Integrand size = 23, antiderivative size = 95 \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {64 a^3 \cos ^{11}(c+d x)}{2145 d (a+a \sin (c+d x))^{11/2}}-\frac {16 a^2 \cos ^{11}(c+d x)}{195 d (a+a \sin (c+d x))^{9/2}}-\frac {2 a \cos ^{11}(c+d x)}{15 d (a+a \sin (c+d x))^{7/2}} \] Output:
-64/2145*a^3*cos(d*x+c)^11/d/(a+a*sin(d*x+c))^(11/2)-16/195*a^2*cos(d*x+c) ^11/d/(a+a*sin(d*x+c))^(9/2)-2/15*a*cos(d*x+c)^11/d/(a+a*sin(d*x+c))^(7/2)
Time = 0.74 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \cos ^{11}(c+d x) \left (263+374 \sin (c+d x)+143 \sin ^2(c+d x)\right )}{2145 d (1+\sin (c+d x))^3 (a (1+\sin (c+d x)))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^10/(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*Cos[c + d*x]^11*(263 + 374*Sin[c + d*x] + 143*Sin[c + d*x]^2))/(2145*d *(1 + Sin[c + d*x])^3*(a*(1 + Sin[c + d*x]))^(5/2))
Time = 0.50 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3153, 3042, 3153, 3042, 3152}
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.
\(\displaystyle \int \frac {\cos ^{10}(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^{10}}{(a \sin (c+d x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {8}{15} a \int \frac {\cos ^{10}(c+d x)}{(\sin (c+d x) a+a)^{7/2}}dx-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{15} a \int \frac {\cos (c+d x)^{10}}{(\sin (c+d x) a+a)^{7/2}}dx-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {8}{15} a \left (\frac {4}{13} a \int \frac {\cos ^{10}(c+d x)}{(\sin (c+d x) a+a)^{9/2}}dx-\frac {2 a \cos ^{11}(c+d x)}{13 d (a \sin (c+d x)+a)^{9/2}}\right )-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{15} a \left (\frac {4}{13} a \int \frac {\cos (c+d x)^{10}}{(\sin (c+d x) a+a)^{9/2}}dx-\frac {2 a \cos ^{11}(c+d x)}{13 d (a \sin (c+d x)+a)^{9/2}}\right )-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {8}{15} a \left (-\frac {8 a^2 \cos ^{11}(c+d x)}{143 d (a \sin (c+d x)+a)^{11/2}}-\frac {2 a \cos ^{11}(c+d x)}{13 d (a \sin (c+d x)+a)^{9/2}}\right )-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}}\) |
Input:
Int[Cos[c + d*x]^10/(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*a*Cos[c + d*x]^11)/(15*d*(a + a*Sin[c + d*x])^(7/2)) + (8*a*((-8*a^2*C os[c + d*x]^11)/(143*d*(a + a*Sin[c + d*x])^(11/2)) - (2*a*Cos[c + d*x]^11 )/(13*d*(a + a*Sin[c + d*x])^(9/2))))/15
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Time = 2.84 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{6} \left (143 \sin \left (d x +c \right )^{2}+374 \sin \left (d x +c \right )+263\right )}{2145 a^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(67\) |
Input:
int(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/2145/a^2*(1+sin(d*x+c))*(sin(d*x+c)-1)^6*(143*sin(d*x+c)^2+374*sin(d*x+ c)+263)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (83) = 166\).
Time = 0.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.12 \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (143 \, \cos \left (d x + c\right )^{8} - 341 \, \cos \left (d x + c\right )^{7} - 736 \, \cos \left (d x + c\right )^{6} + 28 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{4} + 64 \, \cos \left (d x + c\right )^{3} - 128 \, \cos \left (d x + c\right )^{2} + {\left (143 \, \cos \left (d x + c\right )^{7} + 484 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 280 \, \cos \left (d x + c\right )^{4} - 320 \, \cos \left (d x + c\right )^{3} - 384 \, \cos \left (d x + c\right )^{2} - 512 \, \cos \left (d x + c\right ) - 1024\right )} \sin \left (d x + c\right ) + 512 \, \cos \left (d x + c\right ) + 1024\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2145 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/2145*(143*cos(d*x + c)^8 - 341*cos(d*x + c)^7 - 736*cos(d*x + c)^6 + 28 *cos(d*x + c)^5 - 40*cos(d*x + c)^4 + 64*cos(d*x + c)^3 - 128*cos(d*x + c) ^2 + (143*cos(d*x + c)^7 + 484*cos(d*x + c)^6 - 252*cos(d*x + c)^5 - 280*c os(d*x + c)^4 - 320*cos(d*x + c)^3 - 384*cos(d*x + c)^2 - 512*cos(d*x + c) - 1024)*sin(d*x + c) + 512*cos(d*x + c) + 1024)*sqrt(a*sin(d*x + c) + a)/ (a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)
Timed out. \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**10/(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{10}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^10/(a*sin(d*x + c) + a)^(5/2), x)
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {256 \, {\left (143 \, \sqrt {2} \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 330 \, \sqrt {2} \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 195 \, \sqrt {2} \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11}\right )}}{2145 \, a^{3} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:
integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
256/2145*(143*sqrt(2)*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^15 - 330*sqrt (2)*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^13 + 195*sqrt(2)*sqrt(a)*sin(-1 /4*pi + 1/2*d*x + 1/2*c)^11)/(a^3*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{10}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(cos(c + d*x)^10/(a + a*sin(c + d*x))^(5/2),x)
Output:
int(cos(c + d*x)^10/(a + a*sin(c + d*x))^(5/2), x)
\[ \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{10}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**10)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3