3.38 \(\int \frac {a+b x^2}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx\)

Optimal. Leaf size=206 \[ \frac {a \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{c \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {b x \sqrt {c+d x^2}}{d \sqrt {e+f x^2}}-\frac {b \sqrt {e} \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{d \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {531, 418, 492, 411} \[ \frac {a \sqrt {e} \sqrt {c+d x^2} F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{c \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {b x \sqrt {c+d x^2}}{d \sqrt {e+f x^2}}-\frac {b \sqrt {e} \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{d \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.22, size = 131, normalized size = 0.64 \[ -\frac {i \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \left ((a f-b e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+b e E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )\right )}{f \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*(b*e*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + (-(b*e) +
a*f)*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)]))/(Sqrt[d/c]*f*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e}}{d f x^{4} + {\left (d e + c f\right )} x^{2} + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 158, normalized size = 0.77 \[ \frac {\left (a f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+b e \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-b e \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )\right ) \sqrt {\frac {f \,x^{2}+e}{e}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}{\sqrt {-\frac {d}{c}}\, \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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