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

Optimal. Leaf size=43 \[ \frac{2 (d \cos (a+b x))^{5/2}}{5 b d^3}-\frac{2 \sqrt{d \cos (a+b x)}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.0462557, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2565, 14} \[ \frac{2 (d \cos (a+b x))^{5/2}}{5 b d^3}-\frac{2 \sqrt{d \cos (a+b x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\sin ^3(a+b x)}{\sqrt{d \cos (a+b x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-\frac{x^2}{d^2}}{\sqrt{x}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{\sqrt{x}}-\frac{x^{3/2}}{d^2}\right ) \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{2 \sqrt{d \cos (a+b x)}}{b d}+\frac{2 (d \cos (a+b x))^{5/2}}{5 b d^3}\\ \end{align*}

Mathematica [A]  time = 0.178724, size = 57, normalized size = 1.33 \[ \frac{\cos (a+b x) (\cos (2 (a+b x))-9)+8 \cos ^2(a+b x)^{3/4} \sec (a+b x)}{5 b \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[a + b*x]*(-9 + Cos[2*(a + b*x)]) + 8*(Cos[a + b*x]^2)^(3/4)*Sec[a + b*x])/(5*b*Sqrt[d*Cos[a + b*x]])

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Maple [B]  time = 0.033, size = 92, normalized size = 2.1 \begin{align*}{\frac{1}{5\,db} \left ( 8\,\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-8\,\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-8\,\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x)

[Out]

1/5*(8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*sin(1/2*b*x+1/2*a)^4-8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*sin(1/2*
b*x+1/2*a)^2-8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2))/d/b

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Maxima [A]  time = 0.983746, size = 49, normalized size = 1.14 \begin{align*} -\frac{2 \,{\left (5 \, \sqrt{d \cos \left (b x + a\right )} - \frac{\left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}}}{d^{2}}\right )}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

-2/5*(5*sqrt(d*cos(b*x + a)) - (d*cos(b*x + a))^(5/2)/d^2)/(b*d)

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Fricas [A]  time = 1.87462, size = 72, normalized size = 1.67 \begin{align*} \frac{2 \, \sqrt{d \cos \left (b x + a\right )}{\left (\cos \left (b x + a\right )^{2} - 5\right )}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/5*sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 5)/(b*d)

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Sympy [A]  time = 5.35721, size = 63, normalized size = 1.47 \begin{align*} \begin{cases} - \frac{2 \sin ^{2}{\left (a + b x \right )} \sqrt{\cos{\left (a + b x \right )}}}{b \sqrt{d}} - \frac{8 \cos ^{\frac{5}{2}}{\left (a + b x \right )}}{5 b \sqrt{d}} & \text{for}\: b \neq 0 \\\frac{x \sin ^{3}{\left (a \right )}}{\sqrt{d \cos{\left (a \right )}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**3/(d*cos(b*x+a))**(1/2),x)

[Out]

Piecewise((-2*sin(a + b*x)**2*sqrt(cos(a + b*x))/(b*sqrt(d)) - 8*cos(a + b*x)**(5/2)/(5*b*sqrt(d)), Ne(b, 0)),
 (x*sin(a)**3/sqrt(d*cos(a)), True))

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Giac [A]  time = 1.18635, size = 62, normalized size = 1.44 \begin{align*} \frac{2 \,{\left (\sqrt{d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right )^{2} - 5 \, \sqrt{d \cos \left (b x + a\right )} d^{2}\right )}}{5 \, b d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="giac")

[Out]

2/5*(sqrt(d*cos(b*x + a))*d^2*cos(b*x + a)^2 - 5*sqrt(d*cos(b*x + a))*d^2)/(b*d^3)