\(\int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [444]

Optimal result

Integrand size = 29, antiderivative size = 124 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {64 a^3 \cos ^5(c+d x)}{3465 d (a+a \sin (c+d x))^{5/2}}-\frac {16 a^2 \cos ^5(c+d x)}{693 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d} \]

[Out]

-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)-2/11*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2935, 2753, 2752} \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {64 a^3 \cos ^5(c+d x)}{3465 d (a \sin (c+d x)+a)^{5/2}}-\frac {16 a^2 \cos ^5(c+d x)}{693 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}-\frac {2 a \cos ^5(c+d x)}{99 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-64*a^3*Cos[c + d*x]^5)/(3465*d*(a + a*Sin[c + d*x])^(5/2)) - (16*a^2*Cos[c + d*x]^5)/(693*d*(a + a*Sin[c + d
*x])^(3/2)) - (2*a*Cos[c + d*x]^5)/(99*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]
])/(11*d)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

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] + Dist[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]

Rule 2935

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
] + Dist[(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] /; F
reeQ[{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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {1}{11} \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\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}+\frac {1}{99} (8 a) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\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}+\frac {1}{693} \left (32 a^2\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (-3648+1960 \cos (2 (c+d x))-5165 \sin (c+d x)+315 \sin (3 (c+d x)))}{6930 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((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)]))/(6930*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (315 \left (\sin ^{3}\left (d x +c \right )\right )+980 \left (\sin ^{2}\left (d x +c \right )\right )+1055 \sin \left (d x +c \right )+422\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(75\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.22 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} + 350 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{4} - 40 \, \cos \left (d x + c\right )^{3} - 48 \, \cos \left (d x + c\right )^{2} - 64 \, \cos \left (d x + c\right ) - 128\right )} \sin \left (d x + c\right ) + 64 \, \cos \left (d x + c\right ) + 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-2/3465*(315*cos(d*x + c)^6 + 350*cos(d*x + c)^5 - 5*cos(d*x + c)^4 + 8*cos(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)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4*sin(d*x + c), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {32 \, \sqrt {2} {\left (630 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1925 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1980 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 693 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{3465 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-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*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2), x)