Integrand size = 23, antiderivative size = 191 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {16384 a^6 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac {4096 a^5 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {512 a^4 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {128 a^3 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}-\frac {8 a^2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 d} \] Output:
-16384/45045*a^6*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-4096/9009*a^5*cos(d *x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-512/1287*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+ c))^(1/2)-128/429*a^3*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d-8/39*a^2*cos(d *x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-2/15*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2 )/d
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.48 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {2 a^3 \cos ^5(c+d x) \sqrt {a (1+\sin (c+d x))} \left (41735+81815 \sin (c+d x)+86870 \sin ^2(c+d x)+55230 \sin ^3(c+d x)+19635 \sin ^4(c+d x)+3003 \sin ^5(c+d x)\right )}{45045 d (1+\sin (c+d x))^3} \] Input:
Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]
Output:
(-2*a^3*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])]*(41735 + 81815*Sin[c + d *x] + 86870*Sin[c + d*x]^2 + 55230*Sin[c + d*x]^3 + 19635*Sin[c + d*x]^4 + 3003*Sin[c + d*x]^5))/(45045*d*(1 + Sin[c + d*x])^3)
Time = 1.00 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3153, 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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^4 (a \sin (c+d x)+a)^{7/2}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {4}{3} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)^{5/2}dx-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)^{5/2}dx-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {4}{3} a \left (\frac {16}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\right )-\frac {2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(7/2),x]
Output:
(-2*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2))/(15*d) + (4*a*((-2*a*Cos[ c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d) + (16*a*((-2*a*Cos[c + d*x]^ 5*Sqrt[a + a*Sin[c + d*x]])/(11*d) + (12*a*((-2*a*Cos[c + d*x]^5)/(9*d*Sqr t[a + a*Sin[c + d*x]]) + (8*a*((-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))))/9 ))/11))/13))/3
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 = 24.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.51
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{4} \left (\sin \left (d x +c \right )-1\right )^{3} \left (3003 \sin \left (d x +c \right )^{5}+19635 \sin \left (d x +c \right )^{4}+55230 \sin \left (d x +c \right )^{3}+86870 \sin \left (d x +c \right )^{2}+81815 \sin \left (d x +c \right )+41735\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(97\) |
Input:
int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
2/45045*(1+sin(d*x+c))*a^4*(sin(d*x+c)-1)^3*(3003*sin(d*x+c)^5+19635*sin(d *x+c)^4+55230*sin(d*x+c)^3+86870*sin(d*x+c)^2+81815*sin(d*x+c)+41735)/cos( d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.28 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (3003 \, a^{3} \cos \left (d x + c\right )^{8} + 13629 \, a^{3} \cos \left (d x + c\right )^{7} - 17346 \, a^{3} \cos \left (d x + c\right )^{6} - 36932 \, a^{3} \cos \left (d x + c\right )^{5} + 1280 \, a^{3} \cos \left (d x + c\right )^{4} - 2048 \, a^{3} \cos \left (d x + c\right )^{3} + 4096 \, a^{3} \cos \left (d x + c\right )^{2} - 16384 \, a^{3} \cos \left (d x + c\right ) - 32768 \, a^{3} + {\left (3003 \, a^{3} \cos \left (d x + c\right )^{7} - 10626 \, a^{3} \cos \left (d x + c\right )^{6} - 27972 \, a^{3} \cos \left (d x + c\right )^{5} + 8960 \, a^{3} \cos \left (d x + c\right )^{4} + 10240 \, a^{3} \cos \left (d x + c\right )^{3} + 12288 \, a^{3} \cos \left (d x + c\right )^{2} + 16384 \, a^{3} \cos \left (d x + c\right ) + 32768 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")
Output:
2/45045*(3003*a^3*cos(d*x + c)^8 + 13629*a^3*cos(d*x + c)^7 - 17346*a^3*co s(d*x + c)^6 - 36932*a^3*cos(d*x + c)^5 + 1280*a^3*cos(d*x + c)^4 - 2048*a ^3*cos(d*x + c)^3 + 4096*a^3*cos(d*x + c)^2 - 16384*a^3*cos(d*x + c) - 327 68*a^3 + (3003*a^3*cos(d*x + c)^7 - 10626*a^3*cos(d*x + c)^6 - 27972*a^3*c os(d*x + c)^5 + 8960*a^3*cos(d*x + c)^4 + 10240*a^3*cos(d*x + c)^3 + 12288 *a^3*cos(d*x + c)^2 + 16384*a^3*cos(d*x + c) + 32768*a^3)*sin(d*x + c))*sq rt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
Timed out. \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{4} \,d x } \] Input:
integrate(cos(d*x+c)^4*(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)^4, x)
Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {256 \, \sqrt {2} {\left (3003 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 17325 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 40950 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 50050 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 32175 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9009 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{45045 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")
Output:
-256/45045*sqrt(2)*(3003*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)^15 - 17325*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s in(-1/4*pi + 1/2*d*x + 1/2*c)^13 + 40950*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 - 50050*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 + 32175*a^3*sgn(cos(-1/4*p i + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 9009*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)*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \] Input:
int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(7/2),x)
Output:
int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(7/2), x)
\[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{3}d x +3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) \] Input:
int(cos(d*x+c)^4*(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)**4*sin(c + d*x)**3,x ) + 3*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4*sin(c + d*x)**2,x) + 3*in t(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4*sin(c + d*x),x) + int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4,x))