Integrand size = 23, antiderivative size = 95 \[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {64 a^3 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}-\frac {16 a^2 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}} \] Output:
-64/315*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-16/63*a^2*cos(d*x+c)^5/d /(a+a*sin(d*x+c))^(3/2)-2/9*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \cos ^5(c+d x) \sqrt {a (1+\sin (c+d x))} \left (107+110 \sin (c+d x)+35 \sin ^2(c+d x)\right )}{315 d (1+\sin (c+d x))^3} \] Input:
Integrate[Cos[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
Output:
(-2*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])]*(107 + 110*Sin[c + d*x] + 35 *Sin[c + d*x]^2))/(315*d*(1 + Sin[c + d*x])^3)
Time = 0.48 (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.
| \(\displaystyle \int \cos ^4(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^4 \sqrt {a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
Input:
Int[Cos[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
Output:
(-2*a*Cos[c + d*x]^5)/(9*d*Sqrt[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
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 = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \sin \left (d x +c \right )^{2}+110 \sin \left (d x +c \right )+107\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(65\) |
Input:
int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/315*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+110*sin(d*x+c)+10 7)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.39 \[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} + 48 \, \cos \left (d x + c\right )^{2} + 64 \, \cos \left (d x + c\right ) + 128\right )} \sin \left (d x + c\right ) + 64 \, \cos \left (d x + c\right ) + 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-2/315*(35*cos(d*x + c)^5 - 5*cos(d*x + c)^4 + 8*cos(d*x + c)^3 - 16*cos(d *x + c)^2 - (35*cos(d*x + c)^4 + 40*cos(d*x + c)^3 + 48*cos(d*x + c)^2 + 6 4*cos(d*x + c) + 128)*sin(d*x + c) + 64*cos(d*x + c) + 128)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
\[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cos ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a*(sin(c + d*x) + 1))*cos(c + d*x)**4, x)
\[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \,d x } \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4, x)
Time = 0.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (1890 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 420 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 252 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 45 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
1/2520*sqrt(2)*(1890*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2 *d*x + 1/2*c) - 420*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2* d*x + 3/2*c) - 252*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d *x + 5/2*c) + 45*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) + 35*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c))*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:
int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2),x)
Output:
int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2), x)
\[ \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}d x \right ) \] Input:
int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x)
Output:
sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4,x)