3.2.78 \(\int \frac {1}{x^3 (d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [178]

Optimal. Leaf size=183 \[ \frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7} \]

[Out]

2/5*e^2*(-e*x+d)/d^3/(-e^2*x^2+d^2)^(5/2)+1/5*e^2*(-6*e*x+5*d)/d^5/(-e^2*x^2+d^2)^(3/2)-9/2*e^2*arctanh((-e^2*
x^2+d^2)^(1/2)/d)/d^7+2/5*e^2*(-11*e*x+10*d)/d^7/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^6/x^2+2*e*(-e
^2*x^2+d^2)^(1/2)/d^7/x

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Rubi [A]
time = 0.24, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {866, 1819, 1821, 821, 272, 65, 214} \begin {gather*} \frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

(2*e^2*(d - e*x))/(5*d^3*(d^2 - e^2*x^2)^(5/2)) + (e^2*(5*d - 6*e*x))/(5*d^5*(d^2 - e^2*x^2)^(3/2)) + (2*e^2*(
10*d - 11*e*x))/(5*d^7*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^6*x^2) + (2*e*Sqrt[d^2 - e^2*x^2])/(d^7
*x) - (9*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*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 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 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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*} \int \frac {1}{x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {(d-e x)^2}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2+10 d e x-10 e^2 x^2+\frac {8 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2-30 d e x+45 e^2 x^2-\frac {36 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2+30 d e x-60 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {\int \frac {-60 d^3 e+135 d^2 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 \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 )}{2 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 136, normalized size = 0.74 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (5 d^5-10 d^4 e x-94 d^3 e^2 x^2-58 d^2 e^3 x^3+83 d e^4 x^4+64 e^5 x^5\right )}{x^2 (-d+e x) (d+e x)^3}+90 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{10 d^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(5*d^5 - 10*d^4*e*x - 94*d^3*e^2*x^2 - 58*d^2*e^3*x^3 + 83*d*e^4*x^4 + 64*e^5*x^5))/(x^2
*(-d + e*x)*(d + e*x)^3) + 90*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(10*d^7)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(482\) vs. \(2(161)=322\).
time = 0.08, size = 483, normalized size = 2.64

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-4 e x +d \right )}{2 d^{7} x^{2}}+\frac {181 e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{40 d^{7} \left (x +\frac {d}{e}\right )}-\frac {e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{8 d^{7} \left (x -\frac {d}{e}\right )}-\frac {9 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{6} \sqrt {d^{2}}}+\frac {13 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{20 d^{6} \left (x +\frac {d}{e}\right )^{2}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{10 e \,d^{5} \left (x +\frac {d}{e}\right )^{3}}\)	\(251\)
default	\(-\frac {3 e^{2} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{d^{4}}+\frac {-\frac {1}{2 d^{2} x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 e^{2} \left (\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}\right )}{2 d^{2}}}{d^{2}}-\frac {2 e \left (-\frac {1}{d^{2} x \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 e^{2} x}{d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{d^{3}}-\frac {e \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{d^{3}}+\frac {3 e^{2} \left (\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}\right )}{d^{4}}\)	\(483\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-x^2*e^2 + d^2)^(3/2)*(x*e + d)^2*x^3), x)

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Fricas [A]
time = 1.82, size = 201, normalized size = 1.10 \begin {gather*} \frac {54 \, x^{6} e^{6} + 108 \, d x^{5} e^{5} - 108 \, d^{3} x^{3} e^{3} - 54 \, d^{4} x^{2} e^{2} + 45 \, {\left (x^{6} e^{6} + 2 \, d x^{5} e^{5} - 2 \, d^{3} x^{3} e^{3} - d^{4} x^{2} e^{2}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (64 \, x^{5} e^{5} + 83 \, d x^{4} e^{4} - 58 \, d^{2} x^{3} e^{3} - 94 \, d^{3} x^{2} e^{2} - 10 \, d^{4} x e + 5 \, d^{5}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{10 \, {\left (d^{7} x^{6} e^{4} + 2 \, d^{8} x^{5} e^{3} - 2 \, d^{10} x^{3} e - d^{11} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")

[Out]

1/10*(54*x^6*e^6 + 108*d*x^5*e^5 - 108*d^3*x^3*e^3 - 54*d^4*x^2*e^2 + 45*(x^6*e^6 + 2*d*x^5*e^5 - 2*d^3*x^3*e^
3 - d^4*x^2*e^2)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) + (64*x^5*e^5 + 83*d*x^4*e^4 - 58*d^2*x^3*e^3 - 94*d^3*x^2
*e^2 - 10*d^4*x*e + 5*d^5)*sqrt(-x^2*e^2 + d^2))/(d^7*x^6*e^4 + 2*d^8*x^5*e^3 - 2*d^10*x^3*e - d^11*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(e*x+d)**2/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(1/(x**3*(-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)**2), x)

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Giac [C] Result contains complex when optimal does not.
time = 2.41, size = 330, normalized size = 1.80 \begin {gather*} -\frac {1}{40} \, {\left ({\left (\frac {180 \, e^{\left (-6\right )} \log \left (\sqrt {\frac {2 \, d}{x e + d} - 1} + 1\right )}{d^{7} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {180 \, e^{\left (-6\right )} \log \left ({\left | \sqrt {\frac {2 \, d}{x e + d} - 1} - 1 \right |}\right )}{d^{7} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {5 \, e^{\left (-6\right )}}{d^{7} \sqrt {\frac {2 \, d}{x e + d} - 1} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} + \frac {10 \, {\left (5 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {\frac {2 \, d}{x e + d} - 1}\right )} e^{\left (-6\right )}}{d^{7} {\left (\frac {d}{x e + d} - 1\right )}^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {{\left (d^{28} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{24} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} + 15 \, d^{28} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{24} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} + 195 \, d^{28} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{24} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4}\right )} e^{\left (-30\right )}}{d^{35} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{5}}\right )} e^{9} + \frac {2 \, {\left (45 \, e^{3} \log \left (2\right ) - 90 \, e^{3} \log \left (i + 1\right ) + 128 i \, e^{3}\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{7}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")

[Out]

-1/40*((180*e^(-6)*log(sqrt(2*d/(x*e + d) - 1) + 1)/(d^7*sgn(1/(x*e + d))) - 180*e^(-6)*log(abs(sqrt(2*d/(x*e
+ d) - 1) - 1))/(d^7*sgn(1/(x*e + d))) - 5*e^(-6)/(d^7*sqrt(2*d/(x*e + d) - 1)*sgn(1/(x*e + d))) + 10*(5*(2*d/
(x*e + d) - 1)^(3/2) - 3*sqrt(2*d/(x*e + d) - 1))*e^(-6)/(d^7*(d/(x*e + d) - 1)^2*sgn(1/(x*e + d))) - (d^28*(2
*d/(x*e + d) - 1)^(5/2)*e^24*sgn(1/(x*e + d))^4 + 15*d^28*(2*d/(x*e + d) - 1)^(3/2)*e^24*sgn(1/(x*e + d))^4 +
195*d^28*sqrt(2*d/(x*e + d) - 1)*e^24*sgn(1/(x*e + d))^4)*e^(-30)/(d^35*sgn(1/(x*e + d))^5))*e^9 + 2*(45*e^3*l
og(2) - 90*e^3*log(I + 1) + 128*I*e^3)*sgn(1/(x*e + d))/d^7)*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(d^2 - e^2*x^2)^(3/2)*(d + e*x)^2),x)

[Out]

int(1/(x^3*(d^2 - e^2*x^2)^(3/2)*(d + e*x)^2), x)

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