3.3.7 \(\int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [207]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 = {2645, 14} \begin {gather*} \frac {2 \sqrt {d \cos (a+b x)}}{b d^3}+\frac {2}{3 b d (d \cos (a+b x))^{3/2}} \end {gather*}

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 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]]

Rule 2645

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])

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 48, normalized size = 1.12 \begin {gather*} -\frac {2 \left (-4+4 \cos ^2(a+b x)^{3/4}+3 \sin ^2(a+b x)\right )}{3 b d (d \cos (a+b x))^{3/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(84\) vs. \(2(37)=74\).
time = 0.21, size = 85, normalized size = 1.98

method	result	size
default	\(\frac {8 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \left (3 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right )}{3 d^{3} \left (4 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right ) b}\)	\(85\)

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,method=_RETURNVERBOSE)

[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 = 0.32, size = 34, normalized size = 0.79 \begin {gather*} \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 {gather*}

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 = 0.36, size = 38, normalized size = 0.88 \begin {gather*} \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 {gather*}

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 = 5.30, size = 71, normalized size = 1.65 \begin {gather*} \begin {cases} \frac {2 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{3 b \left (d \cos {\left (a + b x \right )}\right )^{\frac {5}{2}}} + \frac {8 \cos ^{3}{\left (a + b x \right )}}{3 b \left (d \cos {\left (a + b x \right )}\right )^{\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 {gather*}

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Mupad [B]
time = 1.06, size = 66, normalized size = 1.53 \begin {gather*} \frac {2\,\sqrt {d\,\cos \left (a+b\,x\right )}\,\left (16\,\cos \left (2\,a+2\,b\,x\right )+3\,\cos \left (4\,a+4\,b\,x\right )+13\right )}{3\,b\,d^3\,\left (4\,\cos \left (2\,a+2\,b\,x\right )+\cos \left (4\,a+4\,b\,x\right )+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(d*cos(a + b*x))^(1/2)*(16*cos(2*a + 2*b*x) + 3*cos(4*a + 4*b*x) + 13))/(3*b*d^3*(4*cos(2*a + 2*b*x) + cos(
4*a + 4*b*x) + 3))

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