Integrand size = 38, antiderivative size = 61 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {(a B+2 b C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(b B+a C) \tan (c+d x)}{d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*(B*a+2*C*b)*arctanh(sin(d*x+c))/d+(B*b+C*a)*tan(d*x+c)/d+1/2*a*B*sec(d *x+c)*tan(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {b C \coth ^{-1}(\sin (c+d x))}{d}+\frac {a B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b B \tan (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
(b*C*ArcCoth[Sin[c + d*x]])/d + (a*B*ArcTanh[Sin[c + d*x]])/(2*d) + (b*B*T an[c + d*x])/d + (a*C*Tan[c + d*x])/d + (a*B*Sec[c + d*x]*Tan[c + d*x])/(2 *d)
Time = 0.65 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3508, 3042, 3447, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x)) (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \sec ^3(c+d x) \left ((a C+b B) \cos (c+d x)+a B+b C \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a C+b B) \sin \left (c+d x+\frac {\pi }{2}\right )+a B+b C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{2} \int (2 (b B+a C)+(a B+2 b C) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 (b B+a C)+(a B+2 b C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{2} \left (2 (a C+b B) \int \sec ^2(c+d x)dx+(a B+2 b C) \int \sec (c+d x)dx\right )+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left ((a B+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{2} \left ((a B+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 (a C+b B) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{2} \left ((a B+2 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 (a C+b B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {(a B+2 b C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 (a C+b B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^ 4,x]
Output:
(a*B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((a*B + 2*b*C)*ArcTanh[Sin[c + d* x]])/d + (2*(b*B + a*C)*Tan[c + d*x])/d)/2
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_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 = 0.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23
| method | result | size |
| derivativedivides | \(\frac {C a \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B b \tan \left (d x +c \right )}{d}\) | \(75\) |
| default | \(\frac {C a \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B b \tan \left (d x +c \right )}{d}\) | \(75\) |
| parts | \(\frac {\left (B b +C a \right ) \tan \left (d x +c \right )}{d}+\frac {B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(76\) |
| parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 B b +2 C a \right ) \sin \left (2 d x +2 c \right )+2 B a \sin \left (d x +c \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(114\) |
| risch | \(-\frac {i \left (B a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-B a \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -2 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C b}{d}-\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C b}{d}\) | \(160\) |
| norman | \(\frac {\frac {\left (B a -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {\left (3 B a -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {2 \left (B a -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (B a +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (B a +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (3 B a +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (B a +2 C b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(250\) |
Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_ RETURNVERBOSE)
Output:
1/d*(C*a*tan(d*x+c)+B*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d *x+c)))+C*b*ln(sec(d*x+c)+tan(d*x+c))+B*b*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {{\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a + 2 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a lgorithm="fricas")
Output:
1/4*((B*a + 2*C*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B*a + 2*C*b)*co s(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a + 2*(C*a + B*b)*cos(d*x + c)) *sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.56 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, C b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a \tan \left (d x + c\right ) - 4 \, B b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a lgorithm="maxima")
Output:
-1/4*(B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l og(sin(d*x + c) - 1)) - 2*C*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*C*a*tan(d*x + c) - 4*B*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (57) = 114\).
Time = 0.14 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.48 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {{\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a lgorithm="giac")
Output:
1/2*((B*a + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (B*a + 2*C*b)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan (1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1/2 *c) + 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^2 - 1)^2)/d
Time = 1.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,a+2\,B\,b+2\,C\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,b-B\,a+2\,C\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B\,a+2\,C\,b\right )}{d} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x )^4,x)
Output:
(tan(c/2 + (d*x)/2)*(B*a + 2*B*b + 2*C*a) - tan(c/2 + (d*x)/2)^3*(2*B*b - B*a + 2*C*a))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (a tanh(tan(c/2 + (d*x)/2))*(B*a + 2*C*b))/d
Time = 0.17 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.59 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a c -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b c -\sin \left (d x +c \right ) a b}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)*a*c - 2*cos(c + d*x)*sin(c + d*x)*b**2 - l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b - 2*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*b*c + log(tan((c + d*x)/2) - 1)*a*b + 2*log(tan((c + d*x) /2) - 1)*b*c + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b + 2*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2*b*c - log(tan((c + d*x)/2) + 1)*a*b - 2*l og(tan((c + d*x)/2) + 1)*b*c - sin(c + d*x)*a*b)/(2*d*(sin(c + d*x)**2 - 1 ))