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]

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

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Rubi [A]  time = 0.09, antiderivative size = 79, 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 = {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 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]

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 ^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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.90, size = 91, normalized size = 1.15 \[ -\frac {2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} - 12 \, a b \sin \left (d x + c\right ) - 8 \, a^{2} - 4 \, b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, {\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) - {\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \]

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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giac [A]  time = 0.86, size = 61, normalized size = 0.77 \[ -\frac {2 \, {\left (\frac {3 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b^{3}} + \frac {6 \, {\left (b \sin \left (d x + c\right ) + a\right )} a - a^{2} + b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{3}}\right )}}{3 \, d} \]

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

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maple [A]  time = 0.39, size = 55, normalized size = 0.70 \[ -\frac {2 \left (-3 b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+12 a b \sin \left (d x +c \right )+8 a^{2}+4 b^{2}\right )}{3 b^{3} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} d} \]

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 = 0.73, size = 64, normalized size = 0.81 \[ -\frac {2 \, {\left (\frac {3 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b^{2}} + \frac {6 \, {\left (b \sin \left (d x + c\right ) + a\right )} a - a^{2} + b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{3 \, b d} \]

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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mupad [B]  time = 11.89, size = 1402, normalized size = 17.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*a^2*b^2*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i)*(a + (b*(cos(d*x) - sin(d*x)*1i)*(cos(c) - s
in(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2))/(3*(a^2*b^5*d - b^7*d + 2*b^7
*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - b^7*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(
4*c)*1i) - a*b^6*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a*b^6*d*(cos(d*x) + sin(d*x)*1i)
*(cos(c) + sin(c)*1i)*4i + 2*a^2*b^5*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - 4*a^4*b^3*d*(co
s(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) + a^3*b^4*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*
c)*1i)*4i + a^2*b^5*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a^3*b^4*d*(cos(d*x) + sin(d*x)*1
i)*(cos(c) + sin(c)*1i)*4i)) - (8*a^4*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i)*(a + (b*(cos(d*x)
- sin(d*x)*1i)*(cos(c) - sin(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2))/(3*
(a^2*b^5*d - b^7*d + 2*b^7*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - b^7*d*(cos(4*d*x) + sin(4
*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a*b^6*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a*b^6*
d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*4i + 2*a^2*b^5*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(
2*c)*1i) - 4*a^4*b^3*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) + a^3*b^4*d*(cos(3*d*x) + sin(3*d
*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a^2*b^5*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a^3*b^
4*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*4i)) - (8*b^4*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2
*c)*1i)*(a + (b*(cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i)*(cos(c) + si
n(c)*1i)*1i)/2)^(1/2))/(3*(a^2*b^5*d - b^7*d + 2*b^7*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) -
b^7*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c)*1i) - a*b^6*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c)
+ sin(3*c)*1i)*4i + a*b^6*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*4i + 2*a^2*b^5*d*(cos(2*d*x) + sin(2
*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - 4*a^4*b^3*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) + a^3*b
^4*d*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*4i + a^2*b^5*d*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*
c) + sin(4*c)*1i) - a^3*b^4*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*4i)) - (a*(cos(d*x) + sin(d*x)*1i)
*(cos(c) + sin(c)*1i)*(a + (b*(cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i
)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2)*8i)/(b^4*d*(cos(2*d*x) + sin(2*d*x)*1i)*(cos(2*c) + sin(2*c)*1i) - b^4*d +
a*b^3*d*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*2i) - (2*(a + (b*(cos(d*x) - sin(d*x)*1i)*(cos(c) - sin
(c)*1i)*1i)/2 - (b*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/2)^(1/2))/(b^3*d)

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sympy [A]  time = 23.93, size = 304, normalized size = 3.85 \[ \begin {cases} \frac {x \cos ^{3}{\relax (c )}}{a^{\frac {5}{2}}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{\frac {5}{2}}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a + b \sin {\relax (c )}\right )^{\frac {5}{2}}} & \text {for}\: d = 0 \\- \frac {16 a^{2}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {24 a b \sin {\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {8 b^{2} \sin ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} - \frac {2 b^{2} \cos ^{2}{\left (c + d x \right )}}{3 a b^{3} d \sqrt {a + b \sin {\left (c + d x \right )}} + 3 b^{4} d \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

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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