3.167 \(\int \sqrt {a+b x^2} \sqrt {c+d x^2} \, dx\)

Optimal. Leaf size=249 \[ \frac {2 c^{3/2} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {x \sqrt {a+b x^2} (a d+b c)}{3 b \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

1/3*(a*d+b*c)*x*(b*x^2+a)^(1/2)/b/(d*x^2+c)^(1/2)+2/3*c^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*Elliptic
F(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*(b*x^2+a)^(1/2)/d^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/
2)/(d*x^2+c)^(1/2)-1/3*(a*d+b*c)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/
c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*(b*x^2+a)^(1/2)/b/d^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+
1/3*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 249, 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} \[ \frac {2 c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {x \sqrt {a+b x^2} (a d+b c)}{3 b \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} (a d+b c) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c + a*d)*x*Sqrt[a + b*x^2])/(3*b*Sqrt[c + d*x^2]) + (x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/3 - (Sqrt[c]*(b*c
+ a*d)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*b*Sqrt[d]*Sqrt[(c*(a + b*x^
2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (2*c^(3/2)*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 -
(b*c)/(a*d)])/(3*Sqrt[d]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + 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 {a+b x^2} \sqrt {c+d x^2} \, dx &=\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2}{3} \int \frac {a c+\frac {1}{2} (b c+a d) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\\ &=\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {1}{3} (2 a c) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx+\frac {1}{3} (b c+a d) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\\ &=\frac {(b c+a d) x \sqrt {a+b x^2}}{3 b \sqrt {c+d x^2}}+\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}+\frac {2 c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}-\frac {(c (b c+a d)) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 b}\\ &=\frac {(b c+a d) x \sqrt {a+b x^2}}{3 b \sqrt {c+d x^2}}+\frac {1}{3} x \sqrt {a+b x^2} \sqrt {c+d x^2}-\frac {\sqrt {c} (b c+a d) \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {2 c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 198, normalized size = 0.80 \[ \frac {-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-b c) \operatorname {EllipticF}\left (i \sinh ^{-1}\left (x \sqrt {\frac {b}{a}}\right ),\frac {a d}{b c}\right )+d x \sqrt {\frac {b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d+b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{3 d \sqrt {\frac {b}{a}} \sqrt {a+b x^2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]*Sqrt[c + d*x^2],x]

[Out]

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2) - I*c*(b*c + a*d)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*A
rcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*c*(-(b*c) + a*d)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*Arc
Sinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(3*Sqrt[b/a]*d*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*x^2 + a)*sqrt(d*x^2 + c), x)

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maple [A]  time = 0.02, size = 328, normalized size = 1.32 \[ \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (\sqrt {-\frac {b}{a}}\, b \,d^{2} x^{5}+\sqrt {-\frac {b}{a}}\, a \,d^{2} x^{3}+\sqrt {-\frac {b}{a}}\, b c d \,x^{3}+\sqrt {-\frac {b}{a}}\, a c d x +\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{3 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right ) \sqrt {-\frac {b}{a}}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*((-1/a*b)^(1/2)*x^5*b*d^2+(-1/a*b)^(1/2)*x^3*a*d^2+(-1/a*b)^(1/2)*x^3*b*c*
d+a*c*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*d-((b*x^2+a)/a)^(1/2
)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b*c^2+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2
)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*c*d+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)
^(1/2)*x,(a/b/c*d)^(1/2))*b*c^2+(-1/a*b)^(1/2)*x*a*c*d)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)/(-1/a*b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^2 + a)*sqrt(d*x^2 + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/2)*(d*x**2+c)**(1/2),x)

[Out]

Integral(sqrt(a + b*x**2)*sqrt(c + d*x**2), x)

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