\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^6 (d+e x)^2} \, dx\) [168]

Optimal result

Integrand size = 27, antiderivative size = 140 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^5 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^2} \]

[Out]

-1/5*(-e^2*x^2+d^2)^(3/2)/x^5+1/2*e*(-e^2*x^2+d^2)^(3/2)/d/x^4-7/15*e^2*(-e^2*x^2+d^2)^(3/2)/d^2/x^3-1/4*e^5*a
rctanh((-e^2*x^2+d^2)^(1/2)/d)/d^2+1/4*e^3*(-e^2*x^2+d^2)^(1/2)/d/x^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {866, 1821, 849, 821, 272, 43, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=-\frac {e^5 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}+\frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2} \]

[In]

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

[Out]

(e^3*Sqrt[d^2 - e^2*x^2])/(4*d*x^2) - (d^2 - e^2*x^2)^(3/2)/(5*x^5) + (e*(d^2 - e^2*x^2)^(3/2))/(2*d*x^4) - (7
*e^2*(d^2 - e^2*x^2)^(3/2))/(15*d^2*x^3) - (e^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(4*d^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 849

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)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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 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*} \text {integral}& = \int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^6} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}-\frac {\int \frac {\left (10 d^3 e-7 d^2 e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x^5} \, dx}{5 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}+\frac {\int \frac {\left (28 d^4 e^2-10 d^3 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x^4} \, dx}{20 d^4} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^3 \int \frac {\sqrt {d^2-e^2 x^2}}{x^3} \, dx}{2 d} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^3 \text {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{4 d} \\ & = \frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}+\frac {e^5 \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{8 d} \\ & = \frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^3 \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 )}{4 d} \\ & = \frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-12 d^4+30 d^3 e x-16 d^2 e^2 x^2-15 d e^3 x^3+28 e^4 x^4\right )}{60 d^2 x^5}-\frac {\sqrt {d^2} e^5 \log (x)}{4 d^3}+\frac {\sqrt {d^2} e^5 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{4 d^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-12*d^4 + 30*d^3*e*x - 16*d^2*e^2*x^2 - 15*d*e^3*x^3 + 28*e^4*x^4))/(60*d^2*x^5) - (Sqrt
[d^2]*e^5*Log[x])/(4*d^3) + (Sqrt[d^2]*e^5*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(4*d^3)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-28 e^{4} x^{4}+15 d \,e^{3} x^{3}+16 d^{2} e^{2} x^{2}-30 d^{3} e x +12 d^{4}\right )}{60 x^{5} d^{2}}-\frac {e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{4 d \sqrt {d^{2}}}\)	\(110\)
default	\(\text {Expression too large to display}\)	\(1349\)

[In]

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

[Out]

-1/60*(-e^2*x^2+d^2)^(1/2)*(-28*e^4*x^4+15*d*e^3*x^3+16*d^2*e^2*x^2-30*d^3*e*x+12*d^4)/x^5/d^2-1/4/d*e^5/(d^2)
^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.69 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (28 \, e^{4} x^{4} - 15 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 30 \, d^{3} e x - 12 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{60 \, d^{2} x^{5}} \]

[In]

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

[Out]

1/60*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (28*e^4*x^4 - 15*d*e^3*x^3 - 16*d^2*e^2*x^2 + 30*d^3*e*x
- 12*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2*x^5)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.88 (sec) , antiderivative size = 660, normalized size of antiderivative = 4.71 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=d^{2} \left (\begin {cases} \frac {3 i d^{3} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 i d e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 i e^{6} x^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {i e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{3} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 d e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 e^{6} x^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**2*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 +
e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d
**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2/d**2) >
1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/
(-15*d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - e*
*4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), True)) - 2*d*e*Piecewise((-d**2/(4*e*x**5*s
qrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) -
1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1))
- 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/
(e*x))/(8*d**3), True)) + e**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2)
- 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**
2*x**2) + 1)/(3*d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=-\frac {e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{4 \, d^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{5}}{4 \, d^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{4 \, d^{3} x^{2}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{15 \, d^{2} x^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{2 \, d x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{5 \, x^{5}} \]

[In]

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

[Out]

-1/4*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^2 + 1/4*sqrt(-e^2*x^2 + d^2)*e^5/d^3 + 1/4*(-e^
2*x^2 + d^2)^(3/2)*e^3/(d^3*x^2) - 7/15*(-e^2*x^2 + d^2)^(3/2)*e^2/(d^2*x^3) + 1/2*(-e^2*x^2 + d^2)^(3/2)*e/(d
*x^4) - 1/5*(-e^2*x^2 + d^2)^(3/2)/x^5

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.96 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=-\frac {1}{960} \, {\left (\frac {240 \, e^{4} \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} - \frac {240 \, e^{4} \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} + \frac {8 \, {\left (15 \, e^{4} \log \left (2\right ) - 30 \, e^{4} \log \left (i + 1\right ) + 56 i \, e^{4}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} - \frac {15 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 250 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 128 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 70 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 15 \, e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2} {\left (\frac {d}{e x + d} - 1\right )}^{5}}\right )} {\left | e \right |} \]

[In]

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

[Out]

-1/960*(240*e^4*log(sqrt(2*d/(e*x + d) - 1) + 1)*sgn(1/(e*x + d))*sgn(e)/d^2 - 240*e^4*log(abs(sqrt(2*d/(e*x +
d) - 1) - 1))*sgn(1/(e*x + d))*sgn(e)/d^2 + 8*(15*e^4*log(2) - 30*e^4*log(I + 1) + 56*I*e^4)*sgn(1/(e*x + d))
*sgn(e)/d^2 - (15*e^4*(2*d/(e*x + d) - 1)^(9/2)*sgn(1/(e*x + d))*sgn(e) + 250*e^4*(2*d/(e*x + d) - 1)^(7/2)*sg
n(1/(e*x + d))*sgn(e) - 128*e^4*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) + 70*e^4*(2*d/(e*x + d) - 1)
^(3/2)*sgn(1/(e*x + d))*sgn(e) - 15*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))/(d^2*(d/(e*x + d) - 1
)^5))*abs(e)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^6\,{\left (d+e\,x\right )}^2} \,d x \]

[In]

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

[Out]

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