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(d*x+c))^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 24, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ -\frac {2}{3 b d (a+b \sin (c+d x))^{3/2}} \]

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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fricas [B]  time = 0.57, size = 55, normalized size = 2.29 \[ \frac {2 \, \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, {\left (b^{3} d \cos \left (d x + c\right )^{2} - 2 \, a b^{2} d \sin \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} d\right )}} \]

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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giac [A]  time = 0.89, size = 20, normalized size = 0.83 \[ -\frac {2}{3 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b d} \]

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)

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maple [A]  time = 0.02, size = 21, normalized size = 0.88 \[ -\frac {2}{3 b d \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}} \]

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 = 0.47, size = 20, normalized size = 0.83 \[ -\frac {2}{3 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b d} \]

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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mupad [B]  time = 7.25, size = 157, normalized size = 6.54 \[ -\frac {8\,\sqrt {a+b\,\sin \left (c+d\,x\right )}\,\left (2\,a^2+b^2-b^2\,\cos \left (2\,c+2\,d\,x\right )+4\,a\,b\,\sin \left (c+d\,x\right )\right )}{3\,b\,d\,\left (8\,a^4+3\,b^4+24\,a^2\,b^2-4\,b^4\,\cos \left (2\,c+2\,d\,x\right )+b^4\,\cos \left (4\,c+4\,d\,x\right )-8\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )-24\,a^2\,b^2\,\cos \left (2\,c+2\,d\,x\right )+24\,a\,b^3\,\sin \left (c+d\,x\right )+32\,a^3\,b\,\sin \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*(a + b*sin(c + d*x))^(1/2)*(2*a^2 + b^2 - b^2*cos(2*c + 2*d*x) + 4*a*b*sin(c + d*x)))/(3*b*d*(8*a^4 + 3*b^
4 + 24*a^2*b^2 - 4*b^4*cos(2*c + 2*d*x) + b^4*cos(4*c + 4*d*x) - 8*a*b^3*sin(3*c + 3*d*x) - 24*a^2*b^2*cos(2*c
+ 2*d*x) + 24*a*b^3*sin(c + d*x) + 32*a^3*b*sin(c + d*x)))

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sympy [A]  time = 22.74, size = 87, normalized size = 3.62 \[ \begin {cases} \frac {x \cos {\relax (c )}}{a^{\frac {5}{2}}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{a^{\frac {5}{2}} d} & \text {for}\: b = 0 \\\frac {x \cos {\relax (c )}}{\left (a + b \sin {\relax (c )}\right )^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {2}{3 a b d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{2} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

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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