3.3.67 \(\int \frac {\sqrt {-a-b x^2}}{\sqrt {c+d x^2}} \, dx\) [267]

Optimal. Leaf size=203 \[ \frac {x \sqrt {-a-b x^2}}{\sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {-a-b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {-a-b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

x*(-b*x^2-a)^(1/2)/(d*x^2+c)^(1/2)-(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^
2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*(-b*x^2-a)^(1/2)/d^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)
+(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1
/2)*(-b*x^2-a)^(1/2)/d^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {433, 429, 506, 422} \begin {gather*} \frac {\sqrt {c} \sqrt {-a-b x^2} F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {c} \sqrt {-a-b x^2} E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {-a-b x^2}}{\sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-a - b*x^2]/Sqrt[c + d*x^2],x]

[Out]

(x*Sqrt[-a - b*x^2])/Sqrt[c + d*x^2] - (Sqrt[c]*Sqrt[-a - b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b
*c)/(a*d)])/(Sqrt[d]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (Sqrt[c]*Sqrt[-a - b*x^2]*Ellipt
icF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(Sqrt[d]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x
^2])

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sq
rt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 433

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[c + d*x^2])), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rubi steps

\begin {align*} \int \frac {\sqrt {-a-b x^2}}{\sqrt {c+d x^2}} \, dx &=-\left (a \int \frac {1}{\sqrt {-a-b x^2} \sqrt {c+d x^2}} \, dx\right )-b \int \frac {x^2}{\sqrt {-a-b x^2} \sqrt {c+d x^2}} \, dx\\ &=\frac {x \sqrt {-a-b x^2}}{\sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {-a-b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-c \int \frac {\sqrt {-a-b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx\\ &=\frac {x \sqrt {-a-b x^2}}{\sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {-a-b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {-a-b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 89, normalized size = 0.44 \begin {gather*} \frac {\sqrt {-a-b x^2} \sqrt {\frac {c+d x^2}{c}} E\left (\sin ^{-1}\left (\sqrt {-\frac {d}{c}} x\right )|\frac {b c}{a d}\right )}{\sqrt {-\frac {d}{c}} \sqrt {\frac {a+b x^2}{a}} \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-a - b*x^2]/Sqrt[c + d*x^2],x]

[Out]

(Sqrt[-a - b*x^2]*Sqrt[(c + d*x^2)/c]*EllipticE[ArcSin[Sqrt[-(d/c)]*x], (b*c)/(a*d)])/(Sqrt[-(d/c)]*Sqrt[(a +
b*x^2)/a]*Sqrt[c + d*x^2])

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Maple [A]
time = 0.08, size = 104, normalized size = 0.51

method	result	size
default	\(\frac {\sqrt {-b \,x^{2}-a}\, \sqrt {d \,x^{2}+c}\, a \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {b c}{a d}}\right )}{\left (b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c \right ) \sqrt {-\frac {d}{c}}}\)	\(104\)
elliptic	\(\frac {\sqrt {-\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (-\frac {a \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1-\frac {-a d -b c}{a d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {-b d \,x^{4}-a d \,x^{2}-c \,x^{2} b -a c}}+\frac {a \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1-\frac {-a d -b c}{a d}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1-\frac {-a d -b c}{a d}}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {-b d \,x^{4}-a d \,x^{2}-c \,x^{2} b -a c}}\right )}{\sqrt {-b \,x^{2}-a}\, \sqrt {d \,x^{2}+c}}\)	\(266\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^2-a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-b*x^2-a)^(1/2)*(d*x^2+c)^(1/2)*a*((d*x^2+c)/c)^(1/2)*((b*x^2+a)/a)^(1/2)*EllipticE(x*(-d/c)^(1/2),(b*c/a/d)^
(1/2))/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/(-d/c)^(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((-b*x^2-a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-b*x^2 - a)/sqrt(d*x^2 + c), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^2-a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- a - b x^{2}}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**2-a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(sqrt(-a - b*x**2)/sqrt(c + d*x**2), 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((-b*x^2-a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-b*x^2 - a)/sqrt(d*x^2 + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {-b\,x^2-a}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- a - b*x^2)^(1/2)/(c + d*x^2)^(1/2),x)

[Out]

int((- a - b*x^2)^(1/2)/(c + d*x^2)^(1/2), x)

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