3.2.30 \(\int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx\) [130]

Optimal. Leaf size=50 \[ \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \]

[Out]

2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(a*sin(f*x+e))^(1/2)/f
/cos(f*x+e)^(1/2)/(b*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2681, 2719} \begin {gather*} \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Sin[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

(2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx &=\frac {\sqrt {a \sin (e+f x)} \int \sqrt {\cos (e+f x)} \, dx}{\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\\ &=\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.16, size = 69, normalized size = 1.38 \begin {gather*} \frac {\, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right ) \sqrt {a \sin (e+f x)} \sin (2 (e+f x))}{2 f \cos ^2(e+f x)^{3/4} \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Sin[e + f*x]]/Sqrt[b*Tan[e + f*x]],x]

[Out]

(Hypergeometric2F1[1/4, 1/2, 3/2, Sin[e + f*x]^2]*Sqrt[a*Sin[e + f*x]]*Sin[2*(e + f*x)])/(2*f*(Cos[e + f*x]^2)
^(3/4)*Sqrt[b*Tan[e + f*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.69, size = 327, normalized size = 6.54

method	result	size
default	\(\frac {2 \sqrt {a \sin \left (f x +e \right )}\, \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+\cos \left (f x +e \right )\right )}{f \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \cos \left (f x +e \right )}\)	\(327\)
risch	\(-\frac {i \sqrt {2}\, \sqrt {-i a \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) {\mathrm e}^{-i \left (f x +e \right )}}}{f \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}-\frac {i \left (-\frac {2 \left (a b \,{\mathrm e}^{2 i \left (f x +e \right )}+a b \right )}{a b \sqrt {{\mathrm e}^{i \left (f x +e \right )} \left (a b \,{\mathrm e}^{2 i \left (f x +e \right )}+a b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (f x +e \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {a b \,{\mathrm e}^{3 i \left (f x +e \right )}+a \,{\mathrm e}^{i \left (f x +e \right )} b}}\right ) \sqrt {2}\, \sqrt {-i a \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) {\mathrm e}^{-i \left (f x +e \right )}}\, \sqrt {a \,{\mathrm e}^{i \left (f x +e \right )} b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}}{f \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(383\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*(a*sin(f*x+e))^(1/2)*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)
-1)/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)-I*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*x+e)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(
1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)-I*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*
x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-cos(f*x+e)^2+cos(f*x+e))/(b*sin(f*x+e)/cos(f*x+e))
^(1/2)/sin(f*x+e)/cos(f*x+e)

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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((a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(f*x + e))/sqrt(b*tan(f*x + e)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 75, normalized size = 1.50 \begin {gather*} -\frac {\sqrt {2} \sqrt {-a b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} \sqrt {-a b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{b f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*sqrt(-a*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + sqrt(
2)*sqrt(-a*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))**(1/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*sin(e + f*x))/sqrt(b*tan(e + f*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((a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(f*x + e))/sqrt(b*tan(f*x + e)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {a\,\sin \left (e+f\,x\right )}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^(1/2)/(b*tan(e + f*x))^(1/2),x)

[Out]

int((a*sin(e + f*x))^(1/2)/(b*tan(e + f*x))^(1/2), x)

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