Integrand size = 19, antiderivative size = 54 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a x}{2}+\frac {a \sin (c+d x)}{d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \] Output:
1/2*a*x+a*sin(d*x+c)/d+1/2*a*cos(d*x+c)*sin(d*x+c)/d-1/3*a*sin(d*x+c)^3/d
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a (c+d x)}{2 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (2 (c+d x))}{4 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x]),x]
Output:
(a*(c + d*x))/(2*d) + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d) + (a*S in[2*(c + d*x)])/(4*d)
Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4274, 3042, 3113, 2009, 3115, 24}
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.
\(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle a \int \cos ^3(c+d x)dx+a \int \cos ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+a \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x]),x]
Output:
a*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (a*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Time = 0.70 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(\frac {a \left (6 d x +\sin \left (3 d x +3 c \right )+3 \sin \left (2 d x +2 c \right )+9 \sin \left (d x +c \right )\right )}{12 d}\) | \(40\) |
risch | \(\frac {a x}{2}+\frac {3 a \sin \left (d x +c \right )}{4 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(48\) |
derivativedivides | \(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(49\) |
default | \(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(49\) |
norman | \(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {a x}{2}+\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(115\) |
Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/12*a*(6*d*x+sin(3*d*x+3*c)+3*sin(2*d*x+2*c)+9*sin(d*x+c))/d
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {3 \, a d x + {\left (2 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 4 \, a\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/6*(3*a*d*x + (2*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + 4*a)*sin(d*x + c)) /d
\[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**3*(a+a*sec(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**3*sec(c + d*x), x) + Integral(cos(c + d*x)**3, x ))
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*a - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a)/d
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} a + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*(d*x + c)*a + 2*(3*a*tan(1/2*d*x + 1/2*c)^5 + 4*a*tan(1/2*d*x + 1/2 *c)^3 + 9*a*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 = 55, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a\,x}{2}+\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cos(c + d*x)^3*(a + a/cos(c + d*x)),x)
Output:
(a*x)/2 + (2*a*sin(c + d*x))/(3*d) + (a*cos(c + d*x)*sin(c + d*x))/(2*d) + (a*cos(c + d*x)^2*sin(c + d*x))/(3*d)
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \left (3 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )+3 d x \right )}{6 d} \] Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c)),x)
Output:
(a*(3*cos(c + d*x)*sin(c + d*x) - 2*sin(c + d*x)**3 + 6*sin(c + d*x) + 3*d *x))/(6*d)