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

Optimal. Leaf size=241 \[ -\frac{\sqrt{2} \sqrt{b x^2+2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right ),1-\frac{3 b}{2 d}\right )}{d^{3/2} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}+\frac{2 \sqrt{2} (3 b-d) \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{3 b d^{3/2} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}+\frac{x \sqrt{b x^2+2} \sqrt{d x^2+3}}{3 d}-\frac{2 x (3 b-d) \sqrt{b x^2+2}}{3 b d \sqrt{d x^2+3}} \]

[Out]

(-2*(3*b - d)*x*Sqrt[2 + b*x^2])/(3*b*d*Sqrt[3 + d*x^2]) + (x*Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2])/(3*d) + (2*Sqrt
[2]*(3*b - d)*Sqrt[2 + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(3*b*d^(3/2)*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)])/(d^(3/2)*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqrt[3 + d*x^2])

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Rubi [A]  time = 0.149874, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {478, 531, 418, 492, 411} \[ -\frac{\sqrt{2} \sqrt{b x^2+2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{d^{3/2} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}+\frac{2 \sqrt{2} (3 b-d) \sqrt{b x^2+2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{3 b d^{3/2} \sqrt{d x^2+3} \sqrt{\frac{b x^2+2}{d x^2+3}}}+\frac{x \sqrt{b x^2+2} \sqrt{d x^2+3}}{3 d}-\frac{2 x (3 b-d) \sqrt{b x^2+2}}{3 b d \sqrt{d x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*b - d)*x*Sqrt[2 + b*x^2])/(3*b*d*Sqrt[3 + d*x^2]) + (x*Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2])/(3*d) + (2*Sqrt
[2]*(3*b - d)*Sqrt[2 + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(3*b*d^(3/2)*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)])/(d^(3/2)*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqrt[3 + d*x^2])

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(m + n*(p + q) + 1)), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

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 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 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]

Rubi steps

\begin{align*} \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}}{3 d}-\frac{\int \frac{6+2 (3 b-d) x^2}{\sqrt{2+b x^2} \sqrt{3+d x^2}} \, dx}{3 d}\\ &=\frac{x \sqrt{2+b x^2} \sqrt{3+d x^2}}{3 d}-\frac{2 \int \frac{1}{\sqrt{2+b x^2} \sqrt{3+d x^2}} \, dx}{d}-\frac{(2 (3 b-d)) \int \frac{x^2}{\sqrt{2+b x^2} \sqrt{3+d x^2}} \, dx}{3 d}\\ &=-\frac{2 (3 b-d) x \sqrt{2+b x^2}}{3 b d \sqrt{3+d x^2}}+\frac{x \sqrt{2+b x^2} \sqrt{3+d x^2}}{3 d}-\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 )}{d^{3/2} \sqrt{\frac{2+b x^2}{3+d x^2}} \sqrt{3+d x^2}}+\frac{(2 (3 b-d)) \int \frac{\sqrt{2+b x^2}}{\left (3+d x^2\right )^{3/2}} \, dx}{b d}\\ &=-\frac{2 (3 b-d) x \sqrt{2+b x^2}}{3 b d \sqrt{3+d x^2}}+\frac{x \sqrt{2+b x^2} \sqrt{3+d x^2}}{3 d}+\frac{2 \sqrt{2} (3 b-d) \sqrt{2+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{3}}\right )|1-\frac{3 b}{2 d}\right )}{3 b d^{3/2} \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 )}{d^{3/2} \sqrt{\frac{2+b x^2}{3+d x^2}} \sqrt{3+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.112274, size = 127, normalized size = 0.53 \[ \frac{-2 i \sqrt{3} (3 b-2 d) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{2}}\right ),\frac{2 d}{3 b}\right )+\sqrt{b} d x \sqrt{b x^2+2} \sqrt{d x^2+3}+2 i \sqrt{3} (3 b-d) E\left (i \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{2}}\right )|\frac{2 d}{3 b}\right )}{3 \sqrt{b} d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 306, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( 3\,bd{x}^{4}+9\,b{x}^{2}+6\,d{x}^{2}+18 \right ) db}\sqrt{b{x}^{2}+2}\sqrt{d{x}^{2}+3} \left ({x}^{5}{b}^{2}d\sqrt{-d}+3\,{x}^{3}{b}^{2}\sqrt{-d}+2\,{x}^{3}bd\sqrt{-d}+3\,\sqrt{2}{\it EllipticF} \left ( 1/3\,x\sqrt{3}\sqrt{-d},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{b}{d}}} \right ) b\sqrt{b{x}^{2}+2}\sqrt{d{x}^{2}+3}-2\,\sqrt{2}{\it EllipticF} \left ( 1/3\,x\sqrt{3}\sqrt{-d},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{b}{d}}} \right ) d\sqrt{b{x}^{2}+2}\sqrt{d{x}^{2}+3}-6\,\sqrt{2}{\it EllipticE} \left ( 1/3\,x\sqrt{3}\sqrt{-d},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{b}{d}}} \right ) b\sqrt{b{x}^{2}+2}\sqrt{d{x}^{2}+3}+2\,\sqrt{2}{\it EllipticE} \left ( 1/3\,x\sqrt{3}\sqrt{-d},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{b}{d}}} \right ) d\sqrt{b{x}^{2}+2}\sqrt{d{x}^{2}+3}+6\,xb\sqrt{-d} \right ){\frac{1}{\sqrt{-d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)*(x^5*b^2*d*(-d)^(1/2)+3*x^3*b^2*(-d)^(1/2)+2*x^3*b*d*(-d)^(1/2)+3*2^(1/2)*
EllipticF(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(1/2))*b*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)-2*2^(1
/2)*EllipticF(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(1/2))*d*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)-6*
2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(1/2))*b*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2
)+2*2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(1/2))*d*(b*x^2+2)^(1/2)*(d*x^2+3)^
(1/2)+6*x*b*(-d)^(1/2))/(b*d*x^4+3*b*x^2+2*d*x^2+6)/d/(-d)^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{2} + 2} x^{2}}{\sqrt{d x^{2} + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + 2} x^{2}}{\sqrt{d x^{2} + 3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{b x^{2} + 2}}{\sqrt{d x^{2} + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{2} + 2} x^{2}}{\sqrt{d x^{2} + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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