\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [465]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {8 a^2 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac {2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

8/315*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)+2/63*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^5/
d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2935, 2753, 2752} \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {8 a^2 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}+\frac {2 a \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(8*a^2*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) + (2*a*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x])^(3
/2)) - (2*Cos[c + d*x]^5)/(9*d*Sqrt[a + a*Sin[c + d*x]])

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)}{9 d \sqrt {a+a \sin (c+d x)}}-\frac {1}{9} \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}-\frac {1}{63} (4 a) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = \frac {8 a^2 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac {2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (87-35 \cos (2 (c+d x))+130 \sin (c+d x))}{315 d \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

-1/315*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(87 - 35*Cos[2*(c + d*x)
] + 130*Sin[c + d*x]))/(d*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \left (\sin ^{2}\left (d x +c \right )\right )+65 \sin \left (d x +c \right )+26\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(64\)

[In]

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

[Out]

2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+65*sin(d*x+c)+26)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) - 16\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

-2/315*(35*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - cos(d*x + c)^3 + 2*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 5*cos
(d*x + c)^3 - 6*cos(d*x + c)^2 - 8*cos(d*x + c) - 16)*sin(d*x + c) - 8*cos(d*x + c) - 16)*sqrt(a*sin(d*x + c)
+ a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

Sympy [F(-1)]

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

[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]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {16 \, \sqrt {2} {\left (70 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 135 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{315 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

16/315*sqrt(2)*(70*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 135*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 6
3*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[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)