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

Optimal. Leaf size=83 \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}-\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d} \]

[Out]

(-2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^3*d) - (2*(a + b
*Sin[c + d*x])^(7/2))/(7*b^3*d)

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Rubi [A]  time = 0.0819508, antiderivative size = 83, 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} \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}-\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d} \]

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)*(a + b*Sin[c + d*x])^(3/2))/(3*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^3*d) - (2*(a + b
*Sin[c + d*x])^(7/2))/(7*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 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+x} \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (-a^2+b^2\right ) \sqrt{a+x}+2 a (a+x)^{3/2}-(a+x)^{5/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d}-\frac{2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.122831, size = 58, normalized size = 0.7 \[ \frac{(a+b \sin (c+d x))^{3/2} \left (-16 a^2+24 a b \sin (c+d x)+15 b^2 \cos (2 (c+d x))+55 b^2\right )}{105 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sin[c + d*x])^(3/2)*(-16*a^2 + 55*b^2 + 15*b^2*Cos[2*(c + d*x)] + 24*a*b*Sin[c + d*x]))/(105*b^3*d)

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Maple [A]  time = 0.295, size = 55, normalized size = 0.7 \begin{align*} -{\frac{-30\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-24\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}-40\,{b}^{2}}{105\,{b}^{3}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}} \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]

-2/105/b^3*(a+b*sin(d*x+c))^(3/2)*(-15*b^2*cos(d*x+c)^2-12*a*b*sin(d*x+c)+8*a^2-20*b^2)/d

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Maxima [A]  time = 0.950959, size = 82, normalized size = 0.99 \begin{align*} -\frac{2 \,{\left (15 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} - 42 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a + 35 \,{\left (a^{2} - b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\right )}}{105 \, 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="maxima")

[Out]

-2/105*(15*(b*sin(d*x + c) + a)^(7/2) - 42*(b*sin(d*x + c) + a)^(5/2)*a + 35*(a^2 - b^2)*(b*sin(d*x + c) + a)^
(3/2))/(b^3*d)

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Fricas [A]  time = 3.13987, size = 192, normalized size = 2.31 \begin{align*} \frac{2 \,{\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{3} + 32 \, a b^{2} +{\left (15 \, b^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{2} b + 20 \, b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{105 \, 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]

2/105*(3*a*b^2*cos(d*x + c)^2 - 8*a^3 + 32*a*b^2 + (15*b^3*cos(d*x + c)^2 + 4*a^2*b + 20*b^3)*sin(d*x + c))*sq
rt(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 = 1.6574, size = 105, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (35 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - \frac{15 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}{b^{2}} + \frac{42 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a}{b^{2}} - \frac{35 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2}}{b^{2}}\right )}}{105 \, 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="giac")

[Out]

2/105*(35*(b*sin(d*x + c) + a)^(3/2) - 15*(b*sin(d*x + c) + a)^(7/2)/b^2 + 42*(b*sin(d*x + c) + a)^(5/2)*a/b^2
- 35*(b*sin(d*x + c) + a)^(3/2)*a^2/b^2)/(b*d)