Integrand size = 23, antiderivative size = 159 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {4096 a^5 \cos ^3(c+d x)}{3465 d (a+a \sin (c+d x))^{3/2}}-\frac {1024 a^4 \cos ^3(c+d x)}{1155 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{231 d}-\frac {32 a^2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{99 d}-\frac {2 a \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 d} \] Output:
-4096/3465*a^5*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-1024/1155*a^4*cos(d*x +c)^3/d/(a+a*sin(d*x+c))^(1/2)-128/231*a^3*cos(d*x+c)^3*(a+a*sin(d*x+c))^( 1/2)/d-32/99*a^2*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/11*a*cos(d*x+c)^3 *(a+a*sin(d*x+c))^(5/2)/d
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.52 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {2 a^3 \cos ^3(c+d x) \sqrt {a (1+\sin (c+d x))} \left (5419+6396 \sin (c+d x)+4530 \sin ^2(c+d x)+1820 \sin ^3(c+d x)+315 \sin ^4(c+d x)\right )}{3465 d (1+\sin (c+d x))^2} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^(7/2),x]
Output:
(-2*a^3*Cos[c + d*x]^3*Sqrt[a*(1 + Sin[c + d*x])]*(5419 + 6396*Sin[c + d*x ] + 4530*Sin[c + d*x]^2 + 1820*Sin[c + d*x]^3 + 315*Sin[c + d*x]^4))/(3465 *d*(1 + Sin[c + d*x])^2)
Time = 0.81 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3153, 3042, 3153, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^2 (a \sin (c+d x)+a)^{7/2}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {16}{11} a \int \cos ^2(c+d x) (\sin (c+d x) a+a)^{5/2}dx-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{11} a \int \cos (c+d x)^2 (\sin (c+d x) a+a)^{5/2}dx-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \int \cos ^2(c+d x) (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \int \cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \left (\frac {8}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \left (\frac {8}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {16}{11} a \left (\frac {4}{3} a \left (\frac {8}{7} a \left (-\frac {8 a^2 \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\right )-\frac {2 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^(7/2),x]
Output:
(-2*a*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*d) + (16*a*((-2*a*Cos [c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(9*d) + (4*a*((-2*a*Cos[c + d*x]^3 *Sqrt[a + a*Sin[c + d*x]])/(7*d) + (8*a*((-8*a^2*Cos[c + d*x]^3)/(15*d*(a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + d* x]])))/7))/3))/11
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 = 1.73 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.55
| method | result | size |
| default | \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{4} \left (\sin \left (d x +c \right )-1\right )^{2} \left (315 \sin \left (d x +c \right )^{4}+1820 \sin \left (d x +c \right )^{3}+4530 \sin \left (d x +c \right )^{2}+6396 \sin \left (d x +c \right )+5419\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(87\) |
Input:
int(cos(d*x+c)^2*(a+a*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/3465*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+1820*sin(d*x +c)^3+4530*sin(d*x+c)^2+6396*sin(d*x+c)+5419)/cos(d*x+c)/(a+a*sin(d*x+c))^ (1/2)/d
Time = 0.08 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.21 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (315 \, a^{3} \cos \left (d x + c\right )^{6} + 1505 \, a^{3} \cos \left (d x + c\right )^{5} - 2150 \, a^{3} \cos \left (d x + c\right )^{4} - 4876 \, a^{3} \cos \left (d x + c\right )^{3} + 512 \, a^{3} \cos \left (d x + c\right )^{2} - 2048 \, a^{3} \cos \left (d x + c\right ) - 4096 \, a^{3} + {\left (315 \, a^{3} \cos \left (d x + c\right )^{5} - 1190 \, a^{3} \cos \left (d x + c\right )^{4} - 3340 \, a^{3} \cos \left (d x + c\right )^{3} + 1536 \, a^{3} \cos \left (d x + c\right )^{2} + 2048 \, a^{3} \cos \left (d x + c\right ) + 4096 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")
Output:
2/3465*(315*a^3*cos(d*x + c)^6 + 1505*a^3*cos(d*x + c)^5 - 2150*a^3*cos(d* x + c)^4 - 4876*a^3*cos(d*x + c)^3 + 512*a^3*cos(d*x + c)^2 - 2048*a^3*cos (d*x + c) - 4096*a^3 + (315*a^3*cos(d*x + c)^5 - 1190*a^3*cos(d*x + c)^4 - 3340*a^3*cos(d*x + c)^3 + 1536*a^3*cos(d*x + c)^2 + 2048*a^3*cos(d*x + c) + 4096*a^3)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*si n(d*x + c) + d)
Timed out. \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(7/2)*cos(d*x + c)^2, x)
Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {64 \, \sqrt {2} {\left (315 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1540 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2970 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2772 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1155 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{3465 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")
Output:
64/3465*sqrt(2)*(315*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 1540*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1 /4*pi + 1/2*d*x + 1/2*c)^9 + 2970*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))* sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 2772*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/ 2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 1155*a^3*sgn(cos(-1/4*pi + 1/2*d* x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)*sqrt(a)/d
Timed out. \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \] Input:
int(cos(c + d*x)^2*(a + a*sin(c + d*x))^(7/2),x)
Output:
int(cos(c + d*x)^2*(a + a*sin(c + d*x))^(7/2), x)
\[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3}d x +3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}d x \right ) \] Input:
int(cos(d*x+c)^2*(a+a*sin(d*x+c))^(7/2),x)
Output:
sqrt(a)*a**3*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**3,x ) + 3*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**2,x) + 3*in t(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x),x) + int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2,x))