3.3.9 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^6 (d+e x)^4} \, dx\) [209]

Optimal. Leaf size=196 \[ -\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {27 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \]

[Out]

27/2*e^5*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^4-8*e^5*(-e*x+d)/d^4/(-e^2*x^2+d^2)^(1/2)-1/5*(-e^2*x^2+d^2)^(1/2)/
x^5+e*(-e^2*x^2+d^2)^(1/2)/d/x^4-13/5*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^3+11/2*e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x^2-6
6/5*e^4*(-e^2*x^2+d^2)^(1/2)/d^4/x

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Rubi [A]
time = 0.48, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 10, 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 {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}+\frac {27 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)^4),x]

[Out]

(-8*e^5*(d - e*x))/(d^4*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(5*x^5) + (e*Sqrt[d^2 - e^2*x^2])/(d*x^4) -
(13*e^2*Sqrt[d^2 - e^2*x^2])/(5*d^2*x^3) + (11*e^3*Sqrt[d^2 - e^2*x^2])/(2*d^3*x^2) - (66*e^4*Sqrt[d^2 - e^2*
x^2])/(5*d^4*x) + (27*e^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*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]

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 {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x^6 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-d^4+4 d^3 e x-7 d^2 e^2 x^2+8 d e^3 x^3-8 e^4 x^4+\frac {8 e^5 x^5}{d}}{x^6 \sqrt {d^2-e^2 x^2}} \, dx}{d^2}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {\int \frac {-20 d^5 e+39 d^4 e^2 x-40 d^3 e^3 x^2+40 d^2 e^4 x^3-40 d e^5 x^4}{x^5 \sqrt {d^2-e^2 x^2}} \, dx}{5 d^4}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {\int \frac {-156 d^6 e^2+220 d^5 e^3 x-160 d^4 e^4 x^2+160 d^3 e^5 x^3}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{20 d^6}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {\int \frac {-660 d^7 e^3+792 d^6 e^4 x-480 d^5 e^5 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{60 d^8}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {\int \frac {-1584 d^8 e^4+1620 d^7 e^5 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{120 d^{10}}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {\left (27 e^5\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^3}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {\left (27 e^5\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^3}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {\left (27 e^3\right ) \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^3}\\ &=-\frac {8 e^5 (d-e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}+\frac {e \sqrt {d^2-e^2 x^2}}{d x^4}-\frac {13 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}+\frac {11 e^3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {66 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {27 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 127, normalized size = 0.65 \begin {gather*} -\frac {\frac {\sqrt {d^2-e^2 x^2} \left (2 d^5-8 d^4 e x+16 d^3 e^2 x^2-29 d^2 e^3 x^3+77 d e^4 x^4+212 e^5 x^5\right )}{x^5 (d+e x)}+270 e^5 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{10 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)^4),x]

[Out]

-1/10*((Sqrt[d^2 - e^2*x^2]*(2*d^5 - 8*d^4*e*x + 16*d^3*e^2*x^2 - 29*d^2*e^3*x^3 + 77*d*e^4*x^4 + 212*e^5*x^5)
)/(x^5*(d + e*x)) + 270*e^5*ArcTanh[(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. \(1996\) vs. \(2(172)=344\).
time = 0.11, size = 1997, normalized size = 10.19

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (132 e^{4} x^{4}-55 d \,e^{3} x^{3}+26 d^{2} x^{2} e^{2}-10 d^{3} e x +2 d^{4}\right )}{10 x^{5} d^{4}}-\frac {8 e^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{4} \left (x +\frac {d}{e}\right )}+\frac {27 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{3} \sqrt {d^{2}}}\)	\(155\)
default	\(\text {Expression too large to display}\)	\(1997\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

56*e^5/d^9*(1/5*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*
e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(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*(-1/5/d^2/x^5*(-e^2*x^2+d^2)^(7/2)-2/
5*e^2/d^2*(-1/3/d^2/x^3*(-e^2*x^2+d^2)^(7/2)-4/3*e^2/d^2*(-1/d^2/x*(-e^2*x^2+d^2)^(7/2)-6*e^2/d^2*(1/6*x*(-e^2
*x^2+d^2)^(5/2)+5/6*d^2*(1/4*x*(-e^2*x^2+d^2)^(3/2)+3/4*d^2*(1/2*x*(-e^2*x^2+d^2)^(1/2)+1/2*d^2/(e^2)^(1/2)*ar
ctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))))))))-4/d^5*e*(-1/4/d^2/x^4*(-e^2*x^2+d^2)^(7/2)-3/4*e^2/d^2*(-1/2/d^
2/x^2*(-e^2*x^2+d^2)^(7/2)-5/2*e^2/d^2*(1/5*(-e^2*x^2+d^2)^(5/2)+d^2*(1/3*(-e^2*x^2+d^2)^(3/2)+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))))))+6*e^3/d^7*(1/d/e/(x+d/e)^3*(-
(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+4*e/d*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+5/3*e/d*(1/5*
(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)
+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2
)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)))))))-20/d^7*e^3*(-1/2/d^2/x^2*(-e^2*x^2+d^2)^(7/2)-5/2*e^2/d^2
*(1/5*(-e^2*x^2+d^2)^(5/2)+d^2*(1/3*(-e^2*x^2+d^2)^(3/2)+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)))))+e^2/d^6*(-1/d/e/(x+d/e)^4*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)-3*e/d
*(1/d/e/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+4*e/d*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))
^(7/2)+5/3*e/d*(1/5*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+
2*d*e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^
2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))))))+35/d^8*e^4*(-1/d^2/x*(-e^2*x^2+d^2)^
(7/2)-6*e^2/d^2*(1/6*x*(-e^2*x^2+d^2)^(5/2)+5/6*d^2*(1/4*x*(-e^2*x^2+d^2)^(3/2)+3/4*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))))))+21*e^4/d^8*(1/3/d/e/(x+d/e)^2*(-(x+d/
e)^2*e^2+2*d*e*(x+d/e))^(7/2)+5/3*e/d*(1/5*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*
e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+
d/e))^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))))+10/d^6*e^2*(-1/
3/d^2/x^3*(-e^2*x^2+d^2)^(7/2)-4/3*e^2/d^2*(-1/d^2/x*(-e^2*x^2+d^2)^(7/2)-6*e^2/d^2*(1/6*x*(-e^2*x^2+d^2)^(5/2
)+5/6*d^2*(1/4*x*(-e^2*x^2+d^2)^(3/2)+3/4*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)))))))-56/d^9*e^5*(1/5*(-e^2*x^2+d^2)^(5/2)+d^2*(1/3*(-e^2*x^2+d^2)^(3/2)+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))))

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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)^(5/2)/x^6/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-x^2*e^2 + d^2)^(5/2)/((x*e + d)^4*x^6), x)

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Fricas [A]
time = 2.28, size = 139, normalized size = 0.71 \begin {gather*} -\frac {80 \, x^{6} e^{6} + 80 \, d x^{5} e^{5} + 135 \, {\left (x^{6} e^{6} + d x^{5} e^{5}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (212 \, x^{5} e^{5} + 77 \, d x^{4} e^{4} - 29 \, d^{2} x^{3} e^{3} + 16 \, d^{3} x^{2} e^{2} - 8 \, d^{4} x e + 2 \, d^{5}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{10 \, {\left (d^{4} x^{6} e + d^{5} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/10*(80*x^6*e^6 + 80*d*x^5*e^5 + 135*(x^6*e^6 + d*x^5*e^5)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) + (212*x^5*e^5
+ 77*d*x^4*e^4 - 29*d^2*x^3*e^3 + 16*d^3*x^2*e^2 - 8*d^4*x*e + 2*d^5)*sqrt(-x^2*e^2 + d^2))/(d^4*x^6*e + d^5*
x^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{x^{6} \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)**(5/2)/x**6/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x**6*(d + e*x)**4), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (163) = 326\).
time = 1.28, size = 426, normalized size = 2.17 \begin {gather*} \frac {27 \, e^{5} \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^{4}} - \frac {x^{5} {\left (\frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{3}}{x} - \frac {45 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e}{x^{2}} + \frac {185 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-1\right )}}{x^{3}} - \frac {870 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-3\right )}}{x^{4}} - \frac {3670 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{\left (-5\right )}}{x^{5}} - e^{5}\right )} e^{10}}{160 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{4} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}} - \frac {\frac {1110 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{16} e^{3}}{x} - \frac {240 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{16} e}{x^{2}} + \frac {55 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{16} e^{\left (-1\right )}}{x^{3}} - \frac {10 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{16} e^{\left (-3\right )}}{x^{4}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{16} e^{\left (-5\right )}}{x^{5}}}{160 \, d^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d)^4,x, algorithm="giac")

[Out]

27/2*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/160*x^5*(9*(d*e + sqrt(-x^2*e^2
+ d^2)*e)*e^3/x - 45*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e/x^2 + 185*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-1)/x^3
- 870*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(-3)/x^4 - 3670*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*e^(-5)/x^5 - e^5)*e
^10/((d*e + sqrt(-x^2*e^2 + d^2)*e)^5*d^4*((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)) - 1/160*(1110*(d*e +
sqrt(-x^2*e^2 + d^2)*e)*d^16*e^3/x - 240*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^16*e/x^2 + 55*(d*e + sqrt(-x^2*e^2
+ d^2)*e)^3*d^16*e^(-1)/x^3 - 10*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^16*e^(-3)/x^4 + (d*e + sqrt(-x^2*e^2 + d^
2)*e)^5*d^16*e^(-5)/x^5)/d^20

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^6\,{\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)^(5/2)/(x^6*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)^4), x)

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