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

Optimal. Leaf size=102 \[ \frac {c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} e \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {539} \[ \frac {c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} e \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^(3/2)*Sqrt[e + f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(a*Sqrt[d]
*e*Sqrt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))])

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right ) \sqrt {e+f x^2}} \, dx &=\frac {c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} e \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 143, normalized size = 1.40 \[ -\frac {i \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \left ((b c-a d) \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+a d F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )\right )}{a b \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 191, normalized size = 1.87 \[ \frac {\left (a d \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-a d \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )+b c \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )\right ) \sqrt {\frac {f \,x^{2}+e}{e}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}}{\sqrt {-\frac {d}{c}}\, \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*d-EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)
^(1/2))*a*d+EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*b*c)*((f*x^2+e)/e)^(1/2)*((d*
x^2+c)/c)^(1/2)/b*(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)/a/(-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 \[ \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )} \sqrt {f x^{2} + e}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {d\,x^2+c}}{\left (b\,x^2+a\right )\,\sqrt {f\,x^2+e}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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