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

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

[Out]

-2/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

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Rubi [A]  time = 0.0422203, 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 = {2668, 32} \[ -\frac{2}{3 b d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

Rule 2668

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]

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

Mathematica [A]  time = 0.0187543, size = 24, normalized size = 1. \[ -\frac{2}{3 b d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(3*b*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.004, size = 21, normalized size = 0.9 \begin{align*} -{\frac{2}{3\,bd} \left ( a+b\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+b*sin(d*x+c))^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.943273, size = 27, normalized size = 1.12 \begin{align*} -\frac{2}{3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/((b*sin(d*x + c) + a)^(3/2)*b*d)

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Fricas [B]  time = 2.26043, size = 130, normalized size = 5.42 \begin{align*} \frac{2 \, \sqrt{b \sin \left (d x + c\right ) + a}}{3 \,{\left (b^{3} d \cos \left (d x + c\right )^{2} - 2 \, a b^{2} d \sin \left (d x + c\right ) -{\left (a^{2} b + b^{3}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 29.4815, size = 87, normalized size = 3.62 \begin{align*} \begin{cases} \frac{x \cos{\left (c \right )}}{a^{\frac{5}{2}}} & \text{for}\: b = 0 \wedge d = 0 \\\frac{\sin{\left (c + d x \right )}}{a^{\frac{5}{2}} d} & \text{for}\: b = 0 \\\frac{x \cos{\left (c \right )}}{\left (a + b \sin{\left (c \right )}\right )^{\frac{5}{2}}} & \text{for}\: d = 0 \\- \frac{2}{3 a b d \sqrt{a + b \sin{\left (c + d x \right )}} + 3 b^{2} d \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*cos(c)/a**(5/2), Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)/(a**(5/2)*d), Eq(b, 0)), (x*cos(c)/(a + b*si
n(c))**(5/2), Eq(d, 0)), (-2/(3*a*b*d*sqrt(a + b*sin(c + d*x)) + 3*b**2*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x
)), True))

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Giac [A]  time = 1.09457, size = 27, normalized size = 1.12 \begin{align*} -\frac{2}{3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/((b*sin(d*x + c) + a)^(3/2)*b*d)