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

Optimal. Leaf size=110 \[ -\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}+\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.219763, antiderivative size = 110, 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, Rules used = {852, 1805, 823, 12, 266, 63, 208} \[ -\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}+\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^4+12 d^3 e x+5 d^2 e^2 x^2}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^4-24 d^3 e x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{15 d^6 e^2}{x \sqrt{d^2-e^2 x^2}} \, dx}{15 d^8 e^2}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^2}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^2}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}-\frac{\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^2 e^2}\\ &=\frac{8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-8 e x}{5 d^3 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 0.148624, size = 76, normalized size = 0.69 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (13 d^2+19 d e x+8 e^2 x^2\right )}{(d+e x)^3}-5 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+5 \log (x)}{5 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(13*d^2 + 19*d*e*x + 8*e^2*x^2))/(d + e*x)^3 + 5*Log[x] - 5*Log[d + Sqrt[d^2 - e^2*x^2]]
)/(5*d^3)

________________________________________________________________________________________

Maple [B]  time = 0.067, size = 196, normalized size = 1.8 \begin{align*}{\frac{1}{{d}^{4}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{1}{{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}{5\,{e}^{4}{d}^{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 ) ^{-4}}+{\frac{2}{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 ) ^{-3}}+{\frac{1}{{e}^{2}{d}^{4}} \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 ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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}\,{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/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.75794, size = 323, normalized size = 2.94 \begin{align*} \frac{13 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 39 \, d^{2} e x + 13 \, d^{3} + 5 \,{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (8 \, e^{2} x^{2} + 19 \, d e x + 13 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (d^{3} e^{3} x^{3} + 3 \, d^{4} e^{2} x^{2} + 3 \, d^{5} e x + d^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(13*e^3*x^3 + 39*d*e^2*x^2 + 39*d^2*e*x + 13*d^3 + 5*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*log(-(d - s
qrt(-e^2*x^2 + d^2))/x) + (8*e^2*x^2 + 19*d*e*x + 13*d^2)*sqrt(-e^2*x^2 + d^2))/(d^3*e^3*x^3 + 3*d^4*e^2*x^2 +
3*d^5*e*x + d^6)

________________________________________________________________________________________

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 \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/(e*x+d)**4,x)

[Out]

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

________________________________________________________________________________________

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/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError