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*(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)

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Rubi [A]  time = 0.0935693, antiderivative size = 79, 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} \[ \frac{2 \left (a^2-b^2\right )}{b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac{4 a \sqrt{a+b \sin (c+d x)}}{b^3 d} \]

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 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)}{(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*}

Mathematica [A]  time = 0.0670612, 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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Maple [A]  time = 0.24, size = 54, normalized size = 0.7 \begin{align*}{\frac{2\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}-8\,{b}^{2}}{3\,{b}^{3}d}{\frac{1}{\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))^(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 = 0.971005, size = 90, normalized size = 1.14 \begin{align*} -\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} \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))^(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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Fricas [A]  time = 2.02746, size = 161, normalized size = 2.04 \begin{align*} \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 )}} \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))^(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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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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.09855, size = 80, normalized size = 1.01 \begin{align*} -\frac{2 \,{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - 6 \, \sqrt{b \sin \left (d x + c\right ) + a} a - \frac{3 \,{\left (a^{2} - b^{2}\right )}}{\sqrt{b \sin \left (d x + c\right ) + a}}\right )}}{3 \, 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))^(3/2),x, algorithm="giac")

[Out]

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