3.3.95 \(\int \frac {\sqrt {2+b x^2}}{\sqrt {3+d x^2}} \, dx\) [295]

Optimal. Leaf size=182 \[ \frac {x \sqrt {2+b x^2}}{\sqrt {3+d x^2}}-\frac {\sqrt {2} \sqrt {2+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}+\frac {\sqrt {2} \sqrt {2+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}} \]

[Out]

x*(b*x^2+2)^(1/2)/(d*x^2+3)^(1/2)-(1/(3*d*x^2+9))^(1/2)*(3*d*x^2+9)^(1/2)*EllipticE(x*d^(1/2)*3^(1/2)/(3*d*x^2
+9)^(1/2),1/2*(4-6*b/d)^(1/2))*2^(1/2)*(b*x^2+2)^(1/2)/d^(1/2)/((b*x^2+2)/(d*x^2+3))^(1/2)/(d*x^2+3)^(1/2)+(1/
(3*d*x^2+9))^(1/2)*(3*d*x^2+9)^(1/2)*EllipticF(x*d^(1/2)*3^(1/2)/(3*d*x^2+9)^(1/2),1/2*(4-6*b/d)^(1/2))*2^(1/2
)*(b*x^2+2)^(1/2)/d^(1/2)/((b*x^2+2)/(d*x^2+3))^(1/2)/(d*x^2+3)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {433, 429, 506, 422} \begin {gather*} \frac {\sqrt {2} \sqrt {b x^2+2} F\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}-\frac {\sqrt {2} \sqrt {b x^2+2} E\left (\text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}+\frac {x \sqrt {b x^2+2}}{\sqrt {d x^2+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + b*x^2]/Sqrt[3 + d*x^2],x]

[Out]

(x*Sqrt[2 + b*x^2])/Sqrt[3 + d*x^2] - (Sqrt[2]*Sqrt[2 + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b
)/(2*d)])/(Sqrt[d]*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqrt[3 + d*x^2]) + (Sqrt[2]*Sqrt[2 + b*x^2]*EllipticF[ArcTan[
(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(Sqrt[d]*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqrt[3 + 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 {2+b x^2}}{\sqrt {3+d x^2}} \, dx &=2 \int \frac {1}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx+b \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx\\ &=\frac {x \sqrt {2+b x^2}}{\sqrt {3+d x^2}}+\frac {\sqrt {2} \sqrt {2+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}-3 \int \frac {\sqrt {2+b x^2}}{\left (3+d x^2\right )^{3/2}} \, dx\\ &=\frac {x \sqrt {2+b x^2}}{\sqrt {3+d x^2}}-\frac {\sqrt {2} \sqrt {2+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}+\frac {\sqrt {2} \sqrt {2+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 37, normalized size = 0.20 \begin {gather*} \frac {\sqrt {2} E\left (\sin ^{-1}\left (\frac {\sqrt {-d} x}{\sqrt {3}}\right )|\frac {3 b}{2 d}\right )}{\sqrt {-d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + b*x^2]/Sqrt[3 + d*x^2],x]

[Out]

(Sqrt[2]*EllipticE[ArcSin[(Sqrt[-d]*x)/Sqrt[3]], (3*b)/(2*d)])/Sqrt[-d]

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Maple [A]
time = 0.00, size = 37, normalized size = 0.20

method	result	size
default	\(\frac {\EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) \sqrt {2}}{\sqrt {-d}}\)	\(37\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}\, \left (\frac {\sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}-\frac {\sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}\right )}{\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+2)^(1/2)/(d*x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

EllipticE(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(b/d)^(1/2))*2^(1/2)/(-d)^(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+2)^(1/2)/(d*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^2 + 2)/sqrt(d*x^2 + 3), 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+2)^(1/2)/(d*x^2+3)^(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 {b x^{2} + 2}}{\sqrt {d x^{2} + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+2)**(1/2)/(d*x**2+3)**(1/2),x)

[Out]

Integral(sqrt(b*x**2 + 2)/sqrt(d*x**2 + 3), 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+2)^(1/2)/(d*x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*x^2 + 2)/sqrt(d*x^2 + 3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b\,x^2+2}}{\sqrt {d\,x^2+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + 2)^(1/2)/(d*x^2 + 3)^(1/2),x)

[Out]

int((b*x^2 + 2)^(1/2)/(d*x^2 + 3)^(1/2), x)

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