3.207 \(\int \frac{\sin ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0517832, 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} \[ \frac{2 \sqrt{d \cos (a+b x)}}{b d^3}+\frac{2}{3 b d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(d*Cos[a + b*x])^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.111302, size = 48, normalized size = 1.12 \[ -\frac{2 \left (3 \sin ^2(a+b x)+4 \cos ^2(a+b x)^{3/4}-4\right )}{3 b d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/(d*Cos[a + b*x])^(5/2),x]

[Out]

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

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Maple [B]  time = 0.089, size = 85, normalized size = 2. \begin{align*}{\frac{8}{3\,{d}^{3}b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( 3\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-3\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) \left ( 4\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/(d*cos(b*x+a))^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.971462, size = 46, normalized size = 1.07 \begin{align*} \frac{2 \,{\left (\frac{1}{\left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}}} + \frac{3 \, \sqrt{d \cos \left (b x + a\right )}}{d^{2}}\right )}}{3 \, 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))^(5/2),x, algorithm="maxima")

[Out]

2/3*(1/(d*cos(b*x + a))^(3/2) + 3*sqrt(d*cos(b*x + a))/d^2)/(b*d)

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Fricas [A]  time = 2.29473, size = 97, normalized size = 2.26 \begin{align*} \frac{2 \, \sqrt{d \cos \left (b x + a\right )}{\left (3 \, \cos \left (b x + a\right )^{2} + 1\right )}}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(d*cos(b*x + a))*(3*cos(b*x + a)^2 + 1)/(b*d^3*cos(b*x + a)^2)

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Sympy [A]  time = 67.2828, size = 63, normalized size = 1.47 \begin{align*} \begin{cases} \frac{2 \sin ^{2}{\left (a + b x \right )}}{3 b d^{\frac{5}{2}} \cos ^{\frac{3}{2}}{\left (a + b x \right )}} + \frac{8 \sqrt{\cos{\left (a + b x \right )}}}{3 b d^{\frac{5}{2}}} & \text{for}\: b \neq 0 \\\frac{x \sin ^{3}{\left (a \right )}}{\left (d \cos{\left (a \right )}\right )^{\frac{5}{2}}} & \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))**(5/2),x)

[Out]

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

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Giac [A]  time = 1.18174, size = 55, normalized size = 1.28 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{d \cos \left (b x + a\right )} + \frac{d}{\sqrt{d \cos \left (b x + a\right )} \cos \left (b x + a\right )}\right )}}{3 \, 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))^(5/2),x, algorithm="giac")

[Out]

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