3.413 \(\int \sqrt {\sin ^4(x) \tan (x)} \, dx\)

Optimal. Leaf size=92 \[ -\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}+\frac {3 \tan ^{-1}\left (\frac {(1-\cot (x)) \csc ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {3 \log \left (\sin (x)+\cos (x)-\sqrt {2} \cot (x) \csc (x) \sqrt {\sin ^4(x) \tan (x)}\right )}{4 \sqrt {2}} \]

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Rubi [B]  time = 0.24, antiderivative size = 204, normalized size of antiderivative = 2.22, number of steps used = 13, number of rules used = 9, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {6719, 288, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (x)}\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {3 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (x)}+1\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {3 \sec ^2(x) \log \left (\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1\right ) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}-\frac {3 \sec ^2(x) \log \left (\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1\right ) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sin[x]^4*Tan[x]],x]

[Out]

-(Cot[x]*Sqrt[Sin[x]^4*Tan[x]])/2 - (3*ArcTan[1 - Sqrt[2]*Sqrt[Tan[x]]]*Sec[x]^2*Sqrt[Sin[x]^4*Tan[x]])/(4*Sqr
t[2]*Tan[x]^(5/2)) + (3*ArcTan[1 + Sqrt[2]*Sqrt[Tan[x]]]*Sec[x]^2*Sqrt[Sin[x]^4*Tan[x]])/(4*Sqrt[2]*Tan[x]^(5/
2)) + (3*Log[1 - Sqrt[2]*Sqrt[Tan[x]] + Tan[x]]*Sec[x]^2*Sqrt[Sin[x]^4*Tan[x]])/(8*Sqrt[2]*Tan[x]^(5/2)) - (3*
Log[1 + Sqrt[2]*Sqrt[Tan[x]] + Tan[x]]*Sec[x]^2*Sqrt[Sin[x]^4*Tan[x]])/(8*Sqrt[2]*Tan[x]^(5/2))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int \sqrt {\sin ^4(x) \tan (x)} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {\frac {x^5}{\left (1+x^2\right )^2}}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {\left (\sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {x^{5/2}}{\left (1+x^2\right )^2} \, dx,x,\tan (x)\right )}{\tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (x)\right )}{4 \tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )}{2 \tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}-\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )}{4 \tan ^{\frac {5}{2}}(x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )}{4 \tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (x)}\right )}{8 \tan ^{\frac {5}{2}}(x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (x)}\right )}{8 \tan ^{\frac {5}{2}}(x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (x)}\right )}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (x)}\right )}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}+\frac {3 \log \left (1-\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}-\frac {3 \log \left (1+\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (x)}\right )}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}-\frac {\left (3 \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (x)}\right )}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}\\ &=-\frac {1}{2} \cot (x) \sqrt {\sin ^4(x) \tan (x)}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (x)}\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (x)}\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{4 \sqrt {2} \tan ^{\frac {5}{2}}(x)}+\frac {3 \log \left (1-\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}-\frac {3 \log \left (1+\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right ) \sec ^2(x) \sqrt {\sin ^4(x) \tan (x)}}{8 \sqrt {2} \tan ^{\frac {5}{2}}(x)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 66, normalized size = 0.72 \[ -\frac {1}{8} \sqrt {\sin (2 x)} \csc ^3(x) \sqrt {\sin ^4(x) \tan (x)} \left (2 \sin (x) \sqrt {\sin (2 x)}+3 \sin ^{-1}(\cos (x)-\sin (x))+3 \log \left (\sin (x)+\sqrt {\sin (2 x)}+\cos (x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sin[x]^4*Tan[x]],x]

[Out]

-1/8*(Csc[x]^3*(3*ArcSin[Cos[x] - Sin[x]] + 3*Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]] + 2*Sin[x]*Sqrt[Sin[2*x]])
*Sqrt[Sin[2*x]]*Sqrt[Sin[x]^4*Tan[x]])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sin ^4(x) \tan (x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sqrt[Sin[x]^4*Tan[x]],x]

[Out]

Could not integrate

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fricas [B]  time = 114.30, size = 1006, normalized size = 10.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)^5/cos(x))^(1/2),x, algorithm="fricas")

[Out]

1/32*(16*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x))*cos(x)*sin(x) - 6*(sqrt(2)*cos(x)^2 - sqrt(2))*arctan
(1/2*(2*cos(x)^4 - 4*cos(x)^2 - 2*(cos(x)^3 - cos(x))*sin(x) + sqrt(2)*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)
/cos(x))*sqrt((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) + 2*(sqrt(2)*cos(x)^2 + sqrt(2)*cos(x)*sin(x))*sqrt((co
s(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)) - sqrt(2)*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)
/cos(x)) + 2)/(cos(x)^4 - 2*cos(x)^2 + (cos(x)^3 - cos(x))*sin(x) + 1)) - 6*(sqrt(2)*cos(x)^2 - sqrt(2))*arcta
n(-1/2*(2*cos(x)^4 - 4*cos(x)^2 - 2*(cos(x)^3 - cos(x))*sin(x) - sqrt(2)*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(
x)/cos(x))*sqrt((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) - 2*(sqrt(2)*cos(x)^2 + sqrt(2)*cos(x)*sin(x))*sqrt((
cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)) + sqrt(2)*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(
x)/cos(x)) + 2)/(cos(x)^4 - 2*cos(x)^2 + (cos(x)^3 - cos(x))*sin(x) + 1)) - 6*(sqrt(2)*cos(x)^2 - sqrt(2))*arc
tan(((sqrt(2)*cos(x)^2 - sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - (2*cos(x)^4
- 3*cos(x)^2 - (sqrt(2)*cos(x)^2 - sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) + 1)
*sqrt((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) + 2*(sqrt(2)*cos(x)^2 + sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 -
2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)))/(cos(x)^2 - 2*(cos(x)^3 - cos(x))*sin(x) - 1)) - 6*(sqrt
(2)*cos(x)^2 - sqrt(2))*arctan(((sqrt(2)*cos(x)^2 - sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*si
n(x)/cos(x)) + (2*cos(x)^4 - 3*cos(x)^2 + (sqrt(2)*cos(x)^2 - sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)
^2 + 1)*sin(x)/cos(x)) + 1)*sqrt((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) - 2*(sqrt(2)*cos(x)^2 + sqrt(2)*cos(
x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)))/(cos(x)^2 - 2*(cos(x)^3 - cos
(x))*sin(x) - 1)) + 3*(sqrt(2)*cos(x)^2 - sqrt(2))*log((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) + 2*(sqrt(2)*c
os(x)^2 + sqrt(2)*cos(x)*sin(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)) - 3*(sqr
t(2)*cos(x)^2 - sqrt(2))*log((cos(x)^2 + 4*(cos(x)^3 - cos(x))*sin(x) - 2*(sqrt(2)*cos(x)^2 + sqrt(2)*cos(x)*s
in(x))*sqrt((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)) - 1)/(cos(x)^2 - 1)))/(cos(x)^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)^5/cos(x))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(sin(x)^5/cos(x)), x)

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maple [C]  time = 0.51, size = 318, normalized size = 3.46

method	result	size
default	\(-\frac {\sqrt {32}\, \left (-1+\cos \relax (x )\right ) \left (-3 i \EllipticPi \left (\sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \relax (x )}{\sin \relax (x )}}\, \sqrt {\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}+3 i \EllipticPi \left (\sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \relax (x )}{\sin \relax (x )}}\, \sqrt {\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}-3 \EllipticPi \left (\sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \relax (x )}{\sin \relax (x )}}\, \sqrt {\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}-3 \EllipticPi \left (\sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \relax (x )}{\sin \relax (x )}}\, \sqrt {\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {-1+\cos \relax (x )-\sin \relax (x )}{\sin \relax (x )}}+2 \left (\cos ^{2}\relax (x )\right ) \sqrt {2}-2 \cos \relax (x ) \sqrt {2}\right ) \left (1+\cos \relax (x )\right )^{2} \sqrt {\frac {\sin ^{5}\relax (x )}{\cos \relax (x )}}}{32 \sin \relax (x )^{5}}\)	\(318\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^5/cos(x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/32*32^(1/2)*(-1+cos(x))*(-3*I*EllipticPi((-(-1+cos(x)-sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos
(x))/sin(x))^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*(-(-1+cos(x)-sin(x))/sin(x))^(1/2)+3*I*EllipticPi((-(-1+c
os(x)-sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(x))/sin(x))^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2
)*(-(-1+cos(x)-sin(x))/sin(x))^(1/2)-3*EllipticPi((-(-1+cos(x)-sin(x))/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((
-1+cos(x))/sin(x))^(1/2)*((-1+cos(x)+sin(x))/sin(x))^(1/2)*(-(-1+cos(x)-sin(x))/sin(x))^(1/2)-3*EllipticPi((-(
-1+cos(x)-sin(x))/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(x))/sin(x))^(1/2)*((-1+cos(x)+sin(x))/sin(x))^
(1/2)*(-(-1+cos(x)-sin(x))/sin(x))^(1/2)+2*cos(x)^2*2^(1/2)-2*cos(x)*2^(1/2))*(1+cos(x))^2*(sin(x)^5/cos(x))^(
1/2)/sin(x)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sin \relax (x)^{5}}{\cos \relax (x)}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)^5/cos(x))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(sin(x)^5/cos(x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\frac {{\sin \relax (x)}^5}{\cos \relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^5/cos(x))^(1/2),x)

[Out]

int((sin(x)^5/cos(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sin ^{5}{\relax (x )}}{\cos {\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)**5/cos(x))**(1/2),x)

[Out]

Integral(sqrt(sin(x)**5/cos(x)), x)

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