3.94 \(\int \frac {\sqrt {2+d x^2} \sqrt {3+f x^2}}{a+b x^2} \, dx\)

Optimal. Leaf size=298 \[ \frac {3 \sqrt {d x^2+2} (2 b-a d) \Pi \left (1-\frac {3 b}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{\sqrt {2} a b \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+\frac {f x \sqrt {d x^2+2}}{b \sqrt {f x^2+3}}+\frac {3 d \sqrt {d x^2+2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{\sqrt {2} b \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}-\frac {\sqrt {2} \sqrt {f} \sqrt {d x^2+2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{b \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {535, 422, 418, 492, 411, 539} \[ \frac {3 \sqrt {d x^2+2} (2 b-a d) \Pi \left (1-\frac {3 b}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{\sqrt {2} a b \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+\frac {f x \sqrt {d x^2+2}}{b \sqrt {f x^2+3}}+\frac {3 d \sqrt {d x^2+2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{\sqrt {2} b \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}-\frac {\sqrt {2} \sqrt {f} \sqrt {d x^2+2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{b \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*x*Sqrt[2 + d*x^2])/(b*Sqrt[3 + f*x^2]) - (Sqrt[2]*Sqrt[f]*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt
[3]], 1 - (3*d)/(2*f)])/(b*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) + (3*d*Sqrt[2 + d*x^2]*EllipticF[Arc
Tan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(Sqrt[2]*b*Sqrt[f]*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])
+ (3*(2*b - a*d)*Sqrt[2 + d*x^2]*EllipticPi[1 - (3*b)/(a*f), ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(S
qrt[2]*a*b*Sqrt[f]*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

Rule 411

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

Rule 418

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

Rule 422

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[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]

Rule 535

Int[(Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2])/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/b, Int[Sq
rt[e + f*x^2]/Sqrt[c + d*x^2], x], x] + Dist[(b*c - a*d)/b, Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&  !SimplerSqrtQ[-(f/e), -(d/c)]

Rule 539

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

Rubi steps

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

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Mathematica [C]  time = 0.40, size = 134, normalized size = 0.45 \[ \frac {i \left ((a d-2 b) \left ((3 b-a f) \Pi \left (\frac {2 b}{a d};i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+a f F\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )\right )-3 a b d E\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )\right )}{\sqrt {3} a b^2 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(-3*a*b*d*EllipticE[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)] + (-2*b + a*d)*(a*f*EllipticF[I*ArcSinh[(S
qrt[d]*x)/Sqrt[2]], (2*f)/(3*d)] + (3*b - a*f)*EllipticPi[(2*b)/(a*d), I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(
3*d)])))/(Sqrt[3]*a*b^2*Sqrt[d])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 293, normalized size = 0.98 \[ -\frac {\left (a^{2} d f \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )-a^{2} d f \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {3 b}{a f}, \frac {\sqrt {2}\, \sqrt {-d}\, \sqrt {3}}{2 \sqrt {-f}}\right )-3 a b d \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+3 a b d \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {3 b}{a f}, \frac {\sqrt {2}\, \sqrt {-d}\, \sqrt {3}}{2 \sqrt {-f}}\right )-2 a b f \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+2 a b f \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {3 b}{a f}, \frac {\sqrt {2}\, \sqrt {-d}\, \sqrt {3}}{2 \sqrt {-f}}\right )-6 b^{2} \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {3 b}{a f}, \frac {\sqrt {2}\, \sqrt {-d}\, \sqrt {3}}{2 \sqrt {-f}}\right )\right ) \sqrt {2}}{2 \sqrt {-f}\, a \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(EllipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*a^2*d*f-a^2*EllipticPi(1/3*3^(1/2)*(
-f)^(1/2)*x,3*b/a/f,1/2*2^(1/2)*(-d)^(1/2)*3^(1/2)/(-f)^(1/2))*d*f-3*EllipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^
(1/2)*2^(1/2)*(d/f)^(1/2))*d*b*a-2*f*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*a+3
*EllipticPi(1/3*3^(1/2)*(-f)^(1/2)*x,3*b/a/f,1/2*2^(1/2)*(-d)^(1/2)*3^(1/2)/(-f)^(1/2))*d*b*a+2*EllipticPi(1/3
*3^(1/2)*(-f)^(1/2)*x,3*b/a/f,1/2*2^(1/2)*(-d)^(1/2)*3^(1/2)/(-f)^(1/2))*f*b*a-6*EllipticPi(1/3*3^(1/2)*(-f)^(
1/2)*x,3*b/a/f,1/2*2^(1/2)*(-d)^(1/2)*3^(1/2)/(-f)^(1/2))*b^2)*2^(1/2)/a/(-f)^(1/2)/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {d\,x^2+2}\,\sqrt {f\,x^2+3}}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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