\(\int \frac {d+e x}{x (d^2-e^2 x^2)^{7/2}} \, dx\) [27]

Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \]

[Out]

1/5*(e*x+d)/d^2/(-e^2*x^2+d^2)^(5/2)+1/15*(4*e*x+5*d)/d^4/(-e^2*x^2+d^2)^(3/2)-arctanh((-e^2*x^2+d^2)^(1/2)/d)
/d^6+1/15*(8*e*x+15*d)/d^6/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {837, 12, 272, 65, 214} \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

Int[(d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(d + e*x)/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (5*d + 4*e*x)/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (15*d + 8*e*x)/(15*d^
6*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 837

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] +
Dist[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])

Rubi steps \begin{align*} \text {integral}& = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {5 d^3 e^2+4 d^2 e^3 x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^5 e^4+8 d^4 e^5 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {15 d^7 e^6}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{12} e^6} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^5} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^5} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\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 )}{d^5 e^2} \\ & = \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.10 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (23 d^4-8 d^3 e x-27 d^2 e^2 x^2+7 d e^3 x^3+8 e^4 x^4\right )}{(d-e x)^3 (d+e x)^2}-15 \sqrt {d^2} \log (x)+15 \sqrt {d^2} \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^7} \]

[In]

Integrate[(d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

((d*Sqrt[d^2 - e^2*x^2]*(23*d^4 - 8*d^3*e*x - 27*d^2*e^2*x^2 + 7*d*e^3*x^3 + 8*e^4*x^4))/((d - e*x)^3*(d + e*x
)^2) - 15*Sqrt[d^2]*Log[x] + 15*Sqrt[d^2]*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(15*d^7)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.56

method	result	size
default	\(e \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+d \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\)	\(182\)

[In]

int((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

e*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))+d*(
1/5/d^2/(-e^2*x^2+d^2)^(5/2)+1/d^2*(1/3/d^2/(-e^2*x^2+d^2)^(3/2)+1/d^2*(1/d^2/(-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))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (103) = 206\).

Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.09 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {23 \, e^{5} x^{5} - 23 \, d e^{4} x^{4} - 46 \, d^{2} e^{3} x^{3} + 46 \, d^{3} e^{2} x^{2} + 23 \, d^{4} e x - 23 \, d^{5} + 15 \, {\left (e^{5} x^{5} - d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{3} e^{2} x^{2} + d^{4} e x - d^{5}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (8 \, e^{4} x^{4} + 7 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x + 23 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{6} e^{5} x^{5} - d^{7} e^{4} x^{4} - 2 \, d^{8} e^{3} x^{3} + 2 \, d^{9} e^{2} x^{2} + d^{10} e x - d^{11}\right )}} \]

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(23*e^5*x^5 - 23*d*e^4*x^4 - 46*d^2*e^3*x^3 + 46*d^3*e^2*x^2 + 23*d^4*e*x - 23*d^5 + 15*(e^5*x^5 - d*e^4*
x^4 - 2*d^2*e^3*x^3 + 2*d^3*e^2*x^2 + d^4*e*x - d^5)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (8*e^4*x^4 + 7*d*e^3
*x^3 - 27*d^2*e^2*x^2 - 8*d^3*e*x + 23*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^5*x^5 - d^7*e^4*x^4 - 2*d^8*e^3*x^3 +
2*d^9*e^2*x^2 + d^10*e*x - d^11)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 14.97 (sec) , antiderivative size = 2378, normalized size of antiderivative = 20.32 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate((e*x+d)/x/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((-46*I*d**6*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**
7*e**6*x**6) - 15*d**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*
x**6) + 30*d**6*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*I*d**
6*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 70*I*d**4*e**2*x**2
*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 45*d**4*
e**2*x**2*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*d*
*4*e**2*x**2*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 90*I*d**4*e
**2*x**2*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*I*d**2*e*
*4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 4
5*d**2*e**4*x**4*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6)
+ 90*d**2*e**4*x**4*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*I
*d**2*e**4*x**4*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 15*e*
*6*x**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*e**6
*x**6*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 30*I*e**6*x**6*asi
n(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6), Abs(e**2*x**2/d**2) > 1),
(-46*d**6*sqrt(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) -
15*d**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 30*d**
6*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) -
15*I*pi*d**6/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 70*d**4*e**2*x**2*sqr
t(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 45*d**4*e**2*
x**2*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*d**4*e*
*2*x**2*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x
**6) + 45*I*pi*d**4*e**2*x**2/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*d*
*2*e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6)
- 45*d**2*e**4*x**4*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x*
*6) + 90*d**2*e**4*x**4*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4
+ 30*d**7*e**6*x**6) - 45*I*pi*d**2*e**4*x**4/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e*
*6*x**6) + 15*e**6*x**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6
*x**6) - 30*e**6*x**6*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 +
30*d**7*e**6*x**6) + 15*I*pi*e**6*x**6/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6
), True)) + e*Piecewise((-15*I*d**4*x/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x
**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)) + 20*I*d**2*e**2*x**3/(15*d**11*sqrt(-1 + e**2*x**2/d
**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)) - 8*I*e**4*x
**5/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt
(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (15*d**4*x/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*e**2
*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2)) - 20*d**2*e**2*x**3/(15*d**11*sqr
t(1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2
)) + 8*e**4*x**5/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**
4*x**4*sqrt(1 - e**2*x**2/d**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {1}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {4 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {8 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{6}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{5}} \]

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/5*e*x/((-e^2*x^2 + d^2)^(5/2)*d^2) + 1/5/((-e^2*x^2 + d^2)^(5/2)*d) + 4/15*e*x/((-e^2*x^2 + d^2)^(3/2)*d^4)
+ 1/3/((-e^2*x^2 + d^2)^(3/2)*d^3) + 8/15*e*x/(sqrt(-e^2*x^2 + d^2)*d^6) - log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2
+ d^2)*d/abs(x))/d^6 + 1/(sqrt(-e^2*x^2 + d^2)*d^5)

Giac [F]

\[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {e x + d}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x} \,d x } \]

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)/((-e^2*x^2 + d^2)^(7/2)*x), x)

Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {d^2-e^2\,x^2}{3\,d^3}+\frac {{\left (d^2-e^2\,x^2\right )}^2}{d^5}+\frac {1}{5\,d}}{{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{d^6}+\frac {e\,x\,\left (15\,d^4-20\,d^2\,e^2\,x^2+8\,e^4\,x^4\right )}{15\,d^6\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \]

[In]

int((d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

((d^2 - e^2*x^2)/(3*d^3) + (d^2 - e^2*x^2)^2/d^5 + 1/(5*d))/(d^2 - e^2*x^2)^(5/2) - atanh((d^2 - e^2*x^2)^(1/2
)/d)/d^6 + (e*x*(15*d^4 + 8*e^4*x^4 - 20*d^2*e^2*x^2))/(15*d^6*(d^2 - e^2*x^2)^(5/2))