3.507 \(\int \frac{\cos ^3(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=81 \[ -\frac{2 \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^3 d}-\frac{2 (a+b \sin (c+d x))^{5/2}}{5 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{3/2}}{3 b^3 d} \]

[Out]

(-2*(a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^3*d) + (4*a*(a + b*Sin[c + d*x])^(3/2))/(3*b^3*d) - (2*(a + b*Sin
[c + d*x])^(5/2))/(5*b^3*d)

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Rubi [A]  time = 0.0848283, antiderivative size = 81, 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} \[ -\frac{2 \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}{b^3 d}-\frac{2 (a+b \sin (c+d x))^{5/2}}{5 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{3/2}}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0756197, size = 58, normalized size = 0.72 \[ \frac{2 \sqrt{a+b \sin (c+d x)} \left (-8 a^2+4 a b \sin (c+d x)-3 b^2 \sin ^2(c+d x)+15 b^2\right )}{15 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*Sin[c + d*x]]*(-8*a^2 + 15*b^2 + 4*a*b*Sin[c + d*x] - 3*b^2*Sin[c + d*x]^2))/(15*b^3*d)

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Maple [A]  time = 0.169, size = 55, normalized size = 0.7 \begin{align*} -{\frac{-6\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-8\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}-24\,{b}^{2}}{15\,{b}^{3}d}\sqrt{a+b\sin \left ( dx+c \right ) }} \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))^(1/2),x)

[Out]

-2/15/b^3*(a+b*sin(d*x+c))^(1/2)*(-3*b^2*cos(d*x+c)^2-4*a*b*sin(d*x+c)+8*a^2-12*b^2)/d

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Maxima [A]  time = 0.966539, size = 101, normalized size = 1.25 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{b \sin \left (d x + c\right ) + a} - \frac{3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 10 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{2}}{b^{2}}\right )}}{15 \, 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))^(1/2),x, algorithm="maxima")

[Out]

2/15*(15*sqrt(b*sin(d*x + c) + a) - (3*(b*sin(d*x + c) + a)^(5/2) - 10*(b*sin(d*x + c) + a)^(3/2)*a + 15*sqrt(
b*sin(d*x + c) + a)*a^2)/b^2)/(b*d)

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Fricas [A]  time = 1.98215, size = 135, normalized size = 1.67 \begin{align*} \frac{2 \,{\left (3 \, b^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) - 8 \, a^{2} + 12 \, b^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{15 \, 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))^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*b^2*cos(d*x + c)^2 + 4*a*b*sin(d*x + c) - 8*a^2 + 12*b^2)*sqrt(b*sin(d*x + c) + a)/(b^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.0996, size = 97, normalized size = 1.2 \begin{align*} -\frac{2 \,{\left (3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 10 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{2} - 15 \, \sqrt{b \sin \left (d x + c\right ) + a} b^{2}\right )}}{15 \, 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))^(1/2),x, algorithm="giac")

[Out]

-2/15*(3*(b*sin(d*x + c) + a)^(5/2) - 10*(b*sin(d*x + c) + a)^(3/2)*a + 15*sqrt(b*sin(d*x + c) + a)*a^2 - 15*s
qrt(b*sin(d*x + c) + a)*b^2)/(b^3*d)