3.208 \(\int \frac{\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0508097, 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} \[ \frac{2}{5 b d (d \cos (a+b x))^{5/2}}-\frac{2}{b d^3 \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

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

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))^{7/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-\frac{x^2}{d^2}}{x^{7/2}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x^{7/2}}-\frac{1}{d^2 x^{3/2}}\right ) \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac{2}{5 b d (d \cos (a+b x))^{5/2}}-\frac{2}{b d^3 \sqrt{d \cos (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.245879, size = 70, normalized size = 1.63 \[ \frac{2 \tan ^2(a+b x) \left (-4 \sqrt [4]{\cos ^2(a+b x)}+4 \left (\sqrt [4]{\cos ^2(a+b x)}-1\right ) \csc ^2(a+b x)+5\right )}{5 b d^3 \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(2*(5 - 4*(Cos[a + b*x]^2)^(1/4) + 4*(-1 + (Cos[a + b*x]^2)^(1/4))*Csc[a + b*x]^2)*Tan[a + b*x]^2)/(5*b*d^3*Sq
rt[d*Cos[a + b*x]])

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Maple [B]  time = 0.103, size = 98, normalized size = 2.3 \begin{align*}{\frac{8}{5\,{d}^{4}b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( 5\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-5\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) \left ( 8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}-12\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+6\, \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))^(7/2),x)

[Out]

8/5/d^4/(8*sin(1/2*b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x+1/2*a)^2-1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d
)^(1/2)*(5*sin(1/2*b*x+1/2*a)^4-5*sin(1/2*b*x+1/2*a)^2+1)/b

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Maxima [A]  time = 0.962176, size = 50, normalized size = 1.16 \begin{align*} -\frac{2 \,{\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )}}{5 \, \left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}} 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))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.12197, size = 99, normalized size = 2.3 \begin{align*} -\frac{2 \, \sqrt{d \cos \left (b x + a\right )}{\left (5 \, \cos \left (b x + a\right )^{2} - 1\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{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))^(7/2),x, algorithm="fricas")

[Out]

-2/5*sqrt(d*cos(b*x + a))*(5*cos(b*x + a)^2 - 1)/(b*d^4*cos(b*x + a)^3)

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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(sin(b*x+a)**3/(d*cos(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17994, size = 61, normalized size = 1.42 \begin{align*} -\frac{2 \,{\left (5 \, d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )}}{5 \, \sqrt{d \cos \left (b x + a\right )} b d^{6} \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))^(7/2),x, algorithm="giac")

[Out]

-2/5*(5*d^3*cos(b*x + a)^2 - d^3)/(sqrt(d*cos(b*x + a))*b*d^6*cos(b*x + a)^2)