Integrand size = 21, antiderivative size = 63 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {5 a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \] Output:
5/2*a^3*x+4*a^3*sin(d*x+c)/d+3/2*a^3*cos(d*x+c)*sin(d*x+c)/d-1/3*a^3*sin(d *x+c)^3/d
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 (30 c+30 d x+45 \sin (c+d x)+9 \sin (2 (c+d x))+\sin (3 (c+d x)))}{12 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3,x]
Output:
(a^3*(30*c + 30*d*x + 45*Sin[c + d*x] + 9*Sin[2*(c + d*x)] + Sin[3*(c + d* x)]))/(12*d)
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4278, 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.
\(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4278 |
\(\displaystyle \int \left (a^3 \cos ^3(c+d x)+3 a^3 \cos ^2(c+d x)+3 a^3 \cos (c+d x)+a^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {5 a^3 x}{2}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3,x]
Output:
(5*a^3*x)/2 + (4*a^3*Sin[c + d*x])/d + (3*a^3*Cos[c + d*x]*Sin[c + d*x])/( 2*d) - (a^3*Sin[c + d*x]^3)/(3*d)
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Time = 0.66 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(\frac {a^{3} \left (30 d x +45 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )+9 \sin \left (2 d x +2 c \right )\right )}{12 d}\) | \(42\) |
risch | \(\frac {5 a^{3} x}{2}+\frac {15 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\) | \(56\) |
derivativedivides | \(\frac {a^{3} \left (d x +c \right )+3 a^{3} \sin \left (d x +c \right )+3 a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(74\) |
default | \(\frac {a^{3} \left (d x +c \right )+3 a^{3} \sin \left (d x +c \right )+3 a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(74\) |
norman | \(\frac {\frac {5 a^{3} x}{2}+\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {26 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {32 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {10 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {5 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+\frac {5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(217\) |
Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/12*a^3*(30*d*x+45*sin(d*x+c)+sin(3*d*x+3*c)+9*sin(2*d*x+2*c))/d
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 22 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/6*(15*a^3*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 22*a^3)*sin (d*x + c))/d
\[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**3,x)
Output:
a**3*(Integral(3*cos(c + d*x)**3*sec(c + d*x), x) + Integral(3*cos(c + d*x )**3*sec(c + d*x)**2, x) + Integral(cos(c + d*x)**3*sec(c + d*x)**3, x) + Integral(cos(c + d*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \, {\left (d x + c\right )} a^{3} - 36 \, a^{3} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d* x + 2*c))*a^3 - 12*(d*x + c)*a^3 - 36*a^3*sin(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.27 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
1/6*(15*(d*x + c)*a^3 + 2*(15*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^3*tan(1/2* d*x + 1/2*c)^3 + 33*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1) ^3)/d
Time = 10.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {5\,a^3\,x}{2}+\frac {11\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^3*(a + a/cos(c + d*x))^3,x)
Output:
(5*a^3*x)/2 + (11*a^3*sin(c + d*x))/(3*d) + (a^3*cos(c + d*x)^2*sin(c + d* x))/(3*d) + (3*a^3*cos(c + d*x)*sin(c + d*x))/(2*d)
Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^{3} \left (9 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )^{3}+24 \sin \left (d x +c \right )+15 d x \right )}{6 d} \] Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^3,x)
Output:
(a**3*(9*cos(c + d*x)*sin(c + d*x) - 2*sin(c + d*x)**3 + 24*sin(c + d*x) + 15*d*x))/(6*d)