3.528 \(\int \frac{\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \left (a^2-b^2\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}}-\frac{4 a}{b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \sqrt{a+b \sin (c+d x)}}{b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0939744, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2668, 697} \[ \frac{2 \left (a^2-b^2\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}}-\frac{4 a}{b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \sqrt{a+b \sin (c+d x)}}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

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

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 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^2-x^2}{(a+x)^{5/2}} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{-a^2+b^2}{(a+x)^{5/2}}+\frac{2 a}{(a+x)^{3/2}}-\frac{1}{\sqrt{a+x}}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{2 \left (a^2-b^2\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}}-\frac{4 a}{b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \sqrt{a+b \sin (c+d x)}}{b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.0541289, size = 56, normalized size = 0.71 \[ -\frac{2 \left (8 a^2+12 a b \sin (c+d x)+3 b^2 \sin ^2(c+d x)+b^2\right )}{3 b^3 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(8*a^2 + b^2 + 12*a*b*Sin[c + d*x] + 3*b^2*Sin[c + d*x]^2))/(3*b^3*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.241, size = 55, normalized size = 0.7 \begin{align*} -{\frac{-6\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+24\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}+8\,{b}^{2}}{3\,{b}^{3}d} \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)^3/(a+b*sin(d*x+c))^(5/2),x)

[Out]

-2/3/b^3*(-3*b^2*cos(d*x+c)^2+12*a*b*sin(d*x+c)+8*a^2+4*b^2)/(a+b*sin(d*x+c))^(3/2)/d

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Maxima [A]  time = 0.94406, size = 86, normalized size = 1.09 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sqrt{b \sin \left (d x + c\right ) + a}}{b^{2}} + \frac{6 \,{\left (b \sin \left (d x + c\right ) + a\right )} a - a^{2} + b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b^{2}}\right )}}{3 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.36386, size = 216, normalized size = 2.73 \begin{align*} -\frac{2 \,{\left (3 \, b^{2} \cos \left (d x + c\right )^{2} - 12 \, a b \sin \left (d x + c\right ) - 8 \, a^{2} - 4 \, b^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{3 \,{\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) -{\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 30.6728, size = 304, normalized size = 3.85 \begin{align*} \begin{cases} \frac{x \cos ^{3}{\left (c \right )}}{a^{\frac{5}{2}}} & \text{for}\: b = 0 \wedge d = 0 \\\frac{\frac{2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{\sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{\frac{5}{2}}} & \text{for}\: b = 0 \\\frac{x \cos ^{3}{\left (c \right )}}{\left (a + b \sin{\left (c \right )}\right )^{\frac{5}{2}}} & \text{for}\: d = 0 \\- \frac{16 a^{2}}{3 a b^{3} d \sqrt{a + b \sin{\left (c + d x \right )}} + 3 b^{4} d \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )}} - \frac{24 a b \sin{\left (c + d x \right )}}{3 a b^{3} d \sqrt{a + b \sin{\left (c + d x \right )}} + 3 b^{4} d \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )}} - \frac{8 b^{2} \sin ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt{a + b \sin{\left (c + d x \right )}} + 3 b^{4} d \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )}} - \frac{2 b^{2} \cos ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt{a + b \sin{\left (c + d x \right )}} + 3 b^{4} 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)**3/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Piecewise((x*cos(c)**3/a**(5/2), Eq(b, 0) & Eq(d, 0)), ((2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c + d*x)**
2/d)/a**(5/2), Eq(b, 0)), (x*cos(c)**3/(a + b*sin(c))**(5/2), Eq(d, 0)), (-16*a**2/(3*a*b**3*d*sqrt(a + b*sin(
c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)) - 24*a*b*sin(c + d*x)/(3*a*b**3*d*sqrt(a + b*sin(c
+ d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)) - 8*b**2*sin(c + d*x)**2/(3*a*b**3*d*sqrt(a + b*sin
(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)) - 2*b**2*cos(c + d*x)**2/(3*a*b**3*d*sqrt(a + b*s
in(c + d*x)) + 3*b**4*d*sqrt(a + b*sin(c + d*x))*sin(c + d*x)), True))

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Giac [A]  time = 1.11739, size = 78, normalized size = 0.99 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{b \sin \left (d x + c\right ) + a} + \frac{6 \,{\left (b \sin \left (d x + c\right ) + a\right )} a - a^{2} + b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\right )}}{3 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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