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

Optimal. Leaf size=143 \[ -\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

(-8*e*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) - (4*e*(5*d - 8*e*x))/(15*d^2*(d^2 - e^2*x^2)^(3/2)) - (e*(60*d - 7
9*e*x))/(15*d^4*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^4*x) + (4*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^4

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Rubi [A]  time = 0.305805, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {852, 1805, 807, 266, 63, 208} \[ -\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*e*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) - (4*e*(5*d - 8*e*x))/(15*d^2*(d^2 - e^2*x^2)^(3/2)) - (e*(60*d - 7
9*e*x))/(15*d^4*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^4*x) + (4*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^4

Rule 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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^2 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^4+20 d^3 e x-27 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^4-60 d^3 e x+64 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^4+60 d^3 e x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}-\frac{(4 e) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^3}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}-\frac{(2 e) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{d^3}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}+\frac{4 \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 )}{d^3 e}\\ &=-\frac{8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (60 d-79 e x)}{15 d^4 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^4 x}+\frac{4 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^4}\\ \end{align*}

Mathematica [A]  time = 0.226226, size = 92, normalized size = 0.64 \[ -\frac{\frac{\sqrt{d^2-e^2 x^2} \left (149 d^2 e x+15 d^3+222 d e^2 x^2+94 e^3 x^3\right )}{x (d+e x)^3}-60 e \log \left (\sqrt{d^2-e^2 x^2}+d\right )+60 e \log (x)}{15 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d^2 - e^2*x^2]*(15*d^3 + 149*d^2*e*x + 222*d*e^2*x^2 + 94*e^3*x^3))/(x*(d + e*x)^3) + 60*e*Log[x] - 60
*e*Log[d + Sqrt[d^2 - e^2*x^2]])/(15*d^4)

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Maple [B]  time = 0.068, size = 361, normalized size = 2.5 \begin{align*} -4\,{\frac{e\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{{d}^{5}}}+4\,{\frac{e}{{d}^{3}\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }+{\frac{e}{{d}^{5}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{{e}^{2}}{{d}^{4}}\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{1}{5\,{e}^{3}{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}-{\frac{11}{15\,{d}^{4}{e}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}-3\,{\frac{1}{{d}^{5}e} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{1}{{d}^{6}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{2}x}{{d}^{6}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{e}^{2}}{{d}^{4}}\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^2/(e*x+d)^4,x)

[Out]

-4/d^5*e*(-e^2*x^2+d^2)^(1/2)+4/d^3*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+1/d^5*e*(-(
d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+1/d^4*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1
/2))-1/5/e^3/d^3/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)-11/15/e^2/d^4/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*
(d/e+x))^(3/2)-3/d^5/e/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)-1/d^6/x*(-e^2*x^2+d^2)^(3/2)-1/d^6*e^2*x
*(-e^2*x^2+d^2)^(1/2)-1/d^4*e^2/(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 )}^{4} x^{2}}\,{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^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.71911, size = 386, normalized size = 2.7 \begin{align*} -\frac{104 \, e^{4} x^{4} + 312 \, d e^{3} x^{3} + 312 \, d^{2} e^{2} x^{2} + 104 \, d^{3} e x + 60 \,{\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (94 \, e^{3} x^{3} + 222 \, d e^{2} x^{2} + 149 \, d^{2} e x + 15 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{4} e^{3} x^{4} + 3 \, d^{5} e^{2} x^{3} + 3 \, d^{6} e x^{2} + d^{7} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(104*e^4*x^4 + 312*d*e^3*x^3 + 312*d^2*e^2*x^2 + 104*d^3*e*x + 60*(e^4*x^4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2
+ d^3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (94*e^3*x^3 + 222*d*e^2*x^2 + 149*d^2*e*x + 15*d^3)*sqrt(-e^2
*x^2 + d^2))/(d^4*e^3*x^4 + 3*d^5*e^2*x^3 + 3*d^6*e*x^2 + d^7*x)

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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^{2} \left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1692, size = 1, normalized size = 0.01 \begin{align*} +\infty \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

+Infinity