Integrand size = 21, antiderivative size = 50 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(A+C) \sin (c+d x)}{d}-\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {A \sin ^5(c+d x)}{5 d} \] Output:
(A+C)*sin(d*x+c)/d-1/3*(2*A+C)*sin(d*x+c)^3/d+1/5*A*sin(d*x+c)^5/d
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.42 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \sin (c+d x)}{d}+\frac {C \sin (c+d x)}{d}-\frac {2 A \sin ^3(c+d x)}{3 d}-\frac {C \sin ^3(c+d x)}{3 d}+\frac {A \sin ^5(c+d x)}{5 d} \] Input:
Integrate[Cos[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Sin[c + d*x])/d + (C*Sin[c + d*x])/d - (2*A*Sin[c + d*x]^3)/(3*d) - (C* Sin[c + d*x]^3)/(3*d) + (A*Sin[c + d*x]^5)/(5*d)
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4532, 3042, 3492, 290, 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.
\(\displaystyle \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 4532 |
\(\displaystyle \int \cos ^3(c+d x) \left (A \cos ^2(c+d x)+C\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C\right )dx\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle -\frac {\int \left (1-\sin ^2(c+d x)\right ) \left (-A \sin ^2(c+d x)+A+C\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {\int \left (A \sin ^4(c+d x)-(2 A+C) \sin ^2(c+d x)+A \left (\frac {C}{A}+1\right )\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{3} (2 A+C) \sin ^3(c+d x)-(A+C) \sin (c+d x)-\frac {1}{5} A \sin ^5(c+d x)}{d}\) |
Input:
Int[Cos[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]
Output:
-((-((A + C)*Sin[c + d*x]) + ((2*A + C)*Sin[c + d*x]^3)/3 - (A*Sin[c + d*x ]^5)/5)/d)
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(\frac {\left (25 A +20 C \right ) \sin \left (3 d x +3 c \right )+3 A \sin \left (5 d x +5 c \right )+150 \sin \left (d x +c \right ) \left (A +\frac {6 C}{5}\right )}{240 d}\) | \(49\) |
derivativedivides | \(\frac {\frac {A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(54\) |
default | \(\frac {\frac {A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(54\) |
risch | \(\frac {5 A \sin \left (d x +c \right )}{8 d}+\frac {3 C \sin \left (d x +c \right )}{4 d}+\frac {A \sin \left (5 d x +5 c \right )}{80 d}+\frac {5 A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) C}{12 d}\) | \(71\) |
norman | \(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 \left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {4 \left (19 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {4 \left (19 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(157\) |
Input:
int(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/240*((25*A+20*C)*sin(3*d*x+3*c)+3*A*sin(5*d*x+5*c)+150*sin(d*x+c)*(A+6/5 *C))/d
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, A \cos \left (d x + c\right )^{4} + {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, A + 10 \, C\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/15*(3*A*cos(d*x + c)^4 + (4*A + 5*C)*cos(d*x + c)^2 + 8*A + 10*C)*sin(d* x + c)/d
\[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**5*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, A \sin \left (d x + c\right )^{5} - 5 \, {\left (2 \, A + C\right )} \sin \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/15*(3*A*sin(d*x + c)^5 - 5*(2*A + C)*sin(d*x + c)^3 + 15*(A + C)*sin(d*x + c))/d
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, A \sin \left (d x + c\right )^{5} - 10 \, A \sin \left (d x + c\right )^{3} - 5 \, C \sin \left (d x + c\right )^{3} + 15 \, A \sin \left (d x + c\right ) + 15 \, C \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/15*(3*A*sin(d*x + c)^5 - 10*A*sin(d*x + c)^3 - 5*C*sin(d*x + c)^3 + 15*A *sin(d*x + c) + 15*C*sin(d*x + c))/d
Time = 11.46 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {A\,{\sin \left (c+d\,x\right )}^5}{5}+\left (-\frac {2\,A}{3}-\frac {C}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (A+C\right )\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2),x)
Output:
((A*sin(c + d*x)^5)/5 + sin(c + d*x)*(A + C) - sin(c + d*x)^3*((2*A)/3 + C /3))/d
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.02 \[ \int \cos ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{4} a -10 \sin \left (d x +c \right )^{2} a -5 \sin \left (d x +c \right )^{2} c +15 a +15 c \right )}{15 d} \] Input:
int(cos(d*x+c)^5*(A+C*sec(d*x+c)^2),x)
Output:
(sin(c + d*x)*(3*sin(c + d*x)**4*a - 10*sin(c + d*x)**2*a - 5*sin(c + d*x) **2*c + 15*a + 15*c))/(15*d)