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]

64/15*cot(x)*(sin(x)*tan(x))^(1/2)+16/15*(sin(x)*tan(x))^(1/2)*tan(x)-2/5*sin(x)^2*(sin(x)*tan(x))^(1/2)*tan(x
)

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Rubi [A]  time = 0.08, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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 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 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))^{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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.72, size = 38, normalized size = 0.76 \[ -\frac {2 \, {\left (3 \, \cos \relax (x)^{4} - 30 \, \cos \relax (x)^{2} - 5\right )} \sqrt {-\frac {\cos \relax (x)^{2} - 1}{\cos \relax (x)}}}{15 \, \cos \relax (x) \sin \relax (x)} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\sin \relax (x) \tan \relax (x)\right )^{\frac {5}{2}}\,{d x} \]

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)

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maple [B]  time = 0.34, size = 324, normalized size = 6.48 \[ -\frac {\left (-1+\cos \relax (x )\right )^{2} \left (6 \left (\cos ^{4}\relax (x )\right )-15 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-\left (\cos ^{2}\relax (x )\right )+2 \cos \relax (x )-2 \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-1}{\sin \relax (x )^{2}}\right )+15 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-\left (\cos ^{2}\relax (x )\right )+2 \cos \relax (x )-2 \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-1\right )}{\sin \relax (x )^{2}}\right )-15 \cos \relax (x ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-\left (\cos ^{2}\relax (x )\right )+2 \cos \relax (x )-2 \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-1}{\sin \relax (x )^{2}}\right )+15 \cos \relax (x ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-\left (\cos ^{2}\relax (x )\right )+2 \cos \relax (x )-2 \sqrt {-\frac {\cos \relax (x )}{\left (1+\cos \relax (x )\right )^{2}}}-1\right )}{\sin \relax (x )^{2}}\right )-60 \left (\cos ^{2}\relax (x )\right )-10\right ) \cos \relax (x ) \left (1+\cos \relax (x )\right )^{2} \left (-\frac {-1+\cos ^{2}\relax (x )}{\cos \relax (x )}\right )^{\frac {5}{2}} \sqrt {4}}{30 \sin \relax (x )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)*tan(x))^(5/2),x)

[Out]

-1/30*(-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+cos(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)/sin(x)^2)-
15*cos(x)*(-cos(x)/(1+cos(x))^2)^(1/2)*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)+15*cos(x)*(-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)/sin(x)^2)-60*cos(x)^2-10)*cos(x)*(1+cos(x))^2*
(-(-1+cos(x)^2)/cos(x))^(5/2)/sin(x)^9*4^(1/2)

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maxima [B]  time = 0.44, size = 82, normalized size = 1.64 \[ -\frac {32 \, {\left (\frac {5 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {5 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {2 \, \sin \relax (x)^{10}}{{\left (\cos \relax (x) + 1\right )}^{10}} - 2\right )}}{15 \, {\left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\sin \relax (x)\,\mathrm {tan}\relax (x)\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)*tan(x))^(5/2),x)

[Out]

int((sin(x)*tan(x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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