3.314 \(\int \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right ) \, dx\)

Optimal. Leaf size=118 \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{b f^2+2 e^2 x}{2 e f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{8 e^3}+\frac{f \left (b f^2+2 e^2 x\right ) \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}{4 e^2}+d x+\frac{e x^2}{2} \]

[Out]

d*x + (e*x^2)/2 + (f*(b*f^2 + 2*e^2*x)*Sqrt[a + b*x + (e^2*x^2)/f^2])/(4*e^2) +
(f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(b*f^2 + 2*e^2*x)/(2*e*f*Sqrt[a + b*x + (e^2*x^
2)/f^2])])/(8*e^3)

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Rubi [A]  time = 0.129191, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115 \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{b f^2+2 e^2 x}{2 e f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{8 e^3}+\frac{f \left (b f^2+2 e^2 x\right ) \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}{4 e^2}+d x+\frac{e x^2}{2} \]

Antiderivative was successfully verified.

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

[Out]

d*x + (e*x^2)/2 + (f*(b*f^2 + 2*e^2*x)*Sqrt[a + b*x + (e^2*x^2)/f^2])/(4*e^2) +
(f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(b*f^2 + 2*e^2*x)/(2*e*f*Sqrt[a + b*x + (e^2*x^
2)/f^2])])/(8*e^3)

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ e \int x\, dx + \int d\, dx + \frac{f \left (b f^{2} + 2 e^{2} x\right ) \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}}{4 e^{2}} + \frac{f^{2} \left (4 a e^{2} - b^{2} f^{2}\right ) \operatorname{atanh}{\left (\frac{b f^{2} + 2 e^{2} x}{2 e f \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}} \right )}}{8 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

e*Integral(x, x) + Integral(d, x) + f*(b*f**2 + 2*e**2*x)*sqrt(a + b*x + e**2*x*
*2/f**2)/(4*e**2) + f**2*(4*a*e**2 - b**2*f**2)*atanh((b*f**2 + 2*e**2*x)/(2*e*f
*sqrt(a + b*x + e**2*x**2/f**2)))/(8*e**3)

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Mathematica [A]  time = 0.400028, size = 120, normalized size = 1.02 \[ \frac{1}{8} \left (\frac{\left (4 a e^2 f^2-b^2 f^4\right ) \log \left (2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2\right )}{e^3}+4 f x \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+\frac{2 b f^3 \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}}{e^2}+8 d x+4 e x^2\right ) \]

Antiderivative was successfully verified.

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

[Out]

(8*d*x + 4*e*x^2 + (2*b*f^3*Sqrt[a + x*(b + (e^2*x)/f^2)])/e^2 + 4*f*x*Sqrt[a +
x*(b + (e^2*x)/f^2)] + ((4*a*e^2*f^2 - b^2*f^4)*Log[b*f^2 + 2*e*(e*x + f*Sqrt[a
+ x*(b + (e^2*x)/f^2)])])/e^3)/8

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Maple [A]  time = 0.007, size = 173, normalized size = 1.5 \[ dx+{\frac{e{x}^{2}}{2}}+{\frac{{f}^{3}b}{4\,{e}^{2}}\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}}}+{\frac{fx}{2}\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}}}+{\frac{fa}{2}\ln \left ({1 \left ({\frac{b}{2}}+{\frac{{e}^{2}x}{{f}^{2}}} \right ){\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}}+\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ){\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}}-{\frac{{f}^{3}{b}^{2}}{8\,{e}^{2}}\ln \left ({1 \left ({\frac{b}{2}}+{\frac{{e}^{2}x}{{f}^{2}}} \right ){\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}}+\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ){\frac{1}{\sqrt{{\frac{{e}^{2}}{{f}^{2}}}}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

d*x+1/2*e*x^2+1/4*f^3/e^2*(a+b*x+e^2*x^2/f^2)^(1/2)*b+1/2*f*(a+b*x+e^2*x^2/f^2)^
(1/2)*x+1/2*f*ln((1/2*b+e^2*x/f^2)/(1/f^2*e^2)^(1/2)+(a+b*x+e^2*x^2/f^2)^(1/2))/
(1/f^2*e^2)^(1/2)*a-1/8*f^3/e^2*ln((1/2*b+e^2*x/f^2)/(1/f^2*e^2)^(1/2)+(a+b*x+e^
2*x^2/f^2)^(1/2))/(1/f^2*e^2)^(1/2)*b^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.288769, size = 166, normalized size = 1.41 \[ \frac{4 \, e^{4} x^{2} + 8 \, d e^{3} x +{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \,{\left (b e f^{3} + 2 \, e^{3} f x\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{8 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [A]  time = 0.273574, size = 150, normalized size = 1.27 \[ \frac{1}{2} \, x^{2} e + d x + \frac{{\left ({\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} e^{\left (-3\right )}{\rm ln}\left ({\left | -b f^{2} - 2 \,{\left (x e - \sqrt{b f^{2} x + a f^{2} + x^{2} e^{2}}\right )} e \right |}\right ) + 2 \, \sqrt{b f^{2} x + a f^{2} + x^{2} e^{2}}{\left (b f^{2} e^{\left (-2\right )} + 2 \, x\right )}\right )}{\left | f \right |}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/2*x^2*e + d*x + 1/8*((b^2*f^4 - 4*a*f^2*e^2)*e^(-3)*ln(abs(-b*f^2 - 2*(x*e - s
qrt(b*f^2*x + a*f^2 + x^2*e^2))*e)) + 2*sqrt(b*f^2*x + a*f^2 + x^2*e^2)*(b*f^2*e
^(-2) + 2*x))*abs(f)/f