3.3.100 \(\int \frac {\sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}} \, dx\) [300]

Optimal. Leaf size=174 \[ \frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {\log \left (1+\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b}+\frac {\log \left (1+\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{2 \sqrt {2} b} \]

[Out]

-1/2*arctan(-1+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)-1/2*arctan(1+2^(1/2)*cos(b*x+a)^(1/2)/sin(
b*x+a)^(1/2))/b*2^(1/2)-1/4*ln(1+cot(b*x+a)-2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)+1/4*ln(1+cot(
b*x+a)+2^(1/2)*cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2))/b*2^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2655, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}\right )}{\sqrt {2} b}-\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{\sqrt {2} b}-\frac {\log \left (\cot (a+b x)-\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b}+\frac {\log \left (\cot (a+b x)+\frac {\sqrt {2} \sqrt {\cos (a+b x)}}{\sqrt {\sin (a+b x)}}+1\right )}{2 \sqrt {2} b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[a + b*x]]/Sqrt[Sin[a + b*x]],x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - ArcTan[1 + (Sqrt[2]*Sqrt[Cos[a + b*x
]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - Log[1 + Cot[a + b*x] - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/
(2*Sqrt[2]*b) + Log[1 + Cot[a + b*x] + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(2*Sqrt[2]*b)

Rule 210

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

Rule 303

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 631

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 642

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 1176

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 1179

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 2655

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*
Sin[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 55, normalized size = 0.32 \begin {gather*} \frac {2 \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\sin ^2(a+b x)\right ) \sqrt {\sin (a+b x)}}{b \sqrt {\cos (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[a + b*x]]/Sqrt[Sin[a + b*x]],x]

[Out]

(2*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, Sin[a + b*x]^2]*Sqrt[Sin[a + b*x]])/(b*Sqrt[Cos[a +
b*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.11, size = 292, normalized size = 1.68

method	result	size
default	\(-\frac {\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \left (\sin ^{\frac {3}{2}}\left (b x +a \right )\right ) \sqrt {2}}{2 b \sqrt {\cos \left (b x +a \right )}\, \left (-1+\cos \left (b x +a \right )\right )}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/b/cos(b*x+a)^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(
1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*(I*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1
/2*2^(1/2))-I*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*EllipticF(((1-c
os(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1
/2-1/2*I,1/2*2^(1/2))+EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2)))*sin(b*x+
a)^(3/2)/(-1+cos(b*x+a))*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(cos(b*x + a))/sqrt(sin(b*x + a)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1185 vs. \(2 (138) = 276\).
time = 14.64, size = 1185, normalized size = 6.81 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \arctan \left (\frac {{\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 4 \, {\left (b^{2} \cos \left (b x + a\right )^{4} - b^{2} \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {1}{b^{4}}}}{2 \, {\left (2 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}\right ) - \frac {1}{4} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \arctan \left (\frac {{\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 4 \, {\left (b^{2} \cos \left (b x + a\right )^{4} - b^{2} \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {1}{b^{4}}}}{2 \, {\left (2 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}\right ) - \frac {1}{4} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \arctan \left (-\frac {{\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) - \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - \sqrt {4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1} {\left ({\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )}}{2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )}\right ) - \frac {1}{4} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \arctan \left (-\frac {{\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) - \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - \sqrt {4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1} {\left ({\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \sin \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )}}{2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )}\right ) + \frac {1}{8} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \log \left (4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1\right ) - \frac {1}{8} \, \sqrt {2} \frac {1}{b^{4}}^{\frac {1}{4}} \log \left (4 \, b^{2} \sqrt {\frac {1}{b^{4}}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (\sqrt {2} b^{3} \frac {1}{b^{4}}^{\frac {3}{4}} \cos \left (b x + a\right ) + \sqrt {2} b \frac {1}{b^{4}}^{\frac {1}{4}} \sin \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right )} \sqrt {\sin \left (b x + a\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(2)*(b^(-4))^(1/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*co
s(b*x + a))*sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a) + s
qrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1)*sqrt(cos(b*x + a))*sqrt(sin(b
*x + a)) + (sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a)
)*sqrt(sin(b*x + a)) + 2*cos(b*x + a)*sin(b*x + a) - 4*(b^2*cos(b*x + a)^4 - b^2*cos(b*x + a)^2)*sqrt(b^(-4)))
/((2*cos(b*x + a)^3 - cos(b*x + a))*sin(b*x + a))) - 1/4*sqrt(2)*(b^(-4))^(1/4)*arctan(1/2*((sqrt(2)*b^3*(b^(-
4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x +
a) - 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*s
qrt(sin(b*x + a)) + 1)*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + (sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt
(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) - 2*cos(b*x + a)*sin(b*x + a) + 4*(b^
2*cos(b*x + a)^4 - b^2*cos(b*x + a)^2)*sqrt(b^(-4)))/((2*cos(b*x + a)^3 - cos(b*x + a))*sin(b*x + a))) - 1/4*s
qrt(2)*(b^(-4))^(1/4)*arctan(-1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) - sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x
+ a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) - sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*
b^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a))
+ 1)*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))*sq
rt(sin(b*x + a)) + 2*cos(b*x + a)*sin(b*x + a)))/(cos(b*x + a)*sin(b*x + a))) - 1/4*sqrt(2)*(b^(-4))^(1/4)*arc
tan(-1/2*((sqrt(2)*b^3*(b^(-4))^(3/4)*sin(b*x + a) - sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))
*sqrt(sin(b*x + a)) - sqrt(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*
x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) + 1)*((sqrt(2)*b^3*(b^(-
4))^(3/4)*sin(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*cos(b*x + a))*sqrt(cos(b*x + a))*sqrt(sin(b*x + a)) - 2*cos(
b*x + a)*sin(b*x + a)))/(cos(b*x + a)*sin(b*x + a))) + 1/8*sqrt(2)*(b^(-4))^(1/4)*log(4*b^2*sqrt(b^(-4))*cos(b
*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqr
t(cos(b*x + a))*sqrt(sin(b*x + a)) + 1) - 1/8*sqrt(2)*(b^(-4))^(1/4)*log(4*b^2*sqrt(b^(-4))*cos(b*x + a)*sin(b
*x + a) - 2*(sqrt(2)*b^3*(b^(-4))^(3/4)*cos(b*x + a) + sqrt(2)*b*(b^(-4))^(1/4)*sin(b*x + a))*sqrt(cos(b*x + a
))*sqrt(sin(b*x + a)) + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\cos {\left (a + b x \right )}}}{\sqrt {\sin {\left (a + b x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**(1/2)/sin(b*x+a)**(1/2),x)

[Out]

Integral(sqrt(cos(a + b*x))/sqrt(sin(a + b*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(1/2)/sin(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(cos(b*x + a))/sqrt(sin(b*x + a)), x)

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Mupad [B]
time = 1.60, size = 44, normalized size = 0.25 \begin {gather*} -\frac {2\,{\cos \left (a+b\,x\right )}^{3/2}\,\sqrt {\sin \left (a+b\,x\right )}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {3}{4};\ \frac {7}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{3\,b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^(1/2)/sin(a + b*x)^(1/2),x)

[Out]

-(2*cos(a + b*x)^(3/2)*sin(a + b*x)^(1/2)*hypergeom([3/4, 3/4], 7/4, cos(a + b*x)^2))/(3*b*(sin(a + b*x)^2)^(1
/4))

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