\(\int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [518]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2}{b d \sqrt {a+b \sin (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2747, 32} \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2}{b d \sqrt {a+b \sin (c+d x)}} \]

[In]

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

[Out]

-2/(b*d*Sqrt[a + b*Sin[c + d*x]])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2747

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x)^{3/2}} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = -\frac {2}{b d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2}{b d \sqrt {a+b \sin (c+d x)}} \]

[In]

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

[Out]

-2/(b*d*Sqrt[a + b*Sin[c + d*x]])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {2}{b d \sqrt {a +b \sin \left (d x +c \right )}}\)	\(21\)
default	\(-\frac {2}{b d \sqrt {a +b \sin \left (d x +c \right )}}\)	\(21\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b^{2} d \sin \left (d x + c\right ) + a b d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (19) = 38\).

Time = 0.66 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.55 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\begin {cases} \frac {x \cos {\left (c \right )}}{a^{\frac {3}{2}}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{a^{\frac {3}{2}} d} & \text {for}\: b = 0 \\\frac {x \cos {\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{\frac {3}{2}}} & \text {for}\: d = 0 \\- \frac {2}{b d \sqrt {a + b \sin {\left (c + d x \right )}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*cos(c)/a**(3/2), Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)/(a**(3/2)*d), Eq(b, 0)), (x*cos(c)/(a + b*si
n(c))**(3/2), Eq(d, 0)), (-2/(b*d*sqrt(a + b*sin(c + d*x))), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2}{\sqrt {b \sin \left (d x + c\right ) + a} b d} \]

[In]

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

[Out]

-2/(sqrt(b*sin(d*x + c) + a)*b*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {4\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{b\,d\,\left (2\,a^2+4\,a\,b\,\sin \left (c+d\,x\right )+2\,b^2\,{\sin \left (c+d\,x\right )}^2\right )} \]

[In]

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

[Out]

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