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

Optimal. Leaf size=356 \[ -\frac{\sqrt{2} \sqrt{d x^2+2} (-10 a d f+3 b d+2 b f) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right ),1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt{f x^2+3}}-\frac{\sqrt{2} \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \sqrt{f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac{b x \left (d x^2+2\right )^{3/2} \sqrt{f x^2+3}}{5 d} \]

[Out]

((5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*x*Sqrt[2 + d*x^2])/(15*d^2*f*Sqrt[3 + f*x^2]) + ((3*b*d -
4*b*f + 5*a*d*f)*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2])/(15*d*f) + (b*x*(2 + d*x^2)^(3/2)*Sqrt[3 + f*x^2])/(5*d)
- (Sqrt[2]*(5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sq
rt[3]], 1 - (3*d)/(2*f)])/(15*d^2*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(3*b*d + 2
*b*f - 10*a*d*f)*Sqrt[2 + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(5*d*f^(3/2)*Sqrt[(2
+ d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])

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Rubi [A]  time = 0.322451, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{x \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt{f x^2+3}}-\frac{\sqrt{2} \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}-\frac{\sqrt{2} \sqrt{d x^2+2} (-10 a d f+3 b d+2 b f) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \sqrt{f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac{b x \left (d x^2+2\right )^{3/2} \sqrt{f x^2+3}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*x*Sqrt[2 + d*x^2])/(15*d^2*f*Sqrt[3 + f*x^2]) + ((3*b*d -
4*b*f + 5*a*d*f)*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2])/(15*d*f) + (b*x*(2 + d*x^2)^(3/2)*Sqrt[3 + f*x^2])/(5*d)
- (Sqrt[2]*(5*a*d*f*(3*d + 2*f) - 2*b*(9*d^2 - 6*d*f + 4*f^2))*Sqrt[2 + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sq
rt[3]], 1 - (3*d)/(2*f)])/(15*d^2*f^(3/2)*Sqrt[(2 + d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2]) - (Sqrt[2]*(3*b*d + 2
*b*f - 10*a*d*f)*Sqrt[2 + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(5*d*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 \left (a+b x^2\right ) \sqrt{2+d x^2} \sqrt{3+f x^2} \, dx &=\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}+\frac{\int \frac{\sqrt{2+d x^2} \left (-3 (2 b-5 a d)+(3 b d-4 b f+5 a d f) x^2\right )}{\sqrt{3+f x^2}} \, dx}{5 d}\\ &=\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}+\frac{\int \frac{-6 (3 b d+2 b f-10 a d f)+\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x^2}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{15 d f}\\ &=\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{(2 (3 b d+2 b f-10 a d f)) \int \frac{1}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{5 d f}+\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac{x^2}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{15 d f}\\ &=\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt{2+d x^2}}{15 d^2 f \sqrt{3+f x^2}}+\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{\sqrt{2} (3 b d+2 b f-10 a d 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 )}{5 d f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}-\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac{\sqrt{2+d x^2}}{\left (3+f x^2\right )^{3/2}} \, dx}{5 d^2 f}\\ &=\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt{2+d x^2}}{15 d^2 f \sqrt{3+f x^2}}+\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{\sqrt{2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \sqrt{2+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}-\frac{\sqrt{2} (3 b d+2 b f-10 a d 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 )}{5 d f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}\\ \end{align*}

Mathematica [C]  time = 0.323886, size = 186, normalized size = 0.52 \[ \frac{i \sqrt{3} (3 d-2 f) (5 a d f-6 b d+2 b f) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right ),\frac{2 f}{3 d}\right )+i \sqrt{3} \left (2 b \left (9 d^2-6 d f+4 f^2\right )-5 a d f (3 d+2 f)\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right )|\frac{2 f}{3 d}\right )+\sqrt{d} f x \sqrt{d x^2+2} \sqrt{f x^2+3} \left (5 a d f+3 b d \left (f x^2+1\right )+2 b f\right )}{15 d^{3/2} f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*f*x*Sqrt[2 + d*x^2]*Sqrt[3 + f*x^2]*(2*b*f + 5*a*d*f + 3*b*d*(1 + f*x^2)) + I*Sqrt[3]*(-5*a*d*f*(3*d
+ 2*f) + 2*b*(9*d^2 - 6*d*f + 4*f^2))*EllipticE[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)] + I*Sqrt[3]*(3*d
- 2*f)*(-6*b*d + 2*b*f + 5*a*d*f)*EllipticF[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)])/(15*d^(3/2)*f^2)

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Maple [B]  time = 0.023, size = 775, normalized size = 2.2 \begin{align*} \text{result too large to display} \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]

1/15*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)*(3*x^7*b*d^3*f^2*(-f)^(1/2)+5*x^5*a*d^3*f^2*(-f)^(1/2)+12*x^5*b*d^3*f*(-f
)^(1/2)+8*x^5*b*d^2*f^2*(-f)^(1/2)+15*x^3*a*d^3*f*(-f)^(1/2)+10*x^3*a*d^2*f^2*(-f)^(1/2)+15*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))*a*d^2*f*(f*x^2+3)^(1/2)*(d*x^2+2)^(1/2)-10*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))*a*d*f^2*(f*x^2+3)^(1/2)*(d*x^2+2)^(1/2)
+15*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^2*f*(f*x^2+3)^(1/2)*(d*x
^2+2)^(1/2)+10*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^2*(f*x^2+3)
^(1/2)*(d*x^2+2)^(1/2)+9*x^3*b*d^3*(-f)^(1/2)+30*x^3*b*d^2*f*(-f)^(1/2)+4*x^3*b*d*f^2*(-f)^(1/2)+9*2^(1/2)*Ell
ipticF(1/3*x*3^(1/2)*(-f)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/f*d)^(1/2))*b*d^2*(f*x^2+3)^(1/2)*(d*x^2+2)^(1/2)-18*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*f*(f*x^2+3)^(1/2)*(d*x^2+2)^(1
/2)+8*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^2*(f*x^2+3)^(1/2)*(d*x
^2+2)^(1/2)-18*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^2*(f*x^2+3)^(
1/2)*(d*x^2+2)^(1/2)+12*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*f*(f
*x^2+3)^(1/2)*(d*x^2+2)^(1/2)-8*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^2*(f*x^2+3)^(1/2)*(d*x^2+2)^(1/2)+30*x*a*d^2*f*(-f)^(1/2)+18*x*b*d^2*(-f)^(1/2)+12*x*b*d*f*(-f)^(1/2))/(d*
f*x^4+3*d*x^2+2*f*x^2+6)/d^2/f/(-f)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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 ({\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 \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{\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)