3.319 \(\int \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2} \, dx\)

Optimal. Leaf size=370 \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{4 e^2 \left (2 d e-b f^2\right )}-\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{2 d e-b f^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{8 \sqrt{2} e^{7/2}}+\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{8 e^3}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e} \]

[Out]

(3*f^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(8*e
^3) + (f^2*(4*a*e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)
)/(4*e^2*(2*d*e - b*f^2)) + (d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/2)/(5
*e) - (f^2*(4*a*e - (b^2*f^2)/e)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/
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]))
) - (3*f^2*Sqrt[2*d*e - b*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]])/(8*Sqrt[2]*e^
(7/2))

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Rubi [A]  time = 1.01787, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233 \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{4 e^2 \left (2 d e-b f^2\right )}-\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{2 d e-b f^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{8 \sqrt{2} e^{7/2}}+\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{8 e^3}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]  Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2),x]

[Out]

(3*f^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(8*e
^3) + (f^2*(4*a*e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)
)/(4*e^2*(2*d*e - b*f^2)) + (d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/2)/(5
*e) - (f^2*(4*a*e - (b^2*f^2)/e)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/
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]))
) - (3*f^2*Sqrt[2*d*e - b*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]])/(8*Sqrt[2]*e^
(7/2))

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

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))**(3/2),x)

[Out]

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

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Mathematica [A]  time = 0.427944, size = 0, normalized size = 0. \[ \int \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2} \, dx \]

Verification is Not applicable to the result.

[In]  Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2),x]

[Out]

Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2), x]

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \[ \int \left ( d+ex+f\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{{\frac{3}{2}}}\, dx \]

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))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(3/2), x)

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Fricas [A]  time = 0.496602, size = 1, normalized size = 0. \[ \left [-\frac{15 \, \sqrt{\frac{1}{2}}{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{-\frac{b f^{2} - 2 \, d e}{e}} \log \left (-\frac{b f^{2} - 2 \, e^{2} x + 4 \, \sqrt{\frac{1}{2}} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} e \sqrt{-\frac{b f^{2} - 2 \, d e}{e}} - 2 \, e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - 4 \, d e}{b f^{2} + 2 \, e^{2} x + 2 \, e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}\right ) + 2 \,{\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \,{\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \,{\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \,{\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{80 \, e^{3}}, \frac{15 \, \sqrt{\frac{1}{2}}{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{\frac{b f^{2} - 2 \, d e}{e}} \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{\sqrt{\frac{b f^{2} - 2 \, d e}{e}}}\right ) -{\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \,{\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \,{\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \,{\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{40 \, e^{3}}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

[-1/80*(15*sqrt(1/2)*(b^2*f^4 - 4*a*e^2*f^2)*sqrt(-(b*f^2 - 2*d*e)/e)*log(-(b*f^
2 - 2*e^2*x + 4*sqrt(1/2)*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d
)*e*sqrt(-(b*f^2 - 2*d*e)/e) - 2*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - 4*d
*e)/(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*(15*b^2
*f^4 - 16*e^4*x^2 - 8*d^2*e^2 - 2*(5*b*d*e + 24*a*e^2)*f^2 + 2*(b*e^2*f^2 - 18*d
*e^3)*x - 2*(5*b*e*f^3 + 8*e^3*f*x - 2*d*e^2*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))/e^3, 1/40*(15*sq
rt(1/2)*(b^2*f^4 - 4*a*e^2*f^2)*sqrt((b*f^2 - 2*d*e)/e)*arctan(2*sqrt(1/2)*sqrt(
e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)/sqrt((b*f^2 - 2*d*e)/e)) - (1
5*b^2*f^4 - 16*e^4*x^2 - 8*d^2*e^2 - 2*(5*b*d*e + 24*a*e^2)*f^2 + 2*(b*e^2*f^2 -
18*d*e^3)*x - 2*(5*b*e*f^3 + 8*e^3*f*x - 2*d*e^2*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))/e^3]

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

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))**(3/2),x)

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(3/2), x)