3.10.68 \(\int \frac {x^2 \sqrt {2+b x^2}}{\sqrt {3+d x^2}} \, dx\) [968]

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

[Out]

-2/3*(3*b-d)*x*(b*x^2+2)^(1/2)/b/d/(d*x^2+3)^(1/2)+2/3*(3*b-d)*(1/(3*d*x^2+9))^(1/2)*(3*d*x^2+9)^(1/2)*Ellipti
cE(x*d^(1/2)*3^(1/2)/(3*d*x^2+9)^(1/2),1/2*(4-6*b/d)^(1/2))*2^(1/2)*(b*x^2+2)^(1/2)/b/d^(3/2)/((b*x^2+2)/(d*x^
2+3))^(1/2)/(d*x^2+3)^(1/2)-(1/(3*d*x^2+9))^(1/2)*(3*d*x^2+9)^(1/2)*EllipticF(x*d^(1/2)*3^(1/2)/(3*d*x^2+9)^(1
/2),1/2*(4-6*b/d)^(1/2))*2^(1/2)*(b*x^2+2)^(1/2)/d^(3/2)/((b*x^2+2)/(d*x^2+3))^(1/2)/(d*x^2+3)^(1/2)+1/3*x*(b*
x^2+2)^(1/2)*(d*x^2+3)^(1/2)/d

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Rubi [A]
time = 0.10, antiderivative size = 241, normalized size of antiderivative = 1.00, 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 = {489, 545, 429, 506, 422} \begin {gather*} -\frac {\sqrt {2} \sqrt {b x^2+2} F\left (\text {ArcTan}\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 (\text {ArcTan}\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}} \end {gather*}

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 422

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

Rule 429

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

Rule 489

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 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 545

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 {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.82, size = 127, normalized size = 0.53 \begin {gather*} \frac {\sqrt {b} d x \sqrt {2+b x^2} \sqrt {3+d x^2}+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 )-2 i \sqrt {3} (3 b-2 d) F\left (i \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|\frac {2 d}{3 b}\right )}{3 \sqrt {b} d^2} \end {gather*}

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 = 0.15, size = 306, normalized size = 1.27

method	result	size
risch	\(\frac {x \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}{3 d}-\frac {2 \left (\frac {3 \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{2 \sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}-\frac {\left (3 b -d \right ) \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}\, b}\right ) \sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}}{3 d \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}\)	\(259\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}\, \left (\frac {x \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}{3 d}-\frac {\sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{d \sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}-\frac {\left (2-\frac {6 b +4 d}{3 d}\right ) \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}\, b}\right )}{\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}\)	\(268\)
default	\(\frac {\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, \left (b^{2} d \,x^{5} \sqrt {-d}+3 b^{2} x^{3} \sqrt {-d}+2 b d \,x^{3} \sqrt {-d}+3 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}-2 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}-6 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}+2 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}+6 b x \sqrt {-d}\right )}{3 \left (b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6\right ) d \sqrt {-d}\, b}\)	\(306\)

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,method=_RETURNVERBOSE)

[Out]

1/3*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)*(b^2*d*x^5*(-d)^(1/2)+3*b^2*x^3*(-d)^(1/2)+2*b*d*x^3*(-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)*(b/d)^(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)*(b/d)^(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)*(b/d)^(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)*(b/d)^(1/2))*d*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)+6*
b*x*(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {b x^{2} + 2}}{\sqrt {d x^{2} + 3}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,\sqrt {b\,x^2+2}}{\sqrt {d\,x^2+3}} \,d x \end {gather*}

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]

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

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