Integrand size = 21, antiderivative size = 47 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {(a-a \sin (c+d x))^4}{2 a^6 d}+\frac {(a-a \sin (c+d x))^5}{5 a^7 d} \] Output:
-1/2*(a-a*sin(d*x+c))^4/a^6/d+1/5*(a-a*sin(d*x+c))^5/a^7/d
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sin (c+d x) \left (-10+10 \sin (c+d x)-5 \sin ^3(c+d x)+2 \sin ^4(c+d x)\right )}{10 a^2 d} \] Input:
Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]
Output:
-1/10*(Sin[c + d*x]*(-10 + 10*Sin[c + d*x] - 5*Sin[c + d*x]^3 + 2*Sin[c + d*x]^4))/(a^2*d)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91, 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.
\(\displaystyle \int \frac {\cos ^7(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (2 a (a-a \sin (c+d x))^3-(a-a \sin (c+d x))^4\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} (a-a \sin (c+d x))^5-\frac {1}{2} a (a-a \sin (c+d x))^4}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]
Output:
(-1/2*(a*(a - a*Sin[c + d*x])^4) + (a - a*Sin[c + d*x])^5/5)/(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 = 1.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(-\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )}{d \,a^{2}}\) | \(46\) |
default | \(-\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )}{d \,a^{2}}\) | \(46\) |
parallelrisch | \(\frac {-25+20 \cos \left (2 d x +2 c \right )-\sin \left (5 d x +5 c \right )+5 \sin \left (3 d x +3 c \right )+5 \cos \left (4 d x +4 c \right )+70 \sin \left (d x +c \right )}{80 d \,a^{2}}\) | \(63\) |
risch | \(\frac {7 \sin \left (d x +c \right )}{8 a^{2} d}-\frac {\sin \left (5 d x +5 c \right )}{80 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right )}{16 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right )}{16 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right )}{4 d \,a^{2}}\) | \(84\) |
norman | \(\frac {-\frac {2}{3 a d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{17}}{3 d a}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d a}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 a d}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{3 d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{3 d a}-\frac {46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}-\frac {46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}-\frac {406 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d a}-\frac {406 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{15 d a}-\frac {518 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d a}-\frac {518 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{15 d a}-\frac {782 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{15 d a}-\frac {782 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(327\) |
Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/d/a^2*(1/5*sin(d*x+c)^5-1/2*sin(d*x+c)^4+sin(d*x+c)^2-sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {5 \, \cos \left (d x + c\right )^{4} - 2 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right )}{10 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/10*(5*cos(d*x + c)^4 - 2*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - 4)*sin(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1037 vs. \(2 (36) = 72\).
Time = 46.29 (sec) , antiderivative size = 1037, normalized size of antiderivative = 22.06 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**7/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((10*tan(c/2 + d*x/2)**9/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2 *d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 + d* x/2)**8/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 5 0*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*t an(c/2 + d*x/2)**2 + 5*a**2*d) + 40*tan(c/2 + d*x/2)**7/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)* *6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a** 2*d) - 20*tan(c/2 + d*x/2)**6/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*t an(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d *x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) + 28*tan(c/2 + d*x/2) **5/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a* *2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c /2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 + d*x/2)**4/(5*a**2*d*tan(c/2 + d* x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2)**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) + 40*tan(c/2 + d*x/2)**3/(5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c /2 + d*x/2)**8 + 50*a**2*d*tan(c/2 + d*x/2)**6 + 50*a**2*d*tan(c/2 + d*x/2 )**4 + 25*a**2*d*tan(c/2 + d*x/2)**2 + 5*a**2*d) - 20*tan(c/2 + d*x/2)**2/ (5*a**2*d*tan(c/2 + d*x/2)**10 + 25*a**2*d*tan(c/2 + d*x/2)**8 + 50*a**...
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right )}{10 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/10*(2*sin(d*x + c)^5 - 5*sin(d*x + c)^4 + 10*sin(d*x + c)^2 - 10*sin(d* x + c))/(a^2*d)
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right )}{10 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/10*(2*sin(d*x + c)^5 - 5*sin(d*x + c)^4 + 10*sin(d*x + c)^2 - 10*sin(d* x + c))/(a^2*d)
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{a^2}-\frac {{\sin \left (c+d\,x\right )}^2}{a^2}+\frac {{\sin \left (c+d\,x\right )}^4}{2\,a^2}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a^2}}{d} \] Input:
int(cos(c + d*x)^7/(a + a*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)/a^2 - sin(c + d*x)^2/a^2 + sin(c + d*x)^4/(2*a^2) - sin(c + d*x)^5/(5*a^2))/d
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (d x +c \right ) \left (-2 \sin \left (d x +c \right )^{4}+5 \sin \left (d x +c \right )^{3}-10 \sin \left (d x +c \right )+10\right )}{10 a^{2} d} \] Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)*( - 2*sin(c + d*x)**4 + 5*sin(c + d*x)**3 - 10*sin(c + d*x) + 10))/(10*a**2*d)