3.148 \(\int (\sin (x) \tan (x))^{3/2} \, dx\)

Optimal. Leaf size=31 \[ \frac{8}{3} \csc (x) \sqrt{\sin (x) \tan (x)}-\frac{2}{3} \sin (x) \sqrt{\sin (x) \tan (x)} \]

[Out]

(8*Csc[x]*Sqrt[Sin[x]*Tan[x]])/3 - (2*Sin[x]*Sqrt[Sin[x]*Tan[x]])/3

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Rubi [A]  time = 0.0534873, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4400, 2598, 2589} \[ \frac{8}{3} \csc (x) \sqrt{\sin (x) \tan (x)}-\frac{2}{3} \sin (x) \sqrt{\sin (x) \tan (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x]*Tan[x])^(3/2),x]

[Out]

(8*Csc[x]*Sqrt[Sin[x]*Tan[x]])/3 - (2*Sin[x]*Sqrt[Sin[x]*Tan[x]])/3

Rule 4400

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])), Int[uu*vv^(m*p)*ww^(n*p)
, x], x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  !InertTrigFreeQ[w])

Rule 2598

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(b*(a*Sin[
e + f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2
*m, 2*n]

Rule 2589

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*(a*Sin[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rubi steps

\begin{align*} \int (\sin (x) \tan (x))^{3/2} \, dx &=\frac{\sqrt{\sin (x) \tan (x)} \int \sin ^{\frac{3}{2}}(x) \tan ^{\frac{3}{2}}(x) \, dx}{\sqrt{\sin (x)} \sqrt{\tan (x)}}\\ &=-\frac{2}{3} \sin (x) \sqrt{\sin (x) \tan (x)}+\frac{\left (4 \sqrt{\sin (x) \tan (x)}\right ) \int \frac{\tan ^{\frac{3}{2}}(x)}{\sqrt{\sin (x)}} \, dx}{3 \sqrt{\sin (x)} \sqrt{\tan (x)}}\\ &=\frac{8}{3} \csc (x) \sqrt{\sin (x) \tan (x)}-\frac{2}{3} \sin (x) \sqrt{\sin (x) \tan (x)}\\ \end{align*}

Mathematica [A]  time = 0.0409342, size = 23, normalized size = 0.74 \[ \frac{2}{3} \sin (x) \left (4 \csc ^2(x)-1\right ) \sqrt{\sin (x) \tan (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x]*Tan[x])^(3/2),x]

[Out]

(2*(-1 + 4*Csc[x]^2)*Sin[x]*Sqrt[Sin[x]*Tan[x]])/3

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Maple [B]  time = 0.13, size = 587, normalized size = 18.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)*tan(x))^(3/2),x)

[Out]

1/12*4^(1/2)*(-1+cos(x))^2*(3*cos(x)^3*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-2*(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(
1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)-3*cos(x)^3*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(
-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+9*cos(
x)^2*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-2*(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/
(1+cos(x))^2)^(1/2)-1)/sin(x)^2)-9*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2
)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+9*cos(x)*(-cos(x)/(1+cos(x))^2)^(3/2)*ln
(-2*(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)-9*c
os(x)*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(
1+cos(x))^2)^(1/2)-1)/sin(x)^2)+3*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-2*(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-
cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)-3*(-cos(x)/(1+cos(x))^2)^(3/2)*ln(-(2*cos(x)^2*(
-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+4*cos(x)^3+12*cos(x)
)*(1+cos(x))^2*(-(-1+cos(x)^2)/cos(x))^(3/2)/sin(x)^7

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Maxima [B]  time = 1.55491, size = 77, normalized size = 2.48 \begin{align*} -\frac{8 \,{\left (\frac{\sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} - 1\right )}}{3 \,{\left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (-\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(3/2),x, algorithm="maxima")

[Out]

-8/3*(sin(x)^6/(cos(x) + 1)^6 - 1)/((sin(x)/(cos(x) + 1) + 1)^(3/2)*(-sin(x)/(cos(x) + 1) + 1)^(3/2)*(sin(x)^2
/(cos(x) + 1)^2 + 1)^(3/2))

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Fricas [A]  time = 2.41241, size = 76, normalized size = 2.45 \begin{align*} \frac{2 \,{\left (\cos \left (x\right )^{2} + 3\right )} \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{3 \, \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(3/2),x, algorithm="fricas")

[Out]

2/3*(cos(x)^2 + 3)*sqrt(-(cos(x)^2 - 1)/cos(x))/sin(x)

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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(x)*tan(x))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\sin \left (x\right ) \tan \left (x\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(3/2),x, algorithm="giac")

[Out]

integrate((sin(x)*tan(x))^(3/2), x)