3.191 \(\int \frac{\cos (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=24 \[ -\frac{2}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

-2/(3*a*d*(a + a*Sin[c + d*x])^(3/2))

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Rubi [A]  time = 0.0348205, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 32} \[ -\frac{2}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-2/(3*a*d*(a + a*Sin[c + d*x])^(3/2))

Rule 2667

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])

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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{(a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac{2}{3 a d (a+a \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0399404, size = 24, normalized size = 1. \[ -\frac{2}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-2/(3*a*d*(a + a*Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.006, size = 21, normalized size = 0.9 \begin{align*} -{\frac{2}{3\,da} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a+a*sin(d*x+c))^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.935473, size = 27, normalized size = 1.12 \begin{align*} -\frac{2}{3 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/3/((a*sin(d*x + c) + a)^(3/2)*a*d)

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Fricas [B]  time = 2.14845, size = 116, normalized size = 4.83 \begin{align*} \frac{2 \, \sqrt{a \sin \left (d x + c\right ) + a}}{3 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 30.179, size = 65, normalized size = 2.71 \begin{align*} \begin{cases} - \frac{2}{3 a^{2} d \sqrt{a \sin{\left (c + d x \right )} + a} \sin{\left (c + d x \right )} + 3 a^{2} d \sqrt{a \sin{\left (c + d x \right )} + a}} & \text{for}\: d \neq 0 \\\frac{x \cos{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Piecewise((-2/(3*a**2*d*sqrt(a*sin(c + d*x) + a)*sin(c + d*x) + 3*a**2*d*sqrt(a*sin(c + d*x) + a)), Ne(d, 0)),
(x*cos(c)/(a*sin(c) + a)**(5/2), True))

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Giac [A]  time = 1.11838, size = 27, normalized size = 1.12 \begin{align*} -\frac{2}{3 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-2/3/((a*sin(d*x + c) + a)^(3/2)*a*d)