3.3.60 \(\int \frac {\sqrt {a+b x^2}}{\sqrt {-c+d x^2}} \, dx\) [260]

Optimal. Leaf size=88 \[ \frac {\sqrt {c} \sqrt {a+b x^2} \sqrt {1-\frac {d x^2}{c}} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {1+\frac {b x^2}{a}} \sqrt {-c+d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {438, 437, 435} \begin {gather*} \frac {\sqrt {c} \sqrt {a+b x^2} \sqrt {1-\frac {d x^2}{c}} E\left (\text {ArcSin}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {\frac {b x^2}{a}+1} \sqrt {d x^2-c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 438

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2]
, Int[Sqrt[a + b*x^2]/Sqrt[1 + (d/c)*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&  !GtQ[c, 0]

Rubi steps

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

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Mathematica [A]
time = 0.66, size = 88, normalized size = 1.00 \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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(166\) vs. \(2(73)=146\).
time = 0.08, size = 167, normalized size = 1.90

method	result	size
default	\(\frac {\left (-a \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-\frac {a d}{b c}}\right ) d -b c \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-\frac {a d}{b c}}\right )+b c \EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-\frac {a d}{b c}}\right )\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}-c}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {-d \,x^{2}+c}{c}}}{\left (-b d \,x^{4}-a d \,x^{2}+c \,x^{2} b +a c \right ) \sqrt {-\frac {b}{a}}\, d}\)	\(167\)
elliptic	\(\frac {\sqrt {-\left (b \,x^{2}+a \right ) \left (-d \,x^{2}+c \right )}\, \left (\frac {a \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1-\frac {d \,x^{2}}{c}}\, \EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1-\frac {a d -b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}-c \,x^{2} b -a c}}+\frac {b c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1-\frac {d \,x^{2}}{c}}\, \left (\EllipticF \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1-\frac {a d -b c}{c b}}\right )-\EllipticE \left (x \sqrt {-\frac {b}{a}}, \sqrt {-1-\frac {a d -b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}-c \,x^{2} b -a c}\, d}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}-c}}\)	\(264\)

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]

(-a*EllipticF(x*(-b/a)^(1/2),(-a*d/b/c)^(1/2))*d-b*c*EllipticF(x*(-b/a)^(1/2),(-a*d/b/c)^(1/2))+b*c*EllipticE(
x*(-b/a)^(1/2),(-a*d/b/c)^(1/2)))*(b*x^2+a)^(1/2)*(d*x^2-c)^(1/2)*((b*x^2+a)/a)^(1/2)*((-d*x^2+c)/c)^(1/2)/(-b
*d*x^4-a*d*x^2+b*c*x^2+a*c)/(-b/a)^(1/2)/d

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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 = -0.01 \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)/(d*x^2 - c)^(1/2),x)

[Out]

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

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