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

Optimal. Leaf size=153 \[ \frac{d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}-\frac{16 \sqrt{d^2-e^2 x^2}}{5 d^7 x}+\frac{8 d+5 e x}{5 d^6 x \sqrt{d^2-e^2 x^2}}+\frac{6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^7} \]

[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

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Rubi [A]  time = 0.126677, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {823, 807, 266, 63, 208} \[ \frac{d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}-\frac{16 \sqrt{d^2-e^2 x^2}}{5 d^7 x}+\frac{8 d+5 e x}{5 d^6 x \sqrt{d^2-e^2 x^2}}+\frac{6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^7} \]

Antiderivative was successfully verified.

[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 823

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] + Di
st[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])

Rule 807

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 266

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \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{\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 \operatorname{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{\operatorname{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]  time = 0.0778692, size = 147, normalized size = 0.96 \[ \frac{52 d^3 e^2 x^2-87 d^2 e^3 x^3-15 e x (d-e x)^2 (d+e x) \sqrt{d^2-e^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )+38 d^4 e x-15 d^5-33 d e^4 x^4+48 e^5 x^5}{15 d^7 x (d-e x)^2 (d+e x) \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.058, size = 195, normalized size = 1.3 \begin{align*}{\frac{e}{5\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{e}{3\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{e}{{d}^{6}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{e}{{d}^{6}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{1}{dx} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{6\,{e}^{2}x}{5\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,{e}^{2}x}{5\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,{e}^{2}x}{5\,{d}^{7}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.15593, size = 548, normalized size = 3.58 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 20.1131, size = 2409, normalized size = 15.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(Abs(e**2)*Abs(x**2)) > 1), (5*I*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*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*sq
rt(-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*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 + 3
0*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
) - 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)/Abs(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), True))

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Giac [A]  time = 1.18744, size = 255, normalized size = 1.67 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left (3 \,{\left (x{\left (\frac{11 \, x e^{6}}{d^{7}} + \frac{5 \, e^{5}}{d^{6}}\right )} - \frac{25 \, e^{4}}{d^{5}}\right )} x - \frac{35 \, e^{3}}{d^{4}}\right )} x + \frac{45 \, e^{2}}{d^{3}}\right )} x + \frac{23 \, e}{d^{2}}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{e \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 )}{d^{7}} + \frac{x e^{3}}{2 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{7}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{7} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*(((3*(x*(11*x*e^6/d^7 + 5*e^5/d^6) - 25*e^4/d^5)*x - 35*e^3/d^4)*x + 45*e^2/d^3)*x
+ 23*e/d^2)/(x^2*e^2 - d^2)^3 - e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^7 + 1/2*x*e^
3/((d*e + sqrt(-x^2*e^2 + d^2)*e)*d^7) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/(d^7*x)