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

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

[Out]

-2/3*(a+b*sin(d*x+c))^(3/2)/b^3/d+2*(a^2-b^2)/b^3/d/(a+b*sin(d*x+c))^(1/2)+4*a*(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} \[ \frac {2 \left (a^2-b^2\right )}{b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {2 (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac {4 a \sqrt {a+b \sin (c+d x)}}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.72 \[ \frac {16 a^2+8 a b \sin (c+d x)+b^2 \cos (2 (c+d x))-7 b^2}{3 b^3 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2 - 7*b^2 + b^2*Cos[2*(c + d*x)] + 8*a*b*Sin[c + d*x])/(3*b^3*d*Sqrt[a + b*Sin[c + d*x]])

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fricas [A]  time = 0.80, size = 67, normalized size = 0.85 \[ \frac {2 \, {\left (b^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + 8 \, a^{2} - 4 \, b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, {\left (b^{4} d \sin \left (d x + c\right ) + a b^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.56, size = 72, normalized size = 0.91 \[ \frac {2 \, {\left (\frac {3 \, {\left (a^{2} - b^{2}\right )}}{\sqrt {b \sin \left (d x + c\right ) + a} b^{3}} - \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{6} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a b^{6}}{b^{9}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(a^2 - b^2)/(sqrt(b*sin(d*x + c) + a)*b^3) - ((b*sin(d*x + c) + a)^(3/2)*b^6 - 6*sqrt(b*sin(d*x + c) +
a)*a*b^6)/b^9)/d

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maple [A]  time = 0.33, size = 54, normalized size = 0.68 \[ \frac {\frac {2 b^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{3}+\frac {8 a b \sin \left (d x +c \right )}{3}+\frac {16 a^{2}}{3}-\frac {8 b^{2}}{3}}{b^{3} \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 67, normalized size = 0.85 \[ -\frac {2 \, {\left (\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a}{b^{2}} - \frac {3 \, {\left (a^{2} - b^{2}\right )}}{\sqrt {b \sin \left (d x + c\right ) + a} b^{2}}\right )}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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