Integrand size = 29, antiderivative size = 124 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {64 a^3 \cos ^5(c+d x)}{3465 d (a+a \sin (c+d x))^{5/2}}-\frac {16 a^2 \cos ^5(c+d x)}{693 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d} \] Output:
-64/3465*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-16/693*a^2*cos(d*x+c)^5 /d/(a+a*sin(d*x+c))^(3/2)-2/99*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)-2/1 1*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d
Time = 1.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (-3648+1960 \cos (2 (c+d x))-5165 \sin (c+d x)+315 \sin (3 (c+d x)))}{6930 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
Output:
((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(-3648 + 1960*Cos[2*(c + d*x)] - 5165*Sin[c + d*x] + 315*Sin[3*(c + d*x)]))/(693 0*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.65 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3335, 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.
\(\displaystyle \int \sin (c+d x) \cos ^4(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 \sqrt {a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3335 |
\(\displaystyle \frac {1}{11} \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {1}{11} \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {1}{11} \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{11} \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {1}{11} \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
Input:
Int[Cos[c + d*x]^4*Sin[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
Output:
(-2*Cos[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(11*d) + ((-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)/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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
Time = 0.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (315 \sin \left (d x +c \right )^{3}+980 \sin \left (d x +c \right )^{2}+1055 \sin \left (d x +c \right )+422\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(75\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE )
Output:
2/3465*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(315*sin(d*x+c)^3+980*sin(d*x+c)^ 2+1055*sin(d*x+c)+422)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.22 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} + 350 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{4} - 40 \, \cos \left (d x + c\right )^{3} - 48 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) - 128\right )} \sin \left (d x + c\right ) + 64 \, \cos \left (d x + c\right ) + 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fri cas")
Output:
-2/3465*(315*cos(d*x + c)^6 + 350*cos(d*x + c)^5 - 5*cos(d*x + c)^4 + 8*co s(d*x + c)^3 - 16*cos(d*x + c)^2 + (315*cos(d*x + c)^5 - 35*cos(d*x + c)^4 - 40*cos(d*x + c)^3 - 48*cos(d*x + c)^2 - 64*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)
Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**(1/2),x)
Output:
Timed out
\[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="max ima")
Output:
integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4*sin(d*x + c), x)
Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {32 \, \sqrt {2} {\left (630 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1925 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1980 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 693 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{3465 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="gia c")
Output:
-32/3465*sqrt(2)*(630*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/ 2*d*x + 1/2*c)^11 - 1925*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 + 1980*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 693*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p i + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:
int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2),x)
Output:
int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2), x)
\[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )d x \right ) \] Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x)
Output:
sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4*sin(c + d*x),x)