3.1.91 \(\int \frac {(d+e x)^3}{x^3 (d^2-e^2 x^2)^{7/2}} \, dx\) [91]

Optimal. Leaf size=182 \[ \frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

[Out]

4/5*e^2*(e*x+d)/d^2/(-e^2*x^2+d^2)^(5/2)+1/15*e^2*(31*e*x+25*d)/d^4/(-e^2*x^2+d^2)^(3/2)-13/2*e^2*arctanh((-e^
2*x^2+d^2)^(1/2)/d)/d^6+1/15*e^2*(107*e*x+90*d)/d^6/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^5/x^2-3*e*
(-e^2*x^2+d^2)^(1/2)/d^6/x

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Rubi [A]
time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1819, 1821, 821, 272, 65, 214} \begin {gather*} \frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(x^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(4*e^2*(d + e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e^2*(25*d + 31*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(90*d + 107*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^5*x^2) - (3*e*Sqrt[d^2 - e^2*x^2])/(
d^6*x) - (13*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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 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 {(d+e x)^3}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3-15 d^2 e x-20 d e^2 x^2-16 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3+45 d^2 e x+75 d e^2 x^2+62 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3-45 d^2 e x-90 d e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {\int \frac {90 d^4 e+195 d^3 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 \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^5}\\ &=\frac {4 e^2 (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d+31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d+107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 118, normalized size = 0.65 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^4+45 d^3 e x-479 d^2 e^2 x^2+717 d e^3 x^3-304 e^4 x^4\right )}{x^2 (-d+e x)^3}+390 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{30 d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(x^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(15*d^4 + 45*d^3*e*x - 479*d^2*e^2*x^2 + 717*d*e^3*x^3 - 304*e^4*x^4))/(x^2*(-d + e*x)^3
) + 390*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(30*d^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(435\) vs. \(2(160)=320\).
time = 0.08, size = 436, normalized size = 2.40

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (6 e x +d \right )}{2 d^{6} x^{2}}-\frac {107 e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{15 d^{6} \left (x -\frac {d}{e}\right )}-\frac {13 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{5} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{5 e \,d^{4} \left (x -\frac {d}{e}\right )^{3}}+\frac {17 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{15 d^{5} \left (x -\frac {d}{e}\right )^{2}}\)	\(214\)
default	\(e^{3} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+d^{3} \left (-\frac {1}{2 d^{2} x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {7 e^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )}{2 d^{2}}\right )+3 d^{2} e \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )+3 e^{2} d \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\)	\(436\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/x^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(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
^3*(-1/2/d^2/x^2/(-e^2*x^2+d^2)^(5/2)+7/2*e^2/d^2*(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)))))
+3*d^2*e*(-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))))+3*e^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)
)))

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Maxima [A]
time = 0.28, size = 201, normalized size = 1.10 \begin {gather*} \frac {19 \, x e^{3}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {13 \, e^{2}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d} - \frac {3 \, e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} x} - \frac {d}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}} + \frac {76 \, x e^{3}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {13 \, e^{2}}{6 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} - \frac {13 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{6}} + \frac {152 \, x e^{3}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{6}} + \frac {13 \, e^{2}}{2 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/x^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

19/5*x*e^3/((-x^2*e^2 + d^2)^(5/2)*d^2) + 13/10*e^2/((-x^2*e^2 + d^2)^(5/2)*d) - 3*e/((-x^2*e^2 + d^2)^(5/2)*x
) - 1/2*d/((-x^2*e^2 + d^2)^(5/2)*x^2) + 76/15*x*e^3/((-x^2*e^2 + d^2)^(3/2)*d^4) + 13/6*e^2/((-x^2*e^2 + d^2)
^(3/2)*d^3) - 13/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-x^2*e^2 + d^2)*d/abs(x))/d^6 + 152/15*x*e^3/(sqrt(-x^2*e^2 +
d^2)*d^6) + 13/2*e^2/(sqrt(-x^2*e^2 + d^2)*d^5)

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Fricas [A]
time = 2.33, size = 192, normalized size = 1.05 \begin {gather*} \frac {254 \, x^{5} e^{5} - 762 \, d x^{4} e^{4} + 762 \, d^{2} x^{3} e^{3} - 254 \, d^{3} x^{2} e^{2} + 195 \, {\left (x^{5} e^{5} - 3 \, d x^{4} e^{4} + 3 \, d^{2} x^{3} e^{3} - d^{3} x^{2} e^{2}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) - {\left (304 \, x^{4} e^{4} - 717 \, d x^{3} e^{3} + 479 \, d^{2} x^{2} e^{2} - 45 \, d^{3} x e - 15 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{30 \, {\left (d^{6} x^{5} e^{3} - 3 \, d^{7} x^{4} e^{2} + 3 \, d^{8} x^{3} e - d^{9} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/x^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/30*(254*x^5*e^5 - 762*d*x^4*e^4 + 762*d^2*x^3*e^3 - 254*d^3*x^2*e^2 + 195*(x^5*e^5 - 3*d*x^4*e^4 + 3*d^2*x^3
*e^3 - d^3*x^2*e^2)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) - (304*x^4*e^4 - 717*d*x^3*e^3 + 479*d^2*x^2*e^2 - 45*d
^3*x*e - 15*d^4)*sqrt(-x^2*e^2 + d^2))/(d^6*x^5*e^3 - 3*d^7*x^4*e^2 + 3*d^8*x^3*e - d^9*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/x**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**3/(x**3*(-(-d + e*x)*(d + e*x))**(7/2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (154) = 308\).
time = 0.71, size = 353, normalized size = 1.94 \begin {gather*} -\frac {13 \, e^{2} \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 )}{2 \, d^{6}} + \frac {x^{2} {\left (\frac {2782 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-2\right )}}{x^{2}} - \frac {9410 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-4\right )}}{x^{3}} + \frac {13645 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-6\right )}}{x^{4}} - \frac {9285 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{\left (-8\right )}}{x^{5}} + \frac {2580 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} e^{\left (-10\right )}}{x^{6}} - \frac {105 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}}{x} - 15 \, e^{2}\right )} e^{4}}{120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{6} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{5}} - \frac {\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{6} e^{\left (-2\right )}}{x^{2}} + \frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{6}}{x}}{8 \, d^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/x^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-13/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^6 + 1/120*x^2*(2782*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^2*e^(-2)/x^2 - 9410*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-4)/x^3 + 13645*(d*e + sqrt(-x^2*e^2 +
d^2)*e)^4*e^(-6)/x^4 - 9285*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*e^(-8)/x^5 + 2580*(d*e + sqrt(-x^2*e^2 + d^2)*e)^
6*e^(-10)/x^6 - 105*(d*e + sqrt(-x^2*e^2 + d^2)*e)/x - 15*e^2)*e^4/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^6*((d*e
+ sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x - 1)^5) - 1/8*((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^6*e^(-2)/x^2 + 12*(d*e +
sqrt(-x^2*e^2 + d^2)*e)*d^6/x)/d^12

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{x^3\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(x^3*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

int((d + e*x)^3/(x^3*(d^2 - e^2*x^2)^(7/2)), x)

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