\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [474]

Optimal result

Integrand size = 29, antiderivative size = 60 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {6 a \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}} \]

[Out]

6/35*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-2/7*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2935, 2752} \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {6 a \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(6*a*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))

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 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)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac {3}{7} \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = \frac {6 a \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac {2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (2+5 \sin (c+d x))}{35 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x])/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (5 \sin \left (d x +c \right )+2\right )}{35 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(57\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/35/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(5*sin(d*x+c)+2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (52) = 104\).

Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} - 6 \, \cos \left (d x + c\right ) - 12\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) + 12\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*cos(d*x + c)^4 - 8*cos(d*x + c)^3 - 19*cos(d*x + c)^2 + (5*cos(d*x + c)^3 + 13*cos(d*x + c)^2 - 6*cos(
d*x + c) - 12)*sin(d*x + c) + 6*cos(d*x + c) + 12)*sqrt(a*sin(d*x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*
x + c) + a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)/(a*sin(d*x + c) + a)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {8 \, \sqrt {2} {\left (10 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{35 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-8/35*sqrt(2)*(10*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 7*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a^2*
d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int((cos(c + d*x)^4*sin(c + d*x))/(a + a*sin(c + d*x))^(3/2),x)

[Out]

int((cos(c + d*x)^4*sin(c + d*x))/(a + a*sin(c + d*x))^(3/2), x)