3.194 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x (d+e x)^4} \, dx\)
Optimal. Leaf size=110 \[ -\frac {4 e x}{5 d \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{5 d^3 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^3} \]
[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 = 0.22, 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 =
{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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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 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 (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*}
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Mathematica [A] time = 0.14, 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)
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fricas [A] time = 0.83, size = 153, normalized size = 1.39 \[ \frac {13 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 39 \, d^{2} e x + 13 \, d^{3} + 5 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (8 \, e^{2} x^{2} + 19 \, d e x + 13 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{3} e^{3} x^{3} + 3 \, d^{4} e^{2} x^{2} + 3 \, d^{5} e x + d^{6}\right )}} \]
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)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
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 >> Unable to parse Giac output: (6*exp(1)*exp(2)^8+12*(-1/2*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)*exp(2)^8+54*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1
))/x/exp(2))^4*exp(1)^13*exp(2)^2+18*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^11*e
xp(2)^3+48*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^13*exp(2)^2+44*(-1/2*(-2*d*exp
(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^15*exp(2)+60*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*e
xp(1))/x/exp(2))^4*exp(1)^11*exp(2)^3+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)*e
xp(2)^8+18*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^9*exp(2)^4+78*(-1/2*(-2*d*exp(
1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^13*exp(2)^2-14*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*
exp(1))/x/exp(2))^3*exp(1)^11*exp(2)^3-87*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)
^9*exp(2)^4-33*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^7*exp(2)^5+84*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^11*exp(2)^3-48*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(
2))*exp(1))/x/exp(2))^3*exp(1)^9*exp(2)^4-120*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*ex
p(1)^7*exp(2)^5-36*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^5*exp(2)^6-120*(-1/2*(
-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^9*exp(2)^4-96*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*
exp(2))*exp(1))/x/exp(2))^3*exp(1)^7*exp(2)^5+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4
*exp(1)^5*exp(2)^6+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^3*exp(2)^7-204*(-1/
2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^7*exp(2)^5-180*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^5*exp(2)^6-30*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2
))^4*exp(1)^3*exp(2)^7+11*exp(1)^9*exp(2)^4+36*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*e
xp(1)^3*exp(2)^7+12*exp(1)^7*exp(2)^5-60*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^
3*exp(2)^7-20*exp(1)^5*exp(2)^6-30*exp(1)^3*exp(2)^7-12*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^3*e
xp(2)^7/x/exp(2)+72*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^5*exp(2)^6/x/exp(2)+87/2*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^7*exp(2)^5/x/exp(2)-27*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)
^9*exp(2)^4/x/exp(2)-24*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^11*exp(2)^3/x/exp(2))/(-(-1/2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)+(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x-exp(2)
)^3/(3*d^3*exp(1)^11-6*d^3*exp(1)^7*exp(2)^2-6*d^3*exp(1)^5*exp(2)^3+3*d^3*exp(1)^9*exp(2)+6*d^3*exp(1)*exp(2)
^5)+1/2*(-4*exp(1)^9*exp(2)^2+10*exp(1)^7*exp(2)^3+8*exp(1)^5*exp(2)^4-8*exp(1)^3*exp(2)^5-4*exp(1)^11*exp(2)-
16*exp(1)*exp(2)^6)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))
/sqrt(-exp(1)^4+exp(2)^2)/(-d^3*exp(1)^11+2*d^3*exp(1)^7*exp(2)^2+2*d^3*exp(1)^5*exp(2)^3-d^3*exp(1)^9*exp(2)-
2*d^3*exp(1)*exp(2)^5)-exp(2)*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^3/exp(1)^
2
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maple [B] time = 0.01, size = 196, normalized size = 1.78 \[ -\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{4}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{5 \left (x +\frac {d}{e}\right )^{4} d^{2} e^{4}}+\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{5 \left (x +\frac {d}{e}\right )^{3} d^{3} e^{3}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{\left (x +\frac {d}{e}\right )^{2} d^{4} e^{2}} \]
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]
1/5/d^2/e^4/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)+2/5/e^3/d^3/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^
(3/2)+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/d^4/e^2/
(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 \[ \int \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )}^{4} x}\,{d x} \]
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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {d^2-e^2\,x^2}}{x\,{\left (d+e\,x\right )}^4} \,d x \]
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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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x \left (d + e x\right )^{4}}\, dx \]
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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