3.1.67 \(\int \frac {\sqrt {e+f x^2}}{(a+b x^2) \sqrt {c+d x^2}} \, dx\) [67]

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

[Out]

e^(3/2)*(1/(1+f*x^2/e))^(1/2)*(1+f*x^2/e)^(1/2)*EllipticPi(x*f^(1/2)/e^(1/2)/(1+f*x^2/e)^(1/2),1-b*e/a/f,(1-d*
e/c/f)^(1/2))*(d*x^2+c)^(1/2)/a/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 = 0.02, 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 = {553} \begin {gather*} \frac {e^{3/2} \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c \sqrt {f} \sqrt {e+f 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]

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

[Out]

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

Rule 553

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

Rubi steps

\begin {align*} \int \frac {\sqrt {e+f x^2}}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx &=\frac {e^{3/2} \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a 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] Result contains complex when optimal does not.
time = 1.85, size = 143, normalized size = 1.40 \begin {gather*} -\frac {i \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} \left (a f F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+(b e-a f) \Pi \left (\frac {b c}{a d};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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*(a*f*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + (b*e - a*f
)*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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Maple [A]
time = 0.13, size = 191, normalized size = 1.87

method	result	size
default	\(\frac {\left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) a f -\EllipticPi \left (x \sqrt {-\frac {d}{c}}, \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right ) a f +\EllipticPi \left (x \sqrt {-\frac {d}{c}}, \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right ) b e \right ) \sqrt {\frac {f \,x^{2}+e}{e}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}{b a \sqrt {-\frac {d}{c}}\, \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right )}\)	\(191\)
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \left (\frac {f \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{b \sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}-\frac {\sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticPi \left (x \sqrt {-\frac {d}{c}}, \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right ) f}{b \sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}+\frac {\sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \EllipticPi \left (x \sqrt {-\frac {d}{c}}, \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right ) e}{a \sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(325\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*f-EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*a*
f+EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*b*e)*((f*x^2+e)/e)^(1/2)*((d*x^2+c)/c)^(1/2)/b*
(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)/a/(-d/c)^(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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