3.2.94 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x (d+e x)^4} \, dx\) [194]

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

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 110, normalized size of antiderivative = 1.00, 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 = {866, 1819, 837, 12, 272, 65, 214} \begin {gather*} -\frac {4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\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 {gather*}

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 12

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

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 837

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] +
Dist[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 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 (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 {\text {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 {\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^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*}

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Mathematica [A]
time = 0.50, size = 88, normalized size = 0.80 \begin {gather*} \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}+10 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{5 d^3} \end {gather*}

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 + 10*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2
*x^2])/d])/(5*d^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(414\) vs. \(2(96)=192\).
time = 0.07, size = 415, normalized size = 3.77

method	result	size
default	\(-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}}{d^{4}}+\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{3} d^{3} \left (x +\frac {d}{e}\right )^{3}}-\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}}{e^{3} d}-\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}}{e \,d^{3}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}}{d^{4}}\)	\(415\)

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,method=_RETURNVERBOSE)

[Out]

-1/d^4*((-(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)))+1/3/e^3/d^3/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)-1/e^3/d*(-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/e/d^3*(-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))))+1/d^4*((-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/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.87, size = 148, normalized size = 1.35 \begin {gather*} \frac {13 \, x^{3} e^{3} + 39 \, d x^{2} e^{2} + 39 \, d^{2} x e + 13 \, d^{3} + 5 \, {\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (8 \, x^{2} e^{2} + 19 \, d x e + 13 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d^{3} x^{3} e^{3} + 3 \, d^{4} x^{2} e^{2} + 3 \, d^{5} x e + d^{6}\right )}} \end {gather*}

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

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (96) = 192\).
time = 1.65, size = 195, normalized size = 1.77 \begin {gather*} -\frac {\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^{3}} - \frac {2 \, {\left (\frac {45 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + \frac {75 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} + \frac {55 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-6\right )}}{x^{3}} + \frac {20 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-8\right )}}{x^{4}} + 13\right )}}{5 \, d^{3} {\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/(e*x+d)^4,x, algorithm="giac")

[Out]

-log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^3 - 2/5*(45*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^
(-2)/x + 75*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^(-4)/x^2 + 55*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-6)/x^3 + 20*
(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(-8)/x^4 + 13)/(d^3*((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\,{\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*(d + e*x)^4),x)

[Out]

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

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