3.2.78 \(\int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx\) [178]

Optimal. Leaf size=235 \[ \frac {(3 b+2 d) x \sqrt {2+b x^2}}{3 b \sqrt {3+d x^2}}+\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}-\frac {\sqrt {2} (3 b+2 d) \sqrt {2+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{3 b \sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}+\frac {2 \sqrt {2} \sqrt {2+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}} \]

[Out]

1/3*(3*b+2*d)*x*(b*x^2+2)^(1/2)/b/(d*x^2+3)^(1/2)-1/3*(3*b+2*d)*(1/(3*d*x^2+9))^(1/2)*(3*d*x^2+9)^(1/2)*Ellipt
icE(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^(1/2)/((b*x^2+2)/(d*x
^2+3))^(1/2)/(d*x^2+3)^(1/2)+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^(1/2)/((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)

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Rubi [A]
time = 0.09, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {428, 545, 429, 506, 422} \begin {gather*} \frac {2 \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 )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}-\frac {\sqrt {2} (3 b+2 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 \sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}+\frac {x (3 b+2 d) \sqrt {b x^2+2}}{3 b \sqrt {d x^2+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b + 2*d)*x*Sqrt[2 + b*x^2])/(3*b*Sqrt[3 + d*x^2]) + (x*Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2])/3 - (Sqrt[2]*(3*b
+ 2*d)*Sqrt[2 + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2*d)])/(3*b*Sqrt[d]*Sqrt[(2 + b*x^2)/
(3 + d*x^2)]*Sqrt[3 + d*x^2]) + (2*Sqrt[2]*Sqrt[2 + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[3]], 1 - (3*b)/(2
*d)])/(Sqrt[d]*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 428

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

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 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 \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx &=\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}+\frac {2}{3} \int \frac {6+\frac {1}{2} (3 b+2 d) x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx\\ &=\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}+4 \int \frac {1}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx+\frac {1}{3} (3 b+2 d) \int \frac {x^2}{\sqrt {2+b x^2} \sqrt {3+d x^2}} \, dx\\ &=\frac {(3 b+2 d) x \sqrt {2+b x^2}}{3 b \sqrt {3+d x^2}}+\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}+\frac {2 \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 )}{\sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}+\frac {(-3 b-2 d) \int \frac {\sqrt {2+b x^2}}{\left (3+d x^2\right )^{3/2}} \, dx}{b}\\ &=\frac {(3 b+2 d) x \sqrt {2+b x^2}}{3 b \sqrt {3+d x^2}}+\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}-\frac {\sqrt {2} (3 b+2 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 \sqrt {d} \sqrt {\frac {2+b x^2}{3+d x^2}} \sqrt {3+d x^2}}+\frac {2 \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 )}{\sqrt {d} \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 = 0.96, size = 127, normalized size = 0.54 \begin {gather*} \frac {\sqrt {b} d x \sqrt {2+b x^2} \sqrt {3+d x^2}-i \sqrt {3} (3 b+2 d) E\left (i \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|\frac {2 d}{3 b}\right )+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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [A]
time = 0.11, size = 303, normalized size = 1.29

method	result	size
risch	\(\frac {x \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}{3}+\frac {\left (\frac {2 \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}-\frac {\left (3 b +2 d \right ) \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{3 \sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}\, b}\right ) \sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}}{\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}\)	\(252\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+2\right ) \left (d \,x^{2}+3\right )}\, \left (\frac {x \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}{3}+\frac {2 \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}}-\frac {\left (b +\frac {2 d}{3}\right ) \sqrt {3 d \,x^{2}+9}\, \sqrt {2 b \,x^{2}+4}\, \left (\EllipticF \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}\, b}\right )}{\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}}\)	\(253\)
default	\(\frac {\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, \left (b^{2} d \,x^{5} \sqrt {-d}+3 b^{2} x^{3} \sqrt {-d}+2 b d \,x^{3} \sqrt {-d}+3 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}-2 \sqrt {2}\, \EllipticF \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}+3 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}+2 \sqrt {2}\, \EllipticE \left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}+6 b x \sqrt {-d}\right )}{3 \left (b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6\right ) \sqrt {-d}\, b}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*(d*x^2+3)^(1/2)*(b*x^2+2)^(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*(d*x^2+3)^(1/2)*(b*x^2+2)^(1/2)+3*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*(d*x^2+3)^(1/2)*(b*x^2+2)^(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*(d*x^2+3)^(1/2)*(b*x^2+2)^(1/2)+6*
b*x*(-d)^(1/2))/(b*d*x^4+3*b*x^2+2*d*x^2+6)/(-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((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^2 + 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((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 \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((b*x**2+2)**(1/2)*(d*x**2+3)**(1/2),x)

[Out]

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

[Out]

integrate(sqrt(b*x^2 + 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 \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((b*x^2 + 2)^(1/2)*(d*x^2 + 3)^(1/2),x)

[Out]

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

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