Integrand size = 21, antiderivative size = 65 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {(4 A+5 C) \tan (c+d x)}{5 d}+\frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 A+5 C) \tan ^3(c+d x)}{15 d} \] Output:
1/5*(4*A+5*C)*tan(d*x+c)/d+1/5*A*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(4*A+5*C)* tan(d*x+c)^3/d
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {C \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d}+\frac {A \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
Output:
(C*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d + (A*(Tan[c + d*x] + (2*Tan[c + d* x]^3)/3 + Tan[c + d*x]^5/5))/d
Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3491, 3042, 4254, 2009}
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.
\(\displaystyle \int \sec ^6(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle \frac {1}{5} (4 A+5 C) \int \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} (4 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {A \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {A \tan (c+d x) \sec ^4(c+d x)}{5 d}-\frac {(4 A+5 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {A \tan (c+d x) \sec ^4(c+d x)}{5 d}-\frac {(4 A+5 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 d}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
Output:
(A*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) - ((4*A + 5*C)*(-Tan[c + d*x] - Tan[ c + d*x]^3/3))/(5*d)
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
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]
Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
default | \(\frac {-A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
parts | \(-\frac {A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(60\) |
parallelrisch | \(\frac {\left (40 A +50 C \right ) \sin \left (3 d x +3 c \right )+\left (8 A +10 C \right ) \sin \left (5 d x +5 c \right )+80 \left (A +\frac {C}{2}\right ) \sin \left (d x +c \right )}{15 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(85\) |
risch | \(\frac {4 i \left (15 C \,{\mathrm e}^{6 i \left (d x +c \right )}+40 A \,{\mathrm e}^{4 i \left (d x +c \right )}+35 C \,{\mathrm e}^{4 i \left (d x +c \right )}+20 A \,{\mathrm e}^{2 i \left (d x +c \right )}+25 C \,{\mathrm e}^{2 i \left (d x +c \right )}+4 A +5 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(87\) |
norman | \(\frac {-\frac {4 \left (A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {4 \left (A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {2 \left (11 A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}-\frac {2 \left (11 A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{5 d}-\frac {8 \left (19 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}\) | \(180\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method=_RETURNVERBOSE)
Output:
1/d*(-A*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-C*(-2/3-1/3* sec(d*x+c)^2)*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {{\left (2 \, {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, A\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")
Output:
1/15*(2*(4*A + 5*C)*cos(d*x + c)^4 + (4*A + 5*C)*cos(d*x + c)^2 + 3*A)*sin (d*x + c)/(d*cos(d*x + c)^5)
Timed out. \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 \, A \tan \left (d x + c\right )^{5} + 5 \, {\left (2 \, A + C\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")
Output:
1/15*(3*A*tan(d*x + c)^5 + 5*(2*A + C)*tan(d*x + c)^3 + 15*(A + C)*tan(d*x + c))/d
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 \, A \tan \left (d x + c\right )^{5} + 10 \, A \tan \left (d x + c\right )^{3} + 5 \, C \tan \left (d x + c\right )^{3} + 15 \, A \tan \left (d x + c\right ) + 15 \, C \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")
Output:
1/15*(3*A*tan(d*x + c)^5 + 10*A*tan(d*x + c)^3 + 5*C*tan(d*x + c)^3 + 15*A *tan(d*x + c) + 15*C*tan(d*x + c))/d
Time = 40.89 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\left (\frac {2\,A}{3}+\frac {C}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\right )}{d} \] Input:
int((A + C*cos(c + d*x)^2)/cos(c + d*x)^6,x)
Output:
((A*tan(c + d*x)^5)/5 + tan(c + d*x)*(A + C) + tan(c + d*x)^3*((2*A)/3 + C /3))/d
Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\sin \left (d x +c \right ) \left (8 \sin \left (d x +c \right )^{4} a +10 \sin \left (d x +c \right )^{4} c -20 \sin \left (d x +c \right )^{2} a -25 \sin \left (d x +c \right )^{2} c +15 a +15 c \right )}{15 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
Output:
(sin(c + d*x)*(8*sin(c + d*x)**4*a + 10*sin(c + d*x)**4*c - 20*sin(c + d*x )**2*a - 25*sin(c + d*x)**2*c + 15*a + 15*c))/(15*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))