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

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

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Rubi [A]  time = 0.31, 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 = {852, 1805, 807, 266, 63, 208} \begin {gather*} -\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} \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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.21, size = 92, normalized size = 0.64 \begin {gather*} -\frac {-60 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\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}+60 e \log (x)}{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) + 60*e*Log[x]
- 60*e*Log[d + Sqrt[d^2 - e^2*x^2]])/d^4

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IntegrateAlgebraic [A]  time = 0.59, size = 107, normalized size = 0.75 \begin {gather*} \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 )}{15 d^4 x (d+e x)^3}-\frac {8 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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))/(15*d^4*x*(d + e*x)^3) - (8*e*ArcTa
nh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^4

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fricas [A]  time = 0.42, size = 181, normalized size = 1.27 \begin {gather*} -\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 {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*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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giac [A]  time = 0.28, size = 1, normalized size = 0.01 \begin {gather*} +\infty \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]

+Infinity

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maple [B]  time = 0.01, size = 361, normalized size = 2.52 \begin {gather*} \frac {4 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{3}}+\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}-\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x}{d^{6}}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, e}{d^{5}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e}{d^{5}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{5 \left (x +\frac {d}{e}\right )^{4} d^{3} e^{3}}-\frac {11 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{15 \left (x +\frac {d}{e}\right )^{3} d^{4} e^{2}}-\frac {3 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{\left (x +\frac {d}{e}\right )^{2} d^{5} e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{6} x} \end {gather*}

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]

-1/5/d^3/e^3/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)-11/15/d^4/e^2/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^
2)^(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
)/(-e^2*x^2+d^2)^(1/2)*x)-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*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+1/d^4*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*
d*e-(x+d/e)^2*e^2)^(1/2)*x)-3/d^5/e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)

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

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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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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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