Integrand size = 21, antiderivative size = 68 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(a-a \sin (c+d x))^4}{a^5 d}+\frac {4 (a-a \sin (c+d x))^5}{5 a^6 d}-\frac {(a-a \sin (c+d x))^6}{6 a^7 d} \] Output:
-(a-a*sin(d*x+c))^4/a^5/d+4/5*(a-a*sin(d*x+c))^5/a^6/d-1/6*(a-a*sin(d*x+c) )^6/a^7/d
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin (c+d x) \left (-30+15 \sin (c+d x)+20 \sin ^2(c+d x)-15 \sin ^3(c+d x)-6 \sin ^4(c+d x)+5 \sin ^5(c+d x)\right )}{30 a d} \] Input:
Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
-1/30*(Sin[c + d*x]*(-30 + 15*Sin[c + d*x] + 20*Sin[c + d*x]^2 - 15*Sin[c + d*x]^3 - 6*Sin[c + d*x]^4 + 5*Sin[c + d*x]^5))/(a*d)
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 49, 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.
\(\displaystyle \int \frac {\cos ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left ((a-a \sin (c+d x))^5-4 a (a-a \sin (c+d x))^4+4 a^2 (a-a \sin (c+d x))^3\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 (a-a \sin (c+d x))^4-\frac {1}{6} (a-a \sin (c+d x))^6+\frac {4}{5} a (a-a \sin (c+d x))^5}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
(-(a^2*(a - a*Sin[c + d*x])^4) + (4*a*(a - a*Sin[c + d*x])^5)/5 - (a - a*S in[c + d*x])^6/6)/(a^7*d)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 1.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(-\frac {\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{2}+\frac {2 \sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )}{d a}\) | \(68\) |
default | \(-\frac {\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{2}+\frac {2 \sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )}{d a}\) | \(68\) |
risch | \(\frac {5 \sin \left (d x +c \right )}{8 a d}+\frac {\cos \left (6 d x +6 c \right )}{192 a d}+\frac {\sin \left (5 d x +5 c \right )}{80 d a}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {5 \sin \left (3 d x +3 c \right )}{48 d a}+\frac {5 \cos \left (2 d x +2 c \right )}{64 a d}\) | \(101\) |
parallelrisch | \(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+78 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+78 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+15\right )}{15 d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} a}\) | \(137\) |
norman | \(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d a}+\frac {42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5 d a}+\frac {42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{5 d a}+\frac {196 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d a}+\frac {196 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{15 d a}+\frac {212 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d a}+\frac {212 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{15 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(257\) |
Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d/a*(1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5-1/2*sin(d*x+c)^4+2/3*sin(d*x+c)^ 3+1/2*sin(d*x+c)^2-sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \, \cos \left (d x + c\right )^{6} + 2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{30 \, a d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/30*(5*cos(d*x + c)^6 + 2*(3*cos(d*x + c)^4 + 4*cos(d*x + c)^2 + 8)*sin(d *x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (54) = 108\).
Time = 21.05 (sec) , antiderivative size = 1096, normalized size of antiderivative = 16.12 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**7/(a+a*sin(d*x+c)),x)
Output:
Piecewise((30*tan(c/2 + d*x/2)**11/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*t an(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/ 2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*tan(c/2 + d*x/2)**10/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 70*ta n(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)** 10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*t an(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 156*tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225 *a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 100*tan(c/2 + d*x/2)** 6/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan (c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)* *4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 156*tan(c/2 + d*x/2)**5/(15*a* d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d *x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90* a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 70*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*ta...
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{6} - 6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right )}{30 \, a d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/30*(5*sin(d*x + c)^6 - 6*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d* x + c)^3 + 15*sin(d*x + c)^2 - 30*sin(d*x + c))/(a*d)
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{6} - 6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right )}{30 \, a d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/30*(5*sin(d*x + c)^6 - 6*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d* x + c)^3 + 15*sin(d*x + c)^2 - 30*sin(d*x + c))/(a*d)
Time = 31.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {2\,{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{2\,a}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}}{d} \] Input:
int(cos(c + d*x)^7/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)/a - sin(c + d*x)^2/(2*a) - (2*sin(c + d*x)^3)/(3*a) + sin(c + d*x)^4/(2*a) + sin(c + d*x)^5/(5*a) - sin(c + d*x)^6/(6*a))/d
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (-5 \sin \left (d x +c \right )^{5}+6 \sin \left (d x +c \right )^{4}+15 \sin \left (d x +c \right )^{3}-20 \sin \left (d x +c \right )^{2}-15 \sin \left (d x +c \right )+30\right )}{30 a d} \] Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)*( - 5*sin(c + d*x)**5 + 6*sin(c + d*x)**4 + 15*sin(c + d*x)* *3 - 20*sin(c + d*x)**2 - 15*sin(c + d*x) + 30))/(30*a*d)