3.185 \(\int \frac{1}{x (d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=115 \[ \frac{5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d-22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

(4*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d - 11*e*x)/(15*d^2*(d^2 - e^2*x^2)
^(3/2)) + (15*d - 22*e*x)/(15*d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*
x^2]/d]/d^4

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Rubi [A]  time = 0.426337, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259 \[ \frac{5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d-22 e x}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]  Int[1/(x*(d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(4*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d - 11*e*x)/(15*d^2*(d^2 - e^2*x^2)
^(3/2)) + (15*d - 22*e*x)/(15*d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*
x^2]/d]/d^4

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Rubi in Sympy [A]  time = 35.9242, size = 97, normalized size = 0.84 \[ \frac{\sqrt{d^{2} - e^{2} x^{2}}}{5 d^{2} \left (d + e x\right )^{3}} + \frac{7 \sqrt{d^{2} - e^{2} x^{2}}}{15 d^{3} \left (d + e x\right )^{2}} - \frac{\operatorname{atanh}{\left (\frac{\sqrt{d^{2} - e^{2} x^{2}}}{d} \right )}}{d^{4}} + \frac{22 \sqrt{d^{2} - e^{2} x^{2}}}{15 d^{4} \left (d + e x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/x/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

sqrt(d**2 - e**2*x**2)/(5*d**2*(d + e*x)**3) + 7*sqrt(d**2 - e**2*x**2)/(15*d**3
*(d + e*x)**2) - atanh(sqrt(d**2 - e**2*x**2)/d)/d**4 + 22*sqrt(d**2 - e**2*x**2
)/(15*d**4*(d + e*x))

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Mathematica [A]  time = 0.13618, size = 76, normalized size = 0.66 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (32 d^2+51 d e x+22 e^2 x^2\right )}{(d+e x)^3}-15 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+15 \log (x)}{15 d^4} \]

Antiderivative was successfully verified.

[In]  Integrate[1/(x*(d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(32*d^2 + 51*d*e*x + 22*e^2*x^2))/(d + e*x)^3 + 15*Log[x]
- 15*Log[d + Sqrt[d^2 - e^2*x^2]])/(15*d^4)

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Maple [A]  time = 0.017, size = 179, normalized size = 1.6 \[ -{\frac{1}{{d}^{3}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}+{\frac{22}{15\,{d}^{4}e}\sqrt{- \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) } \left ( x+{\frac{d}{e}} \right ) ^{-1}}+{\frac{7}{15\,{e}^{2}{d}^{3}}\sqrt{- \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) } \left ( x+{\frac{d}{e}} \right ) ^{-2}}+{\frac{1}{5\,{d}^{2}{e}^{3}}\sqrt{- \left ( x+{\frac{d}{e}} \right ) ^{2}{e}^{2}+2\,de \left ( x+{\frac{d}{e}} \right ) } \left ( x+{\frac{d}{e}} \right ) ^{-3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/x/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/d^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+22/15/d^4/e/
(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+7/15/e^2/d^3/(x+d/e)^2*(-(x+d/e)^2*
e^2+2*d*e*(x+d/e))^(1/2)+1/5/d^2/e^3/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + d\right )}^{3} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x), x)

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Fricas [A]  time = 0.28651, size = 450, normalized size = 3.91 \[ \frac{54 \, e^{5} x^{5} + 145 \, d e^{4} x^{4} - 5 \, d^{2} e^{3} x^{3} - 270 \, d^{3} e^{2} x^{2} - 180 \, d^{4} e x + 15 \,{\left (e^{5} x^{5} + 5 \, d e^{4} x^{4} + 5 \, d^{2} e^{3} x^{3} - 5 \, d^{3} e^{2} x^{2} - 10 \, d^{4} e x - 4 \, d^{5} -{\left (e^{4} x^{4} - 7 \, d^{2} e^{2} x^{2} - 10 \, d^{3} e x - 4 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) - 5 \,{\left (2 \, e^{4} x^{4} - 19 \, d e^{3} x^{3} - 54 \, d^{2} e^{2} x^{2} - 36 \, d^{3} e x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{4} e^{5} x^{5} + 5 \, d^{5} e^{4} x^{4} + 5 \, d^{6} e^{3} x^{3} - 5 \, d^{7} e^{2} x^{2} - 10 \, d^{8} e x - 4 \, d^{9} -{\left (d^{4} e^{4} x^{4} - 7 \, d^{6} e^{2} x^{2} - 10 \, d^{7} e x - 4 \, d^{8}\right )} \sqrt{-e^{2} x^{2} + d^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x),x, algorithm="fricas")

[Out]

1/15*(54*e^5*x^5 + 145*d*e^4*x^4 - 5*d^2*e^3*x^3 - 270*d^3*e^2*x^2 - 180*d^4*e*x
+ 15*(e^5*x^5 + 5*d*e^4*x^4 + 5*d^2*e^3*x^3 - 5*d^3*e^2*x^2 - 10*d^4*e*x - 4*d^
5 - (e^4*x^4 - 7*d^2*e^2*x^2 - 10*d^3*e*x - 4*d^4)*sqrt(-e^2*x^2 + d^2))*log(-(d
- sqrt(-e^2*x^2 + d^2))/x) - 5*(2*e^4*x^4 - 19*d*e^3*x^3 - 54*d^2*e^2*x^2 - 36*
d^3*e*x)*sqrt(-e^2*x^2 + d^2))/(d^4*e^5*x^5 + 5*d^5*e^4*x^4 + 5*d^6*e^3*x^3 - 5*
d^7*e^2*x^2 - 10*d^8*e*x - 4*d^9 - (d^4*e^4*x^4 - 7*d^6*e^2*x^2 - 10*d^7*e*x - 4
*d^8)*sqrt(-e^2*x^2 + d^2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{x \sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/x/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(1/(x*sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

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GIAC/XCAS [F(-2)]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError