Integrand size = 23, antiderivative size = 97 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {16 (a+a \sin (c+d x))^{3/2}}{3 a^4 d}-\frac {24 (a+a \sin (c+d x))^{5/2}}{5 a^5 d}+\frac {12 (a+a \sin (c+d x))^{7/2}}{7 a^6 d}-\frac {2 (a+a \sin (c+d x))^{9/2}}{9 a^7 d} \] Output:
16/3*(a+a*sin(d*x+c))^(3/2)/a^4/d-24/5*(a+a*sin(d*x+c))^(5/2)/a^5/d+12/7*( a+a*sin(d*x+c))^(7/2)/a^6/d-2/9*(a+a*sin(d*x+c))^(9/2)/a^7/d
Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 (a (1+\sin (c+d x)))^{3/2} \left (-319+321 \sin (c+d x)-165 \sin ^2(c+d x)+35 \sin ^3(c+d x)\right )}{315 a^4 d} \] Input:
Integrate[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*(a*(1 + Sin[c + d*x]))^(3/2)*(-319 + 321*Sin[c + d*x] - 165*Sin[c + d* x]^2 + 35*Sin[c + d*x]^3))/(315*a^4*d)
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3146, 53, 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.
\(\displaystyle \int \frac {\cos ^7(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{(a \sin (c+d x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 \sqrt {\sin (c+d x) a+a}d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (-(\sin (c+d x) a+a)^{7/2}+6 a (\sin (c+d x) a+a)^{5/2}-12 a^2 (\sin (c+d x) a+a)^{3/2}+8 a^3 \sqrt {\sin (c+d x) a+a}\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {16}{3} a^3 (a \sin (c+d x)+a)^{3/2}-\frac {24}{5} a^2 (a \sin (c+d x)+a)^{5/2}-\frac {2}{9} (a \sin (c+d x)+a)^{9/2}+\frac {12}{7} a (a \sin (c+d x)+a)^{7/2}}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^7/(a + a*Sin[c + d*x])^(5/2),x]
Output:
((16*a^3*(a + a*Sin[c + d*x])^(3/2))/3 - (24*a^2*(a + a*Sin[c + d*x])^(5/2 ))/5 + (12*a*(a + a*Sin[c + d*x])^(7/2))/7 - (2*(a + a*Sin[c + d*x])^(9/2) )/9)/(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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 = 0.75 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (35 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-165 \cos \left (d x +c \right )^{2}-356 \sin \left (d x +c \right )+484\right )}{315 a^{4} d}\) | \(57\) |
Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/315/a^4*(a+a*sin(d*x+c))^(3/2)*(35*cos(d*x+c)^2*sin(d*x+c)-165*cos(d*x+c )^2-356*sin(d*x+c)+484)/d
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{4} - 226 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (65 \, \cos \left (d x + c\right )^{2} - 64\right )} \sin \left (d x + c\right ) - 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/315*(35*cos(d*x + c)^4 - 226*cos(d*x + c)^2 + 2*(65*cos(d*x + c)^2 - 64 )*sin(d*x + c) - 128)*sqrt(a*sin(d*x + c) + a)/(a^3*d)
Timed out. \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**7/(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (35 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 270 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 756 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 840 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}\right )}}{315 \, a^{7} d} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
-2/315*(35*(a*sin(d*x + c) + a)^(9/2) - 270*(a*sin(d*x + c) + a)^(7/2)*a + 756*(a*sin(d*x + c) + a)^(5/2)*a^2 - 840*(a*sin(d*x + c) + a)^(3/2)*a^3)/ (a^7*d)
Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {32 \, {\left (35 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 135 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 189 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{315 \, a^{3} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:
integrate(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
-32/315*(35*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 135*sqrt(2) *sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 189*sqrt(2)*sqrt(a)*cos(-1/4*p i + 1/2*d*x + 1/2*c)^5 - 105*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c )^3)/(a^3*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^7}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(cos(c + d*x)^7/(a + a*sin(c + d*x))^(5/2),x)
Output:
int(cos(c + d*x)^7/(a + a*sin(c + d*x))^(5/2), x)
\[ \int \frac {\cos ^7(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(cos(d*x+c)^7/(a+a*sin(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**7)/(sin(c + d*x)**3 + 3 *sin(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3