3.2.96 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)^4} \, dx\) [196]

Optimal. Leaf size=183 \[ \frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {19 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^5} \]

[Out]

8/5*e^2*(-e*x+d)/d/(-e^2*x^2+d^2)^(5/2)+4/15*e^2*(-13*e*x+10*d)/d^3/(-e^2*x^2+d^2)^(3/2)-19/2*e^2*arctanh((-e^
2*x^2+d^2)^(1/2)/d)/d^5+1/15*e^2*(-164*e*x+135*d)/d^5/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^4/x^2+4*
e*(-e^2*x^2+d^2)^(1/2)/d^5/x

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Rubi [A]
time = 0.24, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {866, 1819, 1821, 821, 272, 65, 214} \begin {gather*} \frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {19 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^5}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d^2 - e^2*x^2]/(x^3*(d + e*x)^4),x]

[Out]

(8*e^2*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (4*e^2*(10*d - 13*e*x))/(15*d^3*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(135*d - 164*e*x))/(15*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^4*x^2) + (4*e*Sqrt[d^2 - e^2*x^2])/
(d^5*x) - (19*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^5)

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]

Rule 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps

\begin {align*} \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^4+20 d^3 e x-35 d^2 e^2 x^2+32 d e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^4-60 d^3 e x+120 d^2 e^2 x^2-104 d e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^4+60 d^3 e x-135 d^2 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {\int \frac {-120 d^5 e+285 d^4 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {\left (19 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^4}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {\left (19 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^4}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {19 \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 )}{2 d^4}\\ &=\frac {8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (135 d-164 e x)}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^4 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {19 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^5}\\ \end {align*}

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Mathematica [A]
time = 0.54, size = 116, normalized size = 0.63 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (-15 d^4+75 d^3 e x+713 d^2 e^2 x^2+1059 d e^3 x^3+448 e^4 x^4\right )}{x^2 (d+e x)^3}+570 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{30 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d^2 - e^2*x^2]/(x^3*(d + e*x)^4),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(-15*d^4 + 75*d^3*e*x + 713*d^2*e^2*x^2 + 1059*d*e^3*x^3 + 448*e^4*x^4))/(x^2*(d + e*x)^
3) + 570*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(30*d^5)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(597\) vs. \(2(161)=322\).
time = 0.08, size = 598, normalized size = 3.27

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-8 e x +d \right )}{2 d^{5} x^{2}}+\frac {164 e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 d^{5} \left (x +\frac {d}{e}\right )}-\frac {19 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{4} \sqrt {d^{2}}}+\frac {29 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 d^{4} \left (x +\frac {d}{e}\right )^{2}}+\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{3} e \left (x +\frac {d}{e}\right )^{3}}\)	\(205\)
default	\(-\frac {10 e^{2} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d^{6}}+\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{e \,d^{5} \left (x +\frac {d}{e}\right )^{3}}+\frac {-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{2 d^{2}}}{d^{4}}-\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}}{e \,d^{3}}-\frac {4 e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}\right )}{d^{5}}-\frac {6 e \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{d^{5}}+\frac {10 e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{d^{6}}\)	\(598\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-10*e^2/d^6*((-(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/e/d^5/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)+1/d^4*(-1/2/d^2/x^2*(-e^2*x^2+d^2)^(3/2
)-1/2*e^2/d^2*((-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)))-1/e/d^3
*(-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))-4/d^5*e*(-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))))-6*e/d^5*(-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
0/d^6*e^2*((-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^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-x^2*e^2 + d^2)/((x*e + d)^4*x^3), x)

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Fricas [A]
time = 2.34, size = 189, normalized size = 1.03 \begin {gather*} \frac {398 \, x^{5} e^{5} + 1194 \, d x^{4} e^{4} + 1194 \, d^{2} x^{3} e^{3} + 398 \, d^{3} x^{2} e^{2} + 285 \, {\left (x^{5} e^{5} + 3 \, d x^{4} e^{4} + 3 \, d^{2} x^{3} e^{3} + d^{3} x^{2} e^{2}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (448 \, x^{4} e^{4} + 1059 \, d x^{3} e^{3} + 713 \, d^{2} x^{2} e^{2} + 75 \, d^{3} x e - 15 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{30 \, {\left (d^{5} x^{5} e^{3} + 3 \, d^{6} x^{4} e^{2} + 3 \, d^{7} x^{3} e + d^{8} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(398*x^5*e^5 + 1194*d*x^4*e^4 + 1194*d^2*x^3*e^3 + 398*d^3*x^2*e^2 + 285*(x^5*e^5 + 3*d*x^4*e^4 + 3*d^2*x
^3*e^3 + d^3*x^2*e^2)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) + (448*x^4*e^4 + 1059*d*x^3*e^3 + 713*d^2*x^2*e^2 + 7
5*d^3*x*e - 15*d^4)*sqrt(-x^2*e^2 + d^2))/(d^5*x^5*e^3 + 3*d^6*x^4*e^2 + 3*d^7*x^3*e + d^8*x^2)

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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^{3} \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**3/(e*x+d)**4,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**3*(d + e*x)**4), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (156) = 312\).
time = 1.49, size = 353, normalized size = 1.93 \begin {gather*} -\frac {19 \, e^{2} \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 )}{2 \, d^{5}} - \frac {x^{2} {\left (\frac {4234 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-2\right )}}{x^{2}} + \frac {14330 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-4\right )}}{x^{3}} + \frac {20965 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-6\right )}}{x^{4}} + \frac {14385 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{\left (-8\right )}}{x^{5}} + \frac {4080 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} e^{\left (-10\right )}}{x^{6}} + \frac {165 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}}{x} - 15 \, e^{2}\right )} e^{4}}{120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{5} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} - \frac {\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{5} e^{\left (-2\right )}}{x^{2}} - \frac {16 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{5}}{x}}{8 \, d^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-19/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^5 - 1/120*x^2*(4234*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^2*e^(-2)/x^2 + 14330*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-4)/x^3 + 20965*(d*e + sqrt(-x^2*e^2 +
d^2)*e)^4*e^(-6)/x^4 + 14385*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*e^(-8)/x^5 + 4080*(d*e + sqrt(-x^2*e^2 + d^2)*e
)^6*e^(-10)/x^6 + 165*(d*e + sqrt(-x^2*e^2 + d^2)*e)/x - 15*e^2)*e^4/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^5*((d
*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^5) - 1/8*((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^5*e^(-2)/x^2 - 16*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)*d^5/x)/d^10

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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^3\,{\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^3*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(1/2)/(x^3*(d + e*x)^4), x)

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