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

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

[Out]

-1/3*(-3*a*d*f+6*b*d-2*b*f)*x*(d*x^2+2)^(1/2)/d/f/(f*x^2+3)^(1/2)+1/3*(-3*a*d*f+6*b*d-2*b*f)*(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)*(d*x^2+2)^(1
/2)/d/f^(3/2)/((d*x^2+2)/(f*x^2+3))^(1/2)/(f*x^2+3)^(1/2)-(-a*f+b)*(1/(3*f*x^2+9))^(1/2)*(3*f*x^2+9)^(1/2)*Ell
ipticF(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)*(d*x^2+2)^(1/2)/f^(3/2)/((d*x^2+2)/(f*
x^2+3))^(1/2)/(f*x^2+3)^(1/2)+1/3*b*x*(d*x^2+2)^(1/2)*(f*x^2+3)^(1/2)/f

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Rubi [A]  time = 0.17, antiderivative size = 262, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.22, size = 142, normalized size = 0.54 \[ \frac {i \sqrt {3} (3 d-2 f) (a f-2 b) F\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+i \sqrt {3} (-3 a d f+6 b d-2 b f) E\left (i \sinh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+b \sqrt {d} f x \sqrt {d x^2+2} \sqrt {f x^2+3}}{3 \sqrt {d} f^2} \]

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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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + 2}}{\sqrt {f x^{2} + 3}}, x\right ) \]

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

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)

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maple [A]  time = 0.02, size = 367, normalized size = 1.40 \[ \frac {\sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, \left (\sqrt {-f}\, b \,d^{2} f \,x^{5}+3 \sqrt {-f}\, b \,d^{2} x^{3}+2 \sqrt {-f}\, b d f \,x^{3}+3 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, a d f \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+6 \sqrt {-f}\, b d x -6 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b d \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+3 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b d \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )+2 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b f \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )-2 \sqrt {2}\, \sqrt {d \,x^{2}+2}\, \sqrt {f \,x^{2}+3}\, b f \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-f}\, x}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right )\right )}{3 \left (d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6\right ) \sqrt {-f}\, d f} \]

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*3^(1/2)*(-f)^(1/2)*x,1/2*3
^(1/2)*2^(1/2)*(d/f)^(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*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d*(d*x^2+2)^(1/2)*(f*x^2+3)^
(1/2)+2*2^(1/2)*EllipticE(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*f*(d*x^2+2)^(1/2)*(f*x^2
+3)^(1/2)+3*2^(1/2)*EllipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(1/2))*b*d*(d*x^2+2)^(1/2)*(f
*x^2+3)^(1/2)-2*2^(1/2)*EllipticF(1/3*3^(1/2)*(-f)^(1/2)*x,1/2*3^(1/2)*2^(1/2)*(d/f)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + 2}}{\sqrt {f x^{2} + 3}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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