3.289 \(\int \frac{\sqrt{\sin (x)}}{\sqrt{\cos (x)}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{\sqrt{2}}+\frac{\log \left (\tan (x)-\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{2 \sqrt{2}}-\frac{\log \left (\tan (x)+\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{2 \sqrt{2}} \]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]]]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]]]/S
qrt[2] + Log[1 - (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]] + Tan[x]]/(2*Sqrt[2]) - Log[1 + (Sqrt[2]*Sqrt[Sin[x]])/Sq
rt[Cos[x]] + Tan[x]]/(2*Sqrt[2])

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Rubi [A]  time = 0.0816404, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2574, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{\sqrt{2}}+\frac{\log \left (\tan (x)-\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{2 \sqrt{2}}-\frac{\log \left (\tan (x)+\frac{\sqrt{2} \sqrt{\sin (x)}}{\sqrt{\cos (x)}}+1\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sin[x]]/Sqrt[Cos[x]],x]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]]]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]]]/S
qrt[2] + Log[1 - (Sqrt[2]*Sqrt[Sin[x]])/Sqrt[Cos[x]] + Tan[x]]/(2*Sqrt[2]) - Log[1 + (Sqrt[2]*Sqrt[Sin[x]])/Sq
rt[Cos[x]] + Tan[x]]/(2*Sqrt[2])

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0123281, size = 38, normalized size = 0.31 \[ \frac{2 \sin ^{\frac{3}{2}}(x) \cos ^2(x)^{3/4} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\sin ^2(x)\right )}{3 \cos ^{\frac{3}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sin[x]]/Sqrt[Cos[x]],x]

[Out]

(2*(Cos[x]^2)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, Sin[x]^2]*Sin[x]^(3/2))/(3*Cos[x]^(3/2))

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Maple [C]  time = 0.055, size = 171, normalized size = 1.4 \begin{align*} -{\frac{\sqrt{2}}{-2+2\,\cos \left ( x \right ) } \left ( \sin \left ( x \right ) \right ) ^{{\frac{3}{2}}} \left ( i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \right ) \sqrt{{\frac{-1+\cos \left ( x \right ) }{\sin \left ( x \right ) }}}\sqrt{{\frac{-1+\cos \left ( x \right ) +\sin \left ( x \right ) }{\sin \left ( x \right ) }}}\sqrt{-{\frac{-1+\cos \left ( x \right ) -\sin \left ( x \right ) }{\sin \left ( x \right ) }}}{\frac{1}{\sqrt{\cos \left ( x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sin \left (x\right )}}{\sqrt{\cos \left (x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sin(x))/sqrt(cos(x)), x)

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Fricas [B]  time = 5.7764, size = 1613, normalized size = 13.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*(2*cos(x)^3 - 2*cos(x)^2*sin(x) + sqrt(2)*sqrt(2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt
(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1)*sqrt(cos(x))*sqrt(sin(x)) - sqrt(2)*sqrt(cos(x))*sqrt(sin(x)) - 2
*cos(x))/(cos(x)^3 + cos(x)^2*sin(x) - cos(x))) + 1/4*sqrt(2)*arctan(-1/2*(2*cos(x)^3 - 2*cos(x)^2*sin(x) - sq
rt(2)*sqrt(-2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1)*sqrt(cos(x))*
sqrt(sin(x)) + sqrt(2)*sqrt(cos(x))*sqrt(sin(x)) - 2*cos(x))/(cos(x)^3 + cos(x)^2*sin(x) - cos(x))) - 1/4*sqrt
(2)*arctan(-(sqrt(-2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1)*(sqrt(
2)*sqrt(cos(x))*sqrt(sin(x)) + cos(x) + sin(x)) + sqrt(2)*sqrt(cos(x))*sqrt(sin(x)))/(cos(x) - sin(x))) - 1/4*
sqrt(2)*arctan(-(sqrt(2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1)*(sq
rt(2)*sqrt(cos(x))*sqrt(sin(x)) - cos(x) - sin(x)) + sqrt(2)*sqrt(cos(x))*sqrt(sin(x)))/(cos(x) - sin(x))) - 1
/8*sqrt(2)*log(2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1) + 1/8*sqrt
(2)*log(-2*(sqrt(2)*cos(x) + sqrt(2)*sin(x))*sqrt(cos(x))*sqrt(sin(x)) + 4*cos(x)*sin(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sin{\left (x \right )}}}{\sqrt{\cos{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sin(x))/sqrt(cos(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sin(x))/sqrt(cos(x)), x)