tan(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (9 \, a^{2} b \cos \left (d x + c\right ) + 2 \, a^{3} + 2 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="fricas")
Output:
1/12*(3*(3*a^2*b + 2*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(3*a^2* b + 2*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(9*a^2*b*cos(d*x + c) + 2*a^3 + 2*(2*a^3 + 9*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))**3*sec(d*x+c)**4,x)
Output:
Integral((a + b*cos(c + d*x))**3*sec(c + d*x)**4, x)
Time = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.04 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 9 \, a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a b^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 9*a^2*b*(2*sin(d*x + c)/(s in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*b^ 3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*a*b^2*tan(d*x + c)) /d
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (101) = 202\).
Time = 0.57 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.88 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="giac")
Output:
1/6*(3*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*a^3*tan(1/2*d*x + 1/2*c) ^5 - 9*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4* a^3*tan(1/2*d*x + 1/2*c)^3 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*tan(1 /2*d*x + 1/2*c) + 9*a^2*b*tan(1/2*d*x + 1/2*c) + 18*a*b^2*tan(1/2*d*x + 1/ 2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 42.45 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.44 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {\left (2\,a^3-3\,a^2\,b+6\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,a^3}{3}-12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+3\,a^2\,b+6\,a\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*cos(c + d*x))^3/cos(c + d*x)^4,x)
Output:
(atanh(tan(c/2 + (d*x)/2))*(3*a^2*b + 2*b^3))/d - (tan(c/2 + (d*x)/2)^5*(6 *a*b^2 - 3*a^2*b + 2*a^3) - tan(c/2 + (d*x)/2)^3*(12*a*b^2 + (4*a^3)/3) + tan(c/2 + (d*x)/2)*(6*a*b^2 + 3*a^2*b + 2*a^3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.18 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.88 \[ \int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b +4 \sin \left (d x +c \right )^{3} a^{3}+18 \sin \left (d x +c \right )^{3} a \,b^{2}-6 \sin \left (d x +c \right ) a^{3}-18 \sin \left (d x +c \right ) a \,b^{2}}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x)
Output:
( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b - 6*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 + 9*cos(c + d*x) *log(tan((c + d*x)/2) - 1)*a**2*b + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b - 6*cos(c + d*x)*log(tan((c + d*x) /2) + 1)*b**3 - 9*cos(c + d*x)*sin(c + d*x)*a**2*b + 4*sin(c + d*x)**3*a** 3 + 18*sin(c + d*x)**3*a*b**2 - 6*sin(c + d*x)*a**3 - 18*sin(c + d*x)*a*b* *2)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))