3.2.9 \(\int \frac {\sqrt {c+d x^2}}{\sqrt {a+b x^2} (e+f x^2)^{3/2}} \, dx\) [109]

Optimal. Leaf size=319 \[ \frac {(d e-c f) x \sqrt {a+b x^2}}{e (b e-a f) \sqrt {c+d x^2} \sqrt {e+f x^2}}-\frac {\sqrt {c} \sqrt {d e-c f} \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {e+f x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{e (b e-a f) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {e+f x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{a e \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.31, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {568, 433, 429, 506, 422} \begin {gather*} \frac {c^{3/2} \sqrt {a+b x^2} F\left (\text {ArcTan}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {f x^2+e}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{a e \sqrt {c+d x^2} \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {c} \sqrt {a+b x^2} \sqrt {d e-c f} E\left (\text {ArcTan}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {f x^2+e}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{e \sqrt {c+d x^2} (b e-a f) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \sqrt {a+b x^2} (d e-c f)}{e \sqrt {c+d x^2} \sqrt {e+f x^2} (b e-a f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(Sqrt[a + b*x^2]*(e + f*x^2)^(3/2)),x]

[Out]

((d*e - c*f)*x*Sqrt[a + b*x^2])/(e*(b*e - a*f)*Sqrt[c + d*x^2]*Sqrt[e + f*x^2]) - (Sqrt[c]*Sqrt[d*e - c*f]*Sqr
t[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d*e - c*f]*x)/(Sqrt[c]*Sqrt[e + f*x^2])], -(((b*c - a*d)*e)/(a*(d*e - c*f)
))])/(e*(b*e - a*f)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (c^(3/2)*Sqrt[a + b*x^2]*Elliptic
F[ArcTan[(Sqrt[d*e - c*f]*x)/(Sqrt[c]*Sqrt[e + f*x^2])], -(((b*c - a*d)*e)/(a*(d*e - c*f)))])/(a*e*Sqrt[d*e -
c*f]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + 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 433

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

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 568

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)^(3/2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[
c + d*x^2]*(Sqrt[a*((e + f*x^2)/(e*(a + b*x^2)))]/(a*Sqrt[e + f*x^2]*Sqrt[a*((c + d*x^2)/(c*(a + b*x^2)))])),
Subst[Int[Sqrt[1 - (b*c - a*d)*(x^2/c)]/Sqrt[1 - (b*e - a*f)*(x^2/e)], x], x, x/Sqrt[a + b*x^2]], x] /; FreeQ[
{a, b, c, d, e, f}, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^2}}{\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}} \, dx &=\frac {\left (\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{a \left (e+f x^2\right )}}\right ) \text {Subst}\left (\int \frac {\sqrt {1-\frac {(-d e+c f) x^2}{c}}}{\sqrt {1-\frac {(-b e+a f) x^2}{a}}} \, dx,x,\frac {x}{\sqrt {e+f x^2}}\right )}{e \sqrt {a+b x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}\\ &=\frac {\left (\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{a \left (e+f x^2\right )}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {(-b e+a f) x^2}{a}} \sqrt {1-\frac {(-d e+c f) x^2}{c}}} \, dx,x,\frac {x}{\sqrt {e+f x^2}}\right )}{e \sqrt {a+b x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {\left ((-d e+c f) \sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{a \left (e+f x^2\right )}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {(-b e+a f) x^2}{a}} \sqrt {1-\frac {(-d e+c f) x^2}{c}}} \, dx,x,\frac {x}{\sqrt {e+f x^2}}\right )}{c e \sqrt {a+b x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}\\ &=\frac {(d e-c f) x \sqrt {a+b x^2}}{e (b e-a f) \sqrt {c+d x^2} \sqrt {e+f x^2}}+\frac {c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {e+f x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{a e \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\left (a (-d e+c f) \sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{a \left (e+f x^2\right )}}\right ) \text {Subst}\left (\int \frac {\sqrt {1-\frac {(-b e+a f) x^2}{a}}}{\left (1-\frac {(-d e+c f) x^2}{c}\right )^{3/2}} \, dx,x,\frac {x}{\sqrt {e+f x^2}}\right )}{c e (b e-a f) \sqrt {a+b x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}\\ &=\frac {(d e-c f) x \sqrt {a+b x^2}}{e (b e-a f) \sqrt {c+d x^2} \sqrt {e+f x^2}}-\frac {\sqrt {c} \sqrt {d e-c f} \sqrt {a+b x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {e+f x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{e (b e-a f) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} \sqrt {a+b x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {c} \sqrt {e+f x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{a e \sqrt {d e-c f} \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 [A]
time = 5.02, size = 148, normalized size = 0.46 \begin {gather*} \frac {\sqrt {a} \sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{a \left (e+f x^2\right )}} E\left (\sin ^{-1}\left (\frac {\sqrt {-b e+a f} x}{\sqrt {a} \sqrt {e+f x^2}}\right )|\frac {a (-d e+c f)}{c (-b e+a f)}\right )}{e \sqrt {-b e+a f} \sqrt {a+b x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^2]/(Sqrt[a + b*x^2]*(e + f*x^2)^(3/2)),x]

[Out]

(Sqrt[a]*Sqrt[c + d*x^2]*Sqrt[(e*(a + b*x^2))/(a*(e + f*x^2))]*EllipticE[ArcSin[(Sqrt[-(b*e) + a*f]*x)/(Sqrt[a
]*Sqrt[e + f*x^2])], (a*(-(d*e) + c*f))/(c*(-(b*e) + a*f))])/(e*Sqrt[-(b*e) + a*f]*Sqrt[a + b*x^2]*Sqrt[(e*(c
+ d*x^2))/(c*(e + f*x^2))])

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {d \,x^{2}+c}}{\sqrt {b \,x^{2}+a}\, \left (f \,x^{2}+e \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((d*x^2+c)^(1/2)/(b*x^2+a)^(1/2)/(f*x^2+e)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^2 + c)/(sqrt(b*x^2 + a)*(f*x^2 + e)^(3/2)), x)

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Fricas [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((d*x^2+c)^(1/2)/(b*x^2+a)^(1/2)/(f*x^2+e)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(b*f^2*x^6 + a*f^2*x^4 + (b*x^2 + a)*e^2 + 2*(b*f*x^4
+ a*f*x^2)*e), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d x^{2}}}{\sqrt {a + b x^{2}} \left (e + f x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(1/2)/(b*x**2+a)**(1/2)/(f*x**2+e)**(3/2),x)

[Out]

Integral(sqrt(c + d*x**2)/(sqrt(a + b*x**2)*(e + f*x**2)**(3/2)), 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((d*x^2+c)^(1/2)/(b*x^2+a)^(1/2)/(f*x^2+e)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*x^2 + c)/(sqrt(b*x^2 + a)*(f*x^2 + e)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {d\,x^2+c}}{\sqrt {b\,x^2+a}\,{\left (f\,x^2+e\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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