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

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

[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 = 0.13, 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 = {417, 531, 418, 492, 411} \[ \frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}+\frac {x (3 b+2 d) \sqrt {b x^2+2}}{3 b \sqrt {d x^2+3}}+\frac {2 \sqrt {2} \sqrt {b x^2+2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}-\frac {\sqrt {2} (3 b+2 d) \sqrt {b x^2+2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{3 b \sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}} \]

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 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 417

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 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 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 \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]  time = 0.11, size = 127, normalized size = 0.54 \[ \frac {i \sqrt {3} (3 b-2 d) \operatorname {EllipticF}\left (i \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right ),\frac {2 d}{3 b}\right )+\sqrt {b} d x \sqrt {b x^2+2} \sqrt {d x^2+3}-i \sqrt {3} (3 b+2 d) E\left (i \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|\frac {2 d}{3 b}\right )}{3 \sqrt {b} d} \]

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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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}, x\right ) \]

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]

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 \[ \int \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}\,{d x} \]

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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maple [A]  time = 0.04, size = 303, normalized size = 1.29 \[ \frac {\sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, \left (\sqrt {-d}\, b^{2} d \,x^{5}+3 \sqrt {-d}\, b^{2} x^{3}+2 \sqrt {-d}\, b d \,x^{3}+6 \sqrt {-d}\, b x +3 \sqrt {2}\, \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, b \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-d}\, x}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )+3 \sqrt {2}\, \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, b \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-d}\, x}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )+2 \sqrt {2}\, \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, d \EllipticE \left (\frac {\sqrt {3}\, \sqrt {-d}\, x}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )-2 \sqrt {2}\, \sqrt {b \,x^{2}+2}\, \sqrt {d \,x^{2}+3}\, d \EllipticF \left (\frac {\sqrt {3}\, \sqrt {-d}\, x}{3}, \frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\frac {b}{d}}}{2}\right )\right )}{3 \left (b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6\right ) \sqrt {-d}\, b} \]

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]

1/3*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)*(x^5*b^2*d*(-d)^(1/2)+3*x^3*b^2*(-d)^(1/2)+2*x^3*b*d*(-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)*(1/d*b)^(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)*(1/d*b)^(1/2))*d*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)+3*
2^(1/2)*EllipticE(1/3*x*3^(1/2)*(-d)^(1/2),1/2*2^(1/2)*3^(1/2)*(1/d*b)^(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)*(1/d*b)^(1/2))*d*(b*x^2+2)^(1/2)*(d*x^2+3)^
(1/2)+6*x*b*(-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 \[ \int \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {b\,x^2+2}\,\sqrt {d\,x^2+3} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}\, dx \]

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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