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

Optimal. Leaf size=45 \[ -\frac {2 \sqrt {a \sin (c+d x)+a}}{a^3 d}-\frac {4}{a^2 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 = {2667, 43} \[ -\frac {2 \sqrt {a \sin (c+d x)+a}}{a^3 d}-\frac {4}{a^2 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 30, normalized size = 0.67 \[ -\frac {2 (\sin (c+d x)+3)}{a^2 d \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 41, normalized size = 0.91 \[ -\frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sin \left (d x + c\right ) + 3\right )}}{a^{3} d \sin \left (d x + c\right ) + a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.45, size = 39, normalized size = 0.87 \[ -\frac {2 \, {\left (\frac {\sqrt {a \sin \left (d x + c\right ) + a}}{a^{3}} + \frac {2}{\sqrt {a \sin \left (d x + c\right ) + a} a^{2}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 29, normalized size = 0.64 \[ -\frac {2 \left (3+\sin \left (d x +c \right )\right )}{a^{2} \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x)

[Out]

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

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maxima [A]  time = 1.16, size = 42, normalized size = 0.93 \[ -\frac {2 \, {\left (\frac {\sqrt {a \sin \left (d x + c\right ) + a}}{a^{2}} + \frac {2}{\sqrt {a \sin \left (d x + c\right ) + a} a}\right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^3/(a + a*sin(c + d*x))^(5/2), x)

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sympy [A]  time = 26.90, size = 267, normalized size = 5.93 \[ \begin {cases} \text {NaN} & \text {for}\: \left (c = \frac {3 \pi }{2} \vee c = - d x + \frac {3 \pi }{2}\right ) \wedge \left (c = - d x + \frac {3 \pi }{2} \vee d = 0\right ) \\\frac {x \cos ^{3}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {8 \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{3 a^{3} d \sin ^{2}{\left (c + d x \right )} + 6 a^{3} d \sin {\left (c + d x \right )} + 3 a^{3} d} - \frac {24 \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{3 a^{3} d \sin ^{2}{\left (c + d x \right )} + 6 a^{3} d \sin {\left (c + d x \right )} + 3 a^{3} d} - \frac {2 \sqrt {a \sin {\left (c + d x \right )} + a} \cos ^{2}{\left (c + d x \right )}}{3 a^{3} d \sin ^{2}{\left (c + d x \right )} + 6 a^{3} d \sin {\left (c + d x \right )} + 3 a^{3} d} - \frac {16 \sqrt {a \sin {\left (c + d x \right )} + a}}{3 a^{3} d \sin ^{2}{\left (c + d x \right )} + 6 a^{3} d \sin {\left (c + d x \right )} + 3 a^{3} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Piecewise((nan, (Eq(d, 0) | Eq(c, -d*x + 3*pi/2)) & (Eq(c, 3*pi/2) | Eq(c, -d*x + 3*pi/2))), (x*cos(c)**3/(a*s
in(c) + a)**(5/2), Eq(d, 0)), (-8*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2/(3*a**3*d*sin(c + d*x)**2 + 6*a**3*
d*sin(c + d*x) + 3*a**3*d) - 24*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)/(3*a**3*d*sin(c + d*x)**2 + 6*a**3*d*sin
(c + d*x) + 3*a**3*d) - 2*sqrt(a*sin(c + d*x) + a)*cos(c + d*x)**2/(3*a**3*d*sin(c + d*x)**2 + 6*a**3*d*sin(c
+ d*x) + 3*a**3*d) - 16*sqrt(a*sin(c + d*x) + a)/(3*a**3*d*sin(c + d*x)**2 + 6*a**3*d*sin(c + d*x) + 3*a**3*d)
, True))

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