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

Optimal. Leaf size=262 \[ -\frac{\sqrt{2} \sqrt{d x^2+2} (b-a f) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right ),1-\frac{3 d}{2 f}\right )}{f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{\sqrt{2} \sqrt{d x^2+2} (-3 a d f+6 b d-2 b f) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{3 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}-\frac{x \sqrt{d x^2+2} (-3 a d f+6 b d-2 b f)}{3 d f \sqrt{f x^2+3}}+\frac{b x \sqrt{d x^2+2} \sqrt{f x^2+3}}{3 f} \]

[Out]

-((6*b*d - 2*b*f - 3*a*d*f)*x*Sqrt[2 + d*x^2])/(3*d*f*Sqrt[3 + f*x^2]) + (b*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2])
/(3*f) + (Sqrt[2]*(6*b*d - 2*b*f - 3*a*d*f)*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(
2*f)])/(3*d*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(b - a*f)*Sqrt[2 + d*x^2]*Ellipt
icF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

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Rubi [A]  time = 0.167557, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {528, 531, 418, 492, 411} \[ -\frac{\sqrt{2} \sqrt{d x^2+2} (b-a f) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{\sqrt{2} \sqrt{d x^2+2} (-3 a d f+6 b d-2 b f) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{3 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}-\frac{x \sqrt{d x^2+2} (-3 a d f+6 b d-2 b f)}{3 d f \sqrt{f x^2+3}}+\frac{b x \sqrt{d x^2+2} \sqrt{f x^2+3}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((6*b*d - 2*b*f - 3*a*d*f)*x*Sqrt[2 + d*x^2])/(3*d*f*Sqrt[3 + f*x^2]) + (b*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2])
/(3*f) + (Sqrt[2]*(6*b*d - 2*b*f - 3*a*d*f)*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(
2*f)])/(3*d*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(b - a*f)*Sqrt[2 + d*x^2]*Ellipt
icF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

Rule 528

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

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

Mathematica [C]  time = 0.201503, size = 142, normalized size = 0.54 \[ \frac{i \sqrt{3} (3 d-2 f) (a f-2 b) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right ),\frac{2 f}{3 d}\right )+i \sqrt{3} (-3 a d f+6 b d-2 b f) E\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right )|\frac{2 f}{3 d}\right )+b \sqrt{d} f x \sqrt{d x^2+2} \sqrt{f x^2+3}}{3 \sqrt{d} f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*f*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2] + I*Sqrt[3]*(6*b*d - 2*b*f - 3*a*d*f)*EllipticE[I*ArcSinh[(Sqrt
[d]*x)/Sqrt[2]], (2*f)/(3*d)] + I*Sqrt[3]*(3*d - 2*f)*(-2*b + a*f)*EllipticF[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (
2*f)/(3*d)])/(3*Sqrt[d]*f^2)

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Maple [A]  time = 0.02, size = 367, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( 3\,df{x}^{4}+9\,d{x}^{2}+6\,f{x}^{2}+18 \right ) fd}\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3} \left ({x}^{5}b{d}^{2}f\sqrt{-f}+3\,\sqrt{2}{\it EllipticE} \left ( 1/3\,x\sqrt{3}\sqrt{-f},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{d}{f}}} \right ) adf\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3}+3\,{x}^{3}b{d}^{2}\sqrt{-f}+2\,{x}^{3}bdf\sqrt{-f}-6\,\sqrt{2}{\it EllipticE} \left ( 1/3\,x\sqrt{3}\sqrt{-f},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{d}{f}}} \right ) bd\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3}+2\,\sqrt{2}{\it EllipticE} \left ( 1/3\,x\sqrt{3}\sqrt{-f},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{d}{f}}} \right ) bf\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3}+3\,\sqrt{2}{\it EllipticF} \left ( 1/3\,x\sqrt{3}\sqrt{-f},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{d}{f}}} \right ) bd\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3}-2\,\sqrt{2}{\it EllipticF} \left ( 1/3\,x\sqrt{3}\sqrt{-f},1/2\,\sqrt{2}\sqrt{3}\sqrt{{\frac{d}{f}}} \right ) bf\sqrt{d{x}^{2}+2}\sqrt{f{x}^{2}+3}+6\,xbd\sqrt{-f} \right ){\frac{1}{\sqrt{-f}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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