3.100 \(\int \frac{\sqrt{d^2-e^2 x^2}}{x^4 (d+e x)} \, dx\)

Optimal. Leaf size=114 \[ -\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^3} \]

[Out]

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

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Rubi [A]  time = 0.110491, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {850, 835, 807, 266, 63, 208} \[ -\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d^2 - e^2*x^2]/(x^4*(d + e*x)),x]

[Out]

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

Rule 850

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

Rule 835

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 807

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 266

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 63

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 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{d^2-e^2 x^2}}{x^4 (d+e x)} \, dx &=\int \frac{d-e x}{x^4 \sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{\int \frac{3 d^2 e-2 d e^2 x}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{3 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}+\frac{\int \frac{4 d^3 e^2-3 d^2 e^3 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{6 d^4}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{e^3 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{2 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{e \operatorname{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^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{e \sqrt{d^2-e^2 x^2}}{2 d^2 x^2}-\frac{2 e^2 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.11598, size = 84, normalized size = 0.74 \[ \frac{\left (-2 d^2+3 d e x-4 e^2 x^2\right ) \sqrt{d^2-e^2 x^2}+3 e^3 x^3 \log \left (\sqrt{d^2-e^2 x^2}+d\right )-3 e^3 x^3 \log (x)}{6 d^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d^2 - e^2*x^2]/(x^4*(d + e*x)),x]

[Out]

((-2*d^2 + 3*d*e*x - 4*e^2*x^2)*Sqrt[d^2 - e^2*x^2] - 3*e^3*x^3*Log[x] + 3*e^3*x^3*Log[d + Sqrt[d^2 - e^2*x^2]
])/(6*d^3*x^3)

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Maple [B]  time = 0.069, size = 280, normalized size = 2.5 \begin{align*} -{\frac{{e}^{3}}{2\,{d}^{4}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{e}^{3}}{2\,{d}^{2}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{1}{3\,{d}^{3}{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{3}}{{d}^{4}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{{e}^{4}}{{d}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{e}{2\,{d}^{4}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{2}}{{d}^{5}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{4}x}{{d}^{5}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{e}^{4}}{{d}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)/((e*x + d)*x^4), x)

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Fricas [A]  time = 1.92524, size = 157, normalized size = 1.38 \begin{align*} -\frac{3 \, e^{3} x^{3} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (4 \, e^{2} x^{2} - 3 \, d e x + 2 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \, d^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="fricas")

[Out]

-1/6*(3*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (4*e^2*x^2 - 3*d*e*x + 2*d^2)*sqrt(-e^2*x^2 + d^2))/(d^3*
x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{x^{4} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(1/2)/x**4/(e*x+d),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**4*(d + e*x)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError