3.2.26 \(\int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [126]

Optimal. Leaf size=113 \[ -\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \]

[Out]

-3/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^4-3/2*(-e^2*x^2+d^2)^(1/2)/d^3/x^2+2*e*(-e^2*x^2+d^2)^(1/2)/d^4/x+(
-e^2*x^2+d^2)^(1/2)/d^2/x^2/(e*x+d)

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Rubi [A]
time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {871, 849, 821, 272, 65, 214} \begin {gather*} \frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4}-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 871

Int[(((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[d*(f + g*x)^
(n + 1)*((a + c*x^2)^(p + 1)/(2*a*p*(e*f - d*g)*(d + e*x))), x] + Dist[1/(p*(2*c*d)*(e*f - d*g)), Int[(f + g*x
)^n*(a + c*x^2)^p*(c*e*f*(2*p + 1) - c*d*g*(n + 2*p + 1) + c*e*g*(n + 2*p + 2)*x), x], x] /; FreeQ[{a, c, d, e
, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[n, 0] && ILtQ[n + 2*p, 0] &
&  !IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx &=\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {\int \frac {-3 d e^2+2 e^3 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}+\frac {\int \frac {-4 d^2 e^3+3 d e^4 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{2 d^4 e^2}\\ &=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}+\frac {\left (3 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^3}\\ &=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}+\frac {\left (3 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^3}\\ &=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {3 \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^3}\\ &=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 93, normalized size = 0.82 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (-d^2+d e x+4 e^2 x^2\right )}{x^2 (d+e x)}+6 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-d^2 + d*e*x + 4*e^2*x^2))/(x^2*(d + e*x)) + 6*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e
^2*x^2])/d])/(2*d^4)

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Maple [A]
time = 0.08, size = 183, normalized size = 1.62

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-2 e x +d \right )}{2 d^{4} x^{2}}+\frac {e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{4} \left (x +\frac {d}{e}\right )}-\frac {3 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{3} \sqrt {d^{2}}}\)	\(117\)
default	\(\frac {e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{4} \left (x +\frac {d}{e}\right )}+\frac {-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{2} \sqrt {d^{2}}}}{d}+\frac {e \sqrt {-e^{2} x^{2}+d^{2}}}{d^{4} x}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

e/d^4/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/d*(-1/2/d^2/x^2*(-e^2*x^2+d^2)^(1/2)-1/2*e^2/d^2/(d^2)^(1
/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x))+e*(-e^2*x^2+d^2)^(1/2)/d^4/x-e^2/d^3/(d^2)^(1/2)*ln((2*d
^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.45, size = 108, normalized size = 0.96 \begin {gather*} \frac {2 \, x^{3} e^{3} + 2 \, d x^{2} e^{2} + 3 \, {\left (x^{3} e^{3} + d x^{2} e^{2}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (4 \, x^{2} e^{2} + d x e - d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{2 \, {\left (d^{4} x^{3} e + d^{5} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x^3*e^3 + 2*d*x^2*e^2 + 3*(x^3*e^3 + d*x^2*e^2)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) + (4*x^2*e^2 + d*x*e
- d^2)*sqrt(-x^2*e^2 + d^2))/(d^4*x^3*e + d^5*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (98) = 196\).
time = 1.26, size = 233, normalized size = 2.06 \begin {gather*} -\frac {3 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{2 \, d^{4}} - \frac {x^{2} {\left (\frac {20 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-2\right )}}{x^{2}} + \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}}{x} - e^{2}\right )} e^{4}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}} - \frac {\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{\left (-2\right )}}{x^{2}} - \frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{4}}{x}}{8 \, d^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/8*x^2*(20*(d*e + sqrt(-x^2*e^2
+ d^2)*e)^2*e^(-2)/x^2 + 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)/x - e^2)*e^4/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*(
(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)) - 1/8*((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^(-2)/x^2 - 4*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)*d^4/x)/d^8

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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