\(\int \frac {d+e x}{x^2 (d^2-e^2 x^2)^{7/2}} \, dx\) [28]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7} \]

[Out]

1/5*(e*x+d)/d^2/x/(-e^2*x^2+d^2)^(5/2)+1/15*(5*e*x+6*d)/d^4/x/(-e^2*x^2+d^2)^(3/2)-e*arctanh((-e^2*x^2+d^2)^(1
/2)/d)/d^7+1/5*(5*e*x+8*d)/d^6/x/(-e^2*x^2+d^2)^(1/2)-16/5*(-e^2*x^2+d^2)^(1/2)/d^7/x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 153, 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, 821, 272, 65, 214} \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7}+\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(d + e*x)/(5*d^2*x*(d^2 - e^2*x^2)^(5/2)) + (6*d + 5*e*x)/(15*d^4*x*(d^2 - e^2*x^2)^(3/2)) + (8*d + 5*e*x)/(5*
d^6*x*Sqrt[d^2 - e^2*x^2]) - (16*Sqrt[d^2 - e^2*x^2])/(5*d^7*x) - (e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^7

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 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 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {6 d^3 e^2+5 d^2 e^3 x}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {24 d^5 e^4+15 d^4 e^5 x}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {48 d^7 e^6+15 d^6 e^7 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{12} e^6} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {e \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^6} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {e \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^6} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\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^6 e} \\ & = \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.88 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^5-38 d^4 e x-52 d^3 e^2 x^2+87 d^2 e^3 x^3+33 d e^4 x^4-48 e^5 x^5\right )}{x (-d+e x)^3 (d+e x)^2}+30 e \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^7} \]

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(15*d^5 - 38*d^4*e*x - 52*d^3*e^2*x^2 + 87*d^2*e^3*x^3 + 33*d*e^4*x^4 - 48*e^5*x^5))/(x*
(-d + e*x)^3*(d + e*x)^2) + 30*e*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(15*d^7)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.39

method	result	size
default	\(e \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )+d \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )\)	\(213\)
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{7} x}-\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{6} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{24 d^{6} e \left (x +\frac {d}{e}\right )^{2}}-\frac {23 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{48 d^{7} \left (x +\frac {d}{e}\right )}+\frac {17 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{60 d^{6} e \left (x -\frac {d}{e}\right )^{2}}-\frac {413 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{240 d^{7} \left (x -\frac {d}{e}\right )}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{20 d^{5} e^{2} \left (x -\frac {d}{e}\right )^{3}}\)	\(295\)

[In]

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

[Out]

e*(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))))+d*(-1/d^2/x/(-e^2*x^2+d^2)^(5/2)+6*e^2/d^2*(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))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {23 \, e^{6} x^{6} - 23 \, d e^{5} x^{5} - 46 \, d^{2} e^{4} x^{4} + 46 \, d^{3} e^{3} x^{3} + 23 \, d^{4} e^{2} x^{2} - 23 \, d^{5} e x + 15 \, {\left (e^{6} x^{6} - d e^{5} x^{5} - 2 \, d^{2} e^{4} x^{4} + 2 \, d^{3} e^{3} x^{3} + d^{4} e^{2} x^{2} - d^{5} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (48 \, e^{5} x^{5} - 33 \, d e^{4} x^{4} - 87 \, d^{2} e^{3} x^{3} + 52 \, d^{3} e^{2} x^{2} + 38 \, d^{4} e x - 15 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{7} e^{5} x^{6} - d^{8} e^{4} x^{5} - 2 \, d^{9} e^{3} x^{4} + 2 \, d^{10} e^{2} x^{3} + d^{11} e x^{2} - d^{12} x\right )}} \]

[In]

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

[Out]

1/15*(23*e^6*x^6 - 23*d*e^5*x^5 - 46*d^2*e^4*x^4 + 46*d^3*e^3*x^3 + 23*d^4*e^2*x^2 - 23*d^5*e*x + 15*(e^6*x^6
- d*e^5*x^5 - 2*d^2*e^4*x^4 + 2*d^3*e^3*x^3 + d^4*e^2*x^2 - d^5*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (48*
e^5*x^5 - 33*d*e^4*x^4 - 87*d^2*e^3*x^3 + 52*d^3*e^2*x^2 + 38*d^4*e*x - 15*d^5)*sqrt(-e^2*x^2 + d^2))/(d^7*e^5
*x^6 - d^8*e^4*x^5 - 2*d^9*e^3*x^4 + 2*d^10*e^2*x^3 + d^11*e*x^2 - d^12*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.16 (sec) , antiderivative size = 2404, normalized size of antiderivative = 15.71 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

d*Piecewise((5*d**6*e*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*
e**6*x**6) - 30*d**4*e**3*x**2*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4
+ 5*d**8*e**6*x**6) + 40*d**2*e**5*x**4*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e
**4*x**4 + 5*d**8*e**6*x**6) - 16*e**7*x**6*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**
10*e**4*x**4 + 5*d**8*e**6*x**6), Abs(d**2/(e**2*x**2)) > 1), (5*I*d**6*e*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**1
4 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 30*I*d**4*e**3*x**2*sqrt(-d**2/(e**2*x**2) +
1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40*I*d**2*e**5*x**4*sqrt(-d**2/(
e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 16*I*e**7*x**6*sqrt(
-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6), True)) + e*Pie
cewise((-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*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) + 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) - 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(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*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), 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*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(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), Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {6 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {e}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {8 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {e}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x} + \frac {16 \, e^{2} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{7}} + \frac {e}{\sqrt {-e^{2} x^{2} + d^{2}} d^{6}} \]

[In]

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

[Out]

6/5*e^2*x/((-e^2*x^2 + d^2)^(5/2)*d^3) + 1/5*e/((-e^2*x^2 + d^2)^(5/2)*d^2) + 8/5*e^2*x/((-e^2*x^2 + d^2)^(3/2
)*d^5) + 1/3*e/((-e^2*x^2 + d^2)^(3/2)*d^4) - 1/((-e^2*x^2 + d^2)^(5/2)*d*x) + 16/5*e^2*x/(sqrt(-e^2*x^2 + d^2
)*d^7) - e*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^7 + e/(sqrt(-e^2*x^2 + d^2)*d^6)

Giac [F]

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

[In]

integrate((e*x+d)/x^2/(-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^2), x)

Mupad [B] (verification not implemented)

Time = 12.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {e}{5\,d^2}+\frac {e\,{\left (d^2-e^2\,x^2\right )}^2}{d^6}+\frac {e\,\left (d^2-e^2\,x^2\right )}{3\,d^4}}{{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{d^7}-\frac {d^6-6\,d^4\,e^2\,x^2+8\,d^2\,e^4\,x^4-\frac {16\,e^6\,x^6}{5}}{d^7\,x\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \]

[In]

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

[Out]

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