3.149 \(\int (\sin (x) \tan (x))^{5/2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{2}{5} \sin ^2(x) \tan (x) \sqrt{\sin (x) \tan (x)}+\frac{16}{15} \tan (x) \sqrt{\sin (x) \tan (x)}+\frac{64}{15} \cot (x) \sqrt{\sin (x) \tan (x)} \]

[Out]

(64*Cot[x]*Sqrt[Sin[x]*Tan[x]])/15 + (16*Tan[x]*Sqrt[Sin[x]*Tan[x]])/15 - (2*Sin[x]^2*Tan[x]*Sqrt[Sin[x]*Tan[x
]])/5

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Rubi [A]  time = 0.0753787, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {4400, 2598, 2594, 2589} \[ -\frac{2}{5} \sin ^2(x) \tan (x) \sqrt{\sin (x) \tan (x)}+\frac{16}{15} \tan (x) \sqrt{\sin (x) \tan (x)}+\frac{64}{15} \cot (x) \sqrt{\sin (x) \tan (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x]*Tan[x])^(5/2),x]

[Out]

(64*Cot[x]*Sqrt[Sin[x]*Tan[x]])/15 + (16*Tan[x]*Sqrt[Sin[x]*Tan[x]])/15 - (2*Sin[x]^2*Tan[x]*Sqrt[Sin[x]*Tan[x
]])/5

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 2594

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*(n - 1)), x] - Dist[(b^2*(m + n - 1))/(n - 1), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

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))^{5/2} \, dx &=\frac{\sqrt{\sin (x) \tan (x)} \int \sin ^{\frac{5}{2}}(x) \tan ^{\frac{5}{2}}(x) \, dx}{\sqrt{\sin (x)} \sqrt{\tan (x)}}\\ &=-\frac{2}{5} \sin ^2(x) \tan (x) \sqrt{\sin (x) \tan (x)}+\frac{\left (8 \sqrt{\sin (x) \tan (x)}\right ) \int \sqrt{\sin (x)} \tan ^{\frac{5}{2}}(x) \, dx}{5 \sqrt{\sin (x)} \sqrt{\tan (x)}}\\ &=\frac{16}{15} \tan (x) \sqrt{\sin (x) \tan (x)}-\frac{2}{5} \sin ^2(x) \tan (x) \sqrt{\sin (x) \tan (x)}-\frac{\left (32 \sqrt{\sin (x) \tan (x)}\right ) \int \sqrt{\sin (x)} \sqrt{\tan (x)} \, dx}{15 \sqrt{\sin (x)} \sqrt{\tan (x)}}\\ &=\frac{64}{15} \cot (x) \sqrt{\sin (x) \tan (x)}+\frac{16}{15} \tan (x) \sqrt{\sin (x) \tan (x)}-\frac{2}{5} \sin ^2(x) \tan (x) \sqrt{\sin (x) \tan (x)}\\ \end{align*}

Mathematica [A]  time = 0.0991568, size = 29, normalized size = 0.58 \[ \frac{2}{15} \tan (x) \sqrt{\sin (x) \tan (x)} \left (3 \cos ^2(x)+32 \cot ^2(x)+5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x]*Tan[x])^(5/2),x]

[Out]

(2*(5 + 3*Cos[x]^2 + 32*Cot[x]^2)*Tan[x]*Sqrt[Sin[x]*Tan[x]])/15

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Maple [B]  time = 0.167, size = 324, normalized size = 6.5 \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))^(5/2),x)

[Out]

-1/30*4^(1/2)*(-1+cos(x))^2*(6*cos(x)^4-15*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)*ln(-(2*cos(x)^2*(-cos(x)/(1+c
os(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+15*cos(x)^2*(-cos(x)/(1+cos(x))^
2)^(1/2)*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)/si
n(x)^2)-15*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)^(1/2)+15*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)*(-cos(x)/(1+cos(x))^2)^(1/2)-60*cos(x)^2-10)*cos(x)*(1+co
s(x))^2*(-(-1+cos(x)^2)/cos(x))^(5/2)/sin(x)^9

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Maxima [B]  time = 1.60706, size = 111, normalized size = 2.22 \begin{align*} -\frac{32 \,{\left (\frac{5 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac{5 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac{2 \, \sin \left (x\right )^{10}}{{\left (\cos \left (x\right ) + 1\right )}^{10}} - 2\right )}}{15 \,{\left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (-\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(5/2),x, algorithm="maxima")

[Out]

-32/15*(5*sin(x)^4/(cos(x) + 1)^4 - 5*sin(x)^6/(cos(x) + 1)^6 + 2*sin(x)^10/(cos(x) + 1)^10 - 2)/((sin(x)/(cos
(x) + 1) + 1)^(5/2)*(-sin(x)/(cos(x) + 1) + 1)^(5/2)*(sin(x)^2/(cos(x) + 1)^2 + 1)^(5/2))

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Fricas [A]  time = 2.40271, size = 112, normalized size = 2.24 \begin{align*} -\frac{2 \,{\left (3 \, \cos \left (x\right )^{4} - 30 \, \cos \left (x\right )^{2} - 5\right )} \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{15 \, \cos \left (x\right ) \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*cos(x)^4 - 30*cos(x)^2 - 5)*sqrt(-(cos(x)^2 - 1)/cos(x))/(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))**(5/2),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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)*tan(x))^(5/2),x, algorithm="giac")

[Out]

integrate((sin(x)*tan(x))^(5/2), x)