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]

8/3*csc(x)*(sin(x)*tan(x))^(1/2)-2/3*sin(x)*(sin(x)*tan(x))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 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]

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 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])

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.49, size = 26, normalized size = 0.84 \[ \frac {2 \, {\left (\cos \relax (x)^{2} + 3\right )} \sqrt {-\frac {\cos \relax (x)^{2} - 1}{\cos \relax (x)}}}{3 \, \sin \relax (x)} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\sin \relax (x) \tan \relax (x)\right )^{\frac {3}{2}}\,{d x} \]

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)

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maple [B]  time = 0.26, size = 587, normalized size = 18.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 57, normalized size = 1.84 \[ -\frac {8 \, {\left (\frac {\sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} - 1\right )}}{3 \, {\left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}^{\frac {3}{2}} {\left (-\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (\sin \relax (x)\,\mathrm {tan}\relax (x)\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((sin(x)*tan(x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\sin {\relax (x )} \tan {\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((sin(x)*tan(x))**(3/2), x)

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