3.2.95 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx\) [195]

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

[Out]

-8/5*e*(-e*x+d)/(-e^2*x^2+d^2)^(5/2)-4/15*e*(-8*e*x+5*d)/d^2/(-e^2*x^2+d^2)^(3/2)+4*e*arctanh((-e^2*x^2+d^2)^(
1/2)/d)/d^4-1/15*e*(-79*e*x+60*d)/d^4/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^4/x

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Rubi [A]
time = 0.19, antiderivative size = 143, normalized size of antiderivative = 1.00, 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 = {866, 1819, 821, 272, 65, 214} \begin {gather*} -\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 {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 {gather*}

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 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 866

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 1819

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]

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) \text {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 \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 )}{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*}

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Mathematica [A]
time = 0.51, size = 103, normalized size = 0.72 \begin {gather*} -\frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^3+149 d^2 e x+222 d e^2 x^2+94 e^3 x^3\right )}{x (d+e x)^3}+120 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((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) + 120*e*ArcTa
nh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/d^4

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(498\) vs. \(2(127)=254\).
time = 0.09, size = 499, normalized size = 3.49

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{4} x}-\frac {79 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 d^{4} \left (x +\frac {d}{e}\right )}+\frac {4 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{2} d^{2} \left (x +\frac {d}{e}\right )^{3}}-\frac {19 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e \,d^{3} \left (x +\frac {d}{e}\right )^{2}}\)	\(199\)
default	\(\frac {4 e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d^{5}}-\frac {2 \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{2} d^{4} \left (x +\frac {d}{e}\right )^{3}}+\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}}{e^{2} d^{2}}+\frac {-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}}{d^{4}}+\frac {-\frac {3 \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {3 e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}}{d^{4}}-\frac {4 e \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{d^{5}}\)	\(499\)

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,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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Fricas [A]
time = 2.46, size = 173, normalized size = 1.21 \begin {gather*} -\frac {104 \, x^{4} e^{4} + 312 \, d x^{3} e^{3} + 312 \, d^{2} x^{2} e^{2} + 104 \, d^{3} x e + 60 \, {\left (x^{4} e^{4} + 3 \, d x^{3} e^{3} + 3 \, d^{2} x^{2} e^{2} + d^{3} x e\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (94 \, x^{3} e^{3} + 222 \, d x^{2} e^{2} + 149 \, d^{2} x e + 15 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{4} x^{4} e^{3} + 3 \, d^{5} x^{3} e^{2} + 3 \, d^{6} x^{2} e + d^{7} x\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{2} \left (d + e x\right )^{4}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (130) = 260\).
time = 2.09, size = 287, normalized size = 2.01 \begin {gather*} \frac {4 \, e \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 )}{d^{4}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{4} x} + \frac {x {\left (\frac {491 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{x} + \frac {1690 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-3\right )}}{x^{2}} + \frac {2570 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-5\right )}}{x^{3}} + \frac {1815 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-7\right )}}{x^{4}} + \frac {555 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{\left (-9\right )}}{x^{5}} + 15 \, e\right )} e^{2}}{30 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{4} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

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]

4*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(
-1)/(d^4*x) + 1/30*x*(491*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/x + 1690*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^(-
3)/x^2 + 2570*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-5)/x^3 + 1815*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(-7)/x^4 +
555*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*e^(-9)/x^5 + 15*e)*e^2/((d*e + sqrt(-x^2*e^2 + d^2)*e)*d^4*((d*e + sqrt(
-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^5)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}}{x^2\,{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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