\(\int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx\) [992]

Optimal result

Integrand size = 26, antiderivative size = 110 \[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\frac {x \sqrt {2+b x^2}}{b \sqrt {3+d x^2}}-\frac {\sqrt {2} \sqrt {2+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{b \sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {506, 422} \[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\frac {x \sqrt {b x^2+2}}{b \sqrt {d x^2+3}}-\frac {\sqrt {2} \sqrt {b x^2+2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{b \sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}} \]

[In]

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

[Out]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=-\frac {i \sqrt {3} \left (E\left (i \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|\frac {2 d}{3 b}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right ),\frac {2 d}{3 b}\right )\right )}{\sqrt {b} d} \]

[In]

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

[Out]

((-I)*Sqrt[3]*(EllipticE[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2]], (2*d)/(3*b)] - EllipticF[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2
]], (2*d)/(3*b)]))/(Sqrt[b]*d)

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {\left (-F\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )+E\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )\right ) \sqrt {2}}{b \sqrt {-d}}\)	\(70\)
elliptic	\(-\frac {\sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}\, \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (F\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-E\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, \sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}\, b}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.03 \[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=-\frac {3 \, \sqrt {3} \sqrt {b d} x \sqrt {-\frac {1}{d}} E(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{d}}}{x}\right )\,|\,\frac {2 \, d}{3 \, b}) - 3 \, \sqrt {3} \sqrt {b d} x \sqrt {-\frac {1}{d}} F(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{d}}}{x}\right )\,|\,\frac {2 \, d}{3 \, b}) - \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3} d}{b d^{2} x} \]

[In]

integrate(x^2/(b*x^2+2)^(1/2)/(d*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

-(3*sqrt(3)*sqrt(b*d)*x*sqrt(-1/d)*elliptic_e(arcsin(sqrt(3)*sqrt(-1/d)/x), 2/3*d/b) - 3*sqrt(3)*sqrt(b*d)*x*s
qrt(-1/d)*elliptic_f(arcsin(sqrt(3)*sqrt(-1/d)/x), 2/3*d/b) - sqrt(b*x^2 + 2)*sqrt(d*x^2 + 3)*d)/(b*d^2*x)

Sympy [F]

\[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\int \frac {x^{2}}{\sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx=\int \frac {x^2}{\sqrt {b\,x^2+2}\,\sqrt {d\,x^2+3}} \,d x \]

[In]

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

[Out]

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