\(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^8} \, dx\) [14]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^7 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

[Out]

-1/24*e^3*(-e^2*x^2+d^2)^(3/2)/d^2/x^4-1/7*(-e^2*x^2+d^2)^(5/2)/d/x^7-1/6*e*(-e^2*x^2+d^2)^(5/2)/d^2/x^6-2/35*
e^2*(-e^2*x^2+d^2)^(5/2)/d^3/x^5-1/16*e^7*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^3+1/16*e^5*(-e^2*x^2+d^2)^(1/2)/d^
2/x^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {849, 821, 272, 43, 65, 214} \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {e^7 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}+\frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5} \]

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^8,x]

[Out]

(e^5*Sqrt[d^2 - e^2*x^2])/(16*d^2*x^2) - (e^3*(d^2 - e^2*x^2)^(3/2))/(24*d^2*x^4) - (d^2 - e^2*x^2)^(5/2)/(7*d
*x^7) - (e*(d^2 - e^2*x^2)^(5/2))/(6*d^2*x^6) - (2*e^2*(d^2 - e^2*x^2)^(5/2))/(35*d^3*x^5) - (e^7*ArcTanh[Sqrt
[d^2 - e^2*x^2]/d])/(16*d^3)

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {\int \frac {\left (-7 d^2 e-2 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^7} \, dx}{7 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}+\frac {\int \frac {\left (12 d^3 e^2+7 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx}{42 d^4} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^3 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx}{6 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^3 \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )}{12 d^2} \\ & = -\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^5 \text {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{16 d^2} \\ & = \frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}+\frac {e^7 \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{32 d^2} \\ & = \frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^5 \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 )}{16 d^2} \\ & = \frac {e^5 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {e^3 \left (d^2-e^2 x^2\right )^{3/2}}{24 d^2 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{6 d^2 x^6}-\frac {2 e^2 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^3 x^5}-\frac {e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-240 d^6-280 d^5 e x+384 d^4 e^2 x^2+490 d^3 e^3 x^3-48 d^2 e^4 x^4-105 d e^5 x^5-96 e^6 x^6\right )}{1680 d^3 x^7}-\frac {\sqrt {d^2} e^7 \log (x)}{16 d^4}+\frac {\sqrt {d^2} e^7 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{16 d^4} \]

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^8,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-240*d^6 - 280*d^5*e*x + 384*d^4*e^2*x^2 + 490*d^3*e^3*x^3 - 48*d^2*e^4*x^4 - 105*d*e^5*
x^5 - 96*e^6*x^6))/(1680*d^3*x^7) - (Sqrt[d^2]*e^7*Log[x])/(16*d^4) + (Sqrt[d^2]*e^7*Log[Sqrt[d^2] - Sqrt[d^2
- e^2*x^2]])/(16*d^4)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (96 e^{6} x^{6}+105 d \,e^{5} x^{5}+48 d^{2} e^{4} x^{4}-490 d^{3} x^{3} e^{3}-384 d^{4} e^{2} x^{2}+280 d^{5} e x +240 d^{6}\right )}{1680 x^{7} d^{3}}-\frac {e^{7} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{16 d^{2} \sqrt {d^{2}}}\)	\(132\)
default	\(e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 d^{2} x^{6}}+\frac {e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{4 d^{2} x^{4}}-\frac {e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{2 d^{2} x^{2}}-\frac {3 e^{2} \left (\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3}+d^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )\right )}{2 d^{2}}\right )}{4 d^{2}}\right )}{6 d^{2}}\right )+d \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{7 d^{2} x^{7}}-\frac {2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{35 d^{4} x^{5}}\right )\)	\(225\)

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/1680*(-e^2*x^2+d^2)^(1/2)*(96*e^6*x^6+105*d*e^5*x^5+48*d^2*e^4*x^4-490*d^3*e^3*x^3-384*d^4*e^2*x^2+280*d^5*
e*x+240*d^6)/x^7/d^3-1/16/d^2*e^7/(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 = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.70 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {105 \, e^{7} x^{7} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (96 \, e^{6} x^{6} + 105 \, d e^{5} x^{5} + 48 \, d^{2} e^{4} x^{4} - 490 \, d^{3} e^{3} x^{3} - 384 \, d^{4} e^{2} x^{2} + 280 \, d^{5} e x + 240 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, d^{3} x^{7}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

1/1680*(105*e^7*x^7*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (96*e^6*x^6 + 105*d*e^5*x^5 + 48*d^2*e^4*x^4 - 490*d^
3*e^3*x^3 - 384*d^4*e^2*x^2 + 280*d^5*e*x + 240*d^6)*sqrt(-e^2*x^2 + d^2))/(d^3*x^7)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.68 (sec) , antiderivative size = 1037, normalized size of antiderivative = 6.03 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\text {Too large to display} \]

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**8,x)

[Out]

d**3*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(7*x**6) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(35*d**2*x**4) + 4*e*
*5*sqrt(d**2/(e**2*x**2) - 1)/(105*d**4*x**2) + 8*e**7*sqrt(d**2/(e**2*x**2) - 1)/(105*d**6), Abs(d**2/(e**2*x
**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(7*x**6) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(35*d**2*x**4) + 4
*I*e**5*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**4*x**2) + 8*I*e**7*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**6), True))
+ d**2*e*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e
**3/(48*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*
x))/(16*d**5), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x**5*sqr
t(-d**2/(e**2*x**2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-d**2/(
e**2*x**2) + 1)) - I*e**6*asin(d/(e*x))/(16*d**5), True)) - d*e**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*sqr
t(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)) - e**3*Piecewise((-d**2/(4*e*x**5*sqrt(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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=-\frac {e^{7} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{16 \, d^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{7}}{16 \, d^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7}}{48 \, d^{6}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}}{48 \, d^{6} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}}{24 \, d^{4} x^{4}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}}{35 \, d^{3} x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{6 \, d^{2} x^{6}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{7 \, d x^{7}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

-1/16*e^7*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^3 + 1/16*sqrt(-e^2*x^2 + d^2)*e^7/d^4 + 1/48*(
-e^2*x^2 + d^2)^(3/2)*e^7/d^6 + 1/48*(-e^2*x^2 + d^2)^(5/2)*e^5/(d^6*x^2) - 1/24*(-e^2*x^2 + d^2)^(5/2)*e^3/(d
^4*x^4) - 2/35*(-e^2*x^2 + d^2)^(5/2)*e^2/(d^3*x^5) - 1/6*(-e^2*x^2 + d^2)^(5/2)*e/(d^2*x^6) - 1/7*(-e^2*x^2 +
d^2)^(5/2)/(d*x^7)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (148) = 296\).

Time = 0.29 (sec) , antiderivative size = 518, normalized size of antiderivative = 3.01 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {{\left (15 \, e^{8} + \frac {35 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e^{6}}{x} - \frac {21 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} e^{4}}{x^{2}} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} e^{2}}{x^{3}} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{x^{4}} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{2} x^{5}} + \frac {315 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{6}}{e^{4} x^{6}}\right )} e^{14} x^{7}}{13440 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{7} d^{3} {\left | e \right |}} - \frac {e^{8} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{16 \, d^{3} {\left | e \right |}} - \frac {\frac {315 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{18} e^{12}}{x} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{18} e^{10}}{x^{2}} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d^{18} e^{8}}{x^{3}} - \frac {105 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d^{18} e^{6}}{x^{4}} - \frac {21 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5} d^{18} e^{4}}{x^{5}} + \frac {35 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{6} d^{18} e^{2}}{x^{6}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{7} d^{18}}{x^{7}}}{13440 \, d^{21} e^{6} {\left | e \right |}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

1/13440*(15*e^8 + 35*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e^6/x - 21*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*e^4/
x^2 - 105*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*e^2/x^3 - 105*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/x^4 - 105*
(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^2*x^5) + 315*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^6/(e^4*x^6))*e^14*x^
7/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^7*d^3*abs(e)) - 1/16*e^8*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*ab
s(e))/(e^2*abs(x)))/(d^3*abs(e)) - 1/13440*(315*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^18*e^12/x - 105*(d*e + s
qrt(-e^2*x^2 + d^2)*abs(e))^2*d^18*e^10/x^2 - 105*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*d^18*e^8/x^3 - 105*(d*
e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*d^18*e^6/x^4 - 21*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5*d^18*e^4/x^5 + 35*(
d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^6*d^18*e^2/x^6 + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^7*d^18/x^7)/(d^21*e
^6*abs(e))

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx=\frac {8\,d\,e^2\,\sqrt {d^2-e^2\,x^2}}{35\,x^5}-\frac {d^3\,\sqrt {d^2-e^2\,x^2}}{7\,x^7}-\frac {e^4\,\sqrt {d^2-e^2\,x^2}}{35\,d\,x^3}-\frac {2\,e^6\,\sqrt {d^2-e^2\,x^2}}{35\,d^3\,x}-\frac {e\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{6\,x^6}+\frac {d^2\,e\,\sqrt {d^2-e^2\,x^2}}{16\,x^6}-\frac {e\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{16\,d^2\,x^6}+\frac {e^7\,\mathrm {atan}\left (\frac {\sqrt {d^2-e^2\,x^2}\,1{}\mathrm {i}}{d}\right )\,1{}\mathrm {i}}{16\,d^3} \]

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^8,x)

[Out]

(e^7*atan(((d^2 - e^2*x^2)^(1/2)*1i)/d)*1i)/(16*d^3) - (d^3*(d^2 - e^2*x^2)^(1/2))/(7*x^7) - (e*(d^2 - e^2*x^2
)^(3/2))/(6*x^6) - (e^4*(d^2 - e^2*x^2)^(1/2))/(35*d*x^3) - (2*e^6*(d^2 - e^2*x^2)^(1/2))/(35*d^3*x) + (8*d*e^
2*(d^2 - e^2*x^2)^(1/2))/(35*x^5) + (d^2*e*(d^2 - e^2*x^2)^(1/2))/(16*x^6) - (e*(d^2 - e^2*x^2)^(5/2))/(16*d^2
*x^6)