3.321 \(\int \frac{1}{\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}} \, dx\)

Optimal. Leaf size=244 \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

[Out]

Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]]/e - (f^2*(4*a*e - (b^2*f^2)/e)*S
qrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(2*(2*d*e - b*f^2)*(b*f^2 + 2*e*
(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (f^2*(4*a*e^2 - b^2*f^2)*ArcTanh
[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e -
b*f^2]])/(2*Sqrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

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Rubi [A]  time = 0.628531, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]  Int[1/Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]],x]

[Out]

Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]]/e - (f^2*(4*a*e - (b^2*f^2)/e)*S
qrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(2*(2*d*e - b*f^2)*(b*f^2 + 2*e*
(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (f^2*(4*a*e^2 - b^2*f^2)*ArcTanh
[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e -
b*f^2]])/(2*Sqrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{d + e x + f \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/(d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2)), x)

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Mathematica [A]  time = 1.53682, size = 238, normalized size = 0.98 \[ \frac{f^2 \left (b^2 f^2-4 a e^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2 d e-b f^2}}{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]  Integrate[1/Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]],x]

[Out]

Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]/e + (f^2*(-4*a*e^2 + b^2*f^2)*Sq
rt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/(2*e*(2*d*e - b*f^2)*(b*f^2 + 2*e
*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) + (f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[Sq
rt[2*d*e - b*f^2]/(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)
]])])/(2*Sqrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

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Maple [F]  time = 0.022, size = 0, normalized size = 0. \[ \int{\frac{1}{\sqrt{d+ex+f\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [A]  time = 0.499333, size = 1, normalized size = 0. \[ \left [\frac{{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}} \log \left (\frac{2 \, \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}} e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}}{\left (b f^{2} - 2 \, e^{2} x - 4 \, d e\right )} + 4 \,{\left (b e f^{2} - 2 \, d e^{2}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{b f^{2} + 2 \, e^{2} x + 2 \, e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}\right ) + 4 \,{\left (b^{2} e f^{4} - 6 \, b d e^{2} f^{2} + 8 \, d^{2} e^{3} - 2 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x + 2 \,{\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{8 \,{\left (b^{2} e^{2} f^{4} - 4 \, b d e^{3} f^{2} + 4 \, d^{2} e^{4}\right )}}, \frac{{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{2 \, b e f^{2} - 4 \, d e^{2}} \arctan \left (\frac{b f^{2} - 2 \, d e}{\sqrt{2 \, b e f^{2} - 4 \, d e^{2}} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}\right ) + 2 \,{\left (b^{2} e f^{4} - 6 \, b d e^{2} f^{2} + 8 \, d^{2} e^{3} - 2 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x + 2 \,{\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \,{\left (b^{2} e^{2} f^{4} - 4 \, b d e^{3} f^{2} + 4 \, d^{2} e^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/8*((b^2*f^4 - 4*a*e^2*f^2)*sqrt(-2*b*e*f^2 + 4*d*e^2)*log((2*sqrt(-2*b*e*f^2
+ 4*d*e^2)*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(-2*b*e*f^2 + 4*d*e^2
)*(b*f^2 - 2*e^2*x - 4*d*e) + 4*(b*e*f^2 - 2*d*e^2)*sqrt(e*x + f*sqrt((b*f^2*x +
e^2*x^2 + a*f^2)/f^2) + d))/(b*f^2 + 2*e^2*x + 2*e*f*sqrt((b*f^2*x + e^2*x^2 +
a*f^2)/f^2))) + 4*(b^2*e*f^4 - 6*b*d*e^2*f^2 + 8*d^2*e^3 - 2*(b*e^3*f^2 - 2*d*e^
4)*x + 2*(b*e^2*f^3 - 2*d*e^3*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x
+ f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(b^2*e^2*f^4 - 4*b*d*e^3*f^2 +
4*d^2*e^4), 1/4*((b^2*f^4 - 4*a*e^2*f^2)*sqrt(2*b*e*f^2 - 4*d*e^2)*arctan((b*f^2
- 2*d*e)/(sqrt(2*b*e*f^2 - 4*d*e^2)*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^
2)/f^2) + d))) + 2*(b^2*e*f^4 - 6*b*d*e^2*f^2 + 8*d^2*e^3 - 2*(b*e^3*f^2 - 2*d*e
^4)*x + 2*(b*e^2*f^3 - 2*d*e^3*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*
x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(b^2*e^2*f^4 - 4*b*d*e^3*f^2 +
4*d^2*e^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{d + e x + f \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2)), x)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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