3.163 \(\int \frac{\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)^2} \, dx\)

Optimal. Leaf size=96 \[ d (d-e x) \sqrt{d^2-e^2 x^2}-\frac{1}{3} \left (d^2-e^2 x^2\right )^{3/2}+d^3 \left (-\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )\right )-d^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

d*(d - e*x)*Sqrt[d^2 - e^2*x^2] - (d^2 - e^2*x^2)^(3/2)/3 - d^3*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]] - d^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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Rubi [A]  time = 0.390941, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ d (d-e x) \sqrt{d^2-e^2 x^2}-\frac{1}{3} \left (d^2-e^2 x^2\right )^{3/2}+d^3 \left (-\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )\right )-d^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]  Int[(d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^2),x]

[Out]

d*(d - e*x)*Sqrt[d^2 - e^2*x^2] - (d^2 - e^2*x^2)^(3/2)/3 - d^3*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]] - d^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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Rubi in Sympy [A]  time = 34.9221, size = 92, normalized size = 0.96 \[ - d^{3} \operatorname{atan}{\left (\frac{e x}{\sqrt{d^{2} - e^{2} x^{2}}} \right )} - d^{3} \operatorname{atanh}{\left (\frac{\sqrt{d^{2} - e^{2} x^{2}}}{d} \right )} + d^{2} \sqrt{d^{2} - e^{2} x^{2}} - d e x \sqrt{d^{2} - e^{2} x^{2}} - \frac{\left (d^{2} - e^{2} x^{2}\right )^{\frac{3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((-e**2*x**2+d**2)**(5/2)/x/(e*x+d)**2,x)

[Out]

-d**3*atan(e*x/sqrt(d**2 - e**2*x**2)) - d**3*atanh(sqrt(d**2 - e**2*x**2)/d) +
d**2*sqrt(d**2 - e**2*x**2) - d*e*x*sqrt(d**2 - e**2*x**2) - (d**2 - e**2*x**2)*
*(3/2)/3

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Mathematica [A]  time = 0.0855334, size = 96, normalized size = 1. \[ d^3 \log (x)+\sqrt{d^2-e^2 x^2} \left (\frac{2 d^2}{3}-d e x+\frac{e^2 x^2}{3}\right )-d^3 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+d^3 \left (-\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[(d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^2),x]

[Out]

Sqrt[d^2 - e^2*x^2]*((2*d^2)/3 - d*e*x + (e^2*x^2)/3) - d^3*ArcTan[(e*x)/Sqrt[d^
2 - e^2*x^2]] + d^3*Log[x] - d^3*Log[d + Sqrt[d^2 - e^2*x^2]]

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Maple [B]  time = 0.017, size = 290, normalized size = 3. \[{\frac{1}{5\,{d}^{2}} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{1}{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{d}^{2}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}-{{d}^{4}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{8}{15\,{d}^{2}} \left ( - \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{2\,ex}{3\,d} \left ( - \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) \right ) ^{{\frac{3}{2}}}}-de\sqrt{- \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) }x-{{d}^{3}e\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{1}{3\,{d}^{2}{e}^{2}} \left ( - \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) \right ) ^{{\frac{7}{2}}} \left ( x+{\frac{d}{e}} \right ) ^{-2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((-e^2*x^2+d^2)^(5/2)/x/(e*x+d)^2,x)

[Out]

1/5/d^2*(-e^2*x^2+d^2)^(5/2)+1/3*(-e^2*x^2+d^2)^(3/2)+d^2*(-e^2*x^2+d^2)^(1/2)-d
^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-8/15/d^2*(-(x+d/
e)^2*e^2+2*d*e*(x+d/e))^(5/2)-2/3/d*e*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)*x-d*e
*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)*x-d^3*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(
-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))-1/3/d^2/e^2/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e
*(x+d/e))^(7/2)

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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^2*x^2 + d^2)^(5/2)/((e*x + d)^2*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.300798, size = 392, normalized size = 4.08 \[ \frac{e^{6} x^{6} - 3 \, d e^{5} x^{5} - 3 \, d^{2} e^{4} x^{4} + 15 \, d^{3} e^{3} x^{3} - 12 \, d^{5} e x + 6 \,{\left (3 \, d^{4} e^{2} x^{2} - 4 \, d^{6} -{\left (d^{3} e^{2} x^{2} - 4 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + 3 \,{\left (3 \, d^{4} e^{2} x^{2} - 4 \, d^{6} -{\left (d^{3} e^{2} x^{2} - 4 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + 3 \,{\left (d e^{4} x^{4} - 3 \, d^{2} e^{3} x^{3} + 4 \, d^{4} e x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \,{\left (3 \, d e^{2} x^{2} - 4 \, d^{3} -{\left (e^{2} x^{2} - 4 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^2*x),x, algorithm="fricas")

[Out]

1/3*(e^6*x^6 - 3*d*e^5*x^5 - 3*d^2*e^4*x^4 + 15*d^3*e^3*x^3 - 12*d^5*e*x + 6*(3*
d^4*e^2*x^2 - 4*d^6 - (d^3*e^2*x^2 - 4*d^5)*sqrt(-e^2*x^2 + d^2))*arctan(-(d - s
qrt(-e^2*x^2 + d^2))/(e*x)) + 3*(3*d^4*e^2*x^2 - 4*d^6 - (d^3*e^2*x^2 - 4*d^5)*s
qrt(-e^2*x^2 + d^2))*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + 3*(d*e^4*x^4 - 3*d^2*e
^3*x^3 + 4*d^4*e*x)*sqrt(-e^2*x^2 + d^2))/(3*d*e^2*x^2 - 4*d^3 - (e^2*x^2 - 4*d^
2)*sqrt(-e^2*x^2 + d^2))

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Sympy [A]  time = 19.0629, size = 267, normalized size = 2.78 \[ d^{2} \left (\begin{cases} \frac{d^{2}}{e x \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname{acosh}{\left (\frac{d}{e x} \right )} - \frac{e x}{\sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} & \text{for}\: \left |{\frac{d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac{i d^{2}}{e x \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname{asin}{\left (\frac{d}{e x} \right )} + \frac{i e x}{\sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} & \text{otherwise} \end{cases}\right ) - 2 d e \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e} - \frac{i d x}{2 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{3}}{2 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \left |{\frac{e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e} + \frac{d x \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}}{2} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} \frac{x^{2} \sqrt{d^{2}}}{2} & \text{for}\: e^{2} = 0 \\- \frac{\left (d^{2} - e^{2} x^{2}\right )^{\frac{3}{2}}}{3 e^{2}} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e**2*x**2+d**2)**(5/2)/x/(e*x+d)**2,x)

[Out]

d**2*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/s
qrt(d**2/(e**2*x**2) - 1), Abs(d**2/(e**2*x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/
(e**2*x**2) + 1)) + I*d*asin(d/(e*x)) + I*e*x/sqrt(-d**2/(e**2*x**2) + 1), True)
) - 2*d*e*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d
**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (
d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) + e**2*Piecewise
((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True))

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GIAC/XCAS [A]  time = 4.91639, size = 1, normalized size = 0.01 \[ +\infty \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^2*x),x, algorithm="giac")

[Out]

+Infinity