\(\int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 23, antiderivative size = 47 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {4 \sqrt {a+a \sin (c+d x)}}{a^2 d}-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 a^3 d} \]

[Out]

-2/3*(a+a*sin(d*x+c))^(3/2)/a^3/d+4*(a+a*sin(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2746, 45} \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {4 \sqrt {a \sin (c+d x)+a}}{a^2 d}-\frac {2 (a \sin (c+d x)+a)^{3/2}}{3 a^3 d} \]

[In]

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

[Out]

(4*Sqrt[a + a*Sin[c + d*x]])/(a^2*d) - (2*(a + a*Sin[c + d*x])^(3/2))/(3*a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a-x}{\sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {2 a}{\sqrt {a+x}}-\sqrt {a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {4 \sqrt {a+a \sin (c+d x)}}{a^2 d}-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (-5+\sin (c+d x)) \sqrt {a (1+\sin (c+d x))}}{3 a^2 d} \]

[In]

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

[Out]

(-2*(-5 + Sin[c + d*x])*Sqrt[a*(1 + Sin[c + d*x])])/(3*a^2*d)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +a \sin \left (d x +c \right )}\right )}{d \,a^{3}}\)	\(39\)
default	\(-\frac {2 \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +a \sin \left (d x +c \right )}\right )}{d \,a^{3}}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sin \left (d x + c\right ) - 5\right )}}{3 \, a^{2} d} \]

[In]

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

[Out]

-2/3*sqrt(a*sin(d*x + c) + a)*(sin(d*x + c) - 5)/(a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {a \sin \left (d x + c\right ) + a} a\right )}}{3 \, a^{3} d} \]

[In]

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

[Out]

-2/3*((a*sin(d*x + c) + a)^(3/2) - 6*sqrt(a*sin(d*x + c) + a)*a)/(a^3*d)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {4 \, {\left (\sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{3 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

-4/3*(sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 3*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c))/(a^
2*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[In]

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

[Out]

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