\(\int \frac {(d+e x)^2}{x^4 (d^2-e^2 x^2)^{7/2}} \, dx\) [53]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 27, antiderivative size = 209 \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {7 e^3 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8} \]

[Out]

2/5*e^3*(e*x+d)/d^4/(-e^2*x^2+d^2)^(5/2)+1/15*e^3*(23*e*x+20*d)/d^6/(-e^2*x^2+d^2)^(3/2)-7*e^3*arctanh((-e^2*x
^2+d^2)^(1/2)/d)/d^8+2/15*e^3*(53*e*x+45*d)/d^8/(-e^2*x^2+d^2)^(1/2)-1/3*(-e^2*x^2+d^2)^(1/2)/d^6/x^3-e*(-e^2*
x^2+d^2)^(1/2)/d^7/x^2-14/3*e^2*(-e^2*x^2+d^2)^(1/2)/d^8/x

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1819, 1821, 821, 272, 65, 214} \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {7 e^3 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}} \]

[In]

Int[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(2*e^3*(d + e*x))/(5*d^4*(d^2 - e^2*x^2)^(5/2)) + (e^3*(20*d + 23*e*x))/(15*d^6*(d^2 - e^2*x^2)^(3/2)) + (2*e^
3*(45*d + 53*e*x))/(15*d^8*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*d^6*x^3) - (e*Sqrt[d^2 - e^2*x^2])/(d
^7*x^2) - (14*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^8*x) - (7*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^8

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*} \text {integral}& = \frac {2 e^3 (d+e x)}{5 d^4 \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 {10 e^3 x^3}{d}-\frac {8 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \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 {60 e^3 x^3}{d}+\frac {46 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2-30 d e x-60 e^2 x^2-\frac {90 e^3 x^3}{d}}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}+\frac {\int \frac {90 d^3 e+210 d^2 e^2 x+270 d e^3 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{45 d^8} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {\int \frac {-420 d^4 e^2-630 d^3 e^3 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{90 d^{10}} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {\left (7 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^7} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {\left (7 e^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^7} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {(7 e) \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^7} \\ & = \frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (5 d^6+5 d^5 e x+40 d^4 e^2 x^2-246 d^3 e^3 x^3+122 d^2 e^4 x^4+247 d e^5 x^5-176 e^6 x^6\right )}{x^3 (-d+e x)^3 (d+e x)}-105 \sqrt {d^2} e^3 \log (x)+105 \sqrt {d^2} e^3 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^9} \]

[In]

Integrate[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

((d*Sqrt[d^2 - e^2*x^2]*(5*d^6 + 5*d^5*e*x + 40*d^4*e^2*x^2 - 246*d^3*e^3*x^3 + 122*d^2*e^4*x^4 + 247*d*e^5*x^
5 - 176*e^6*x^6))/(x^3*(-d + e*x)^3*(d + e*x)) - 105*Sqrt[d^2]*e^3*Log[x] + 105*Sqrt[d^2]*e^3*Log[Sqrt[d^2] -
Sqrt[d^2 - e^2*x^2]])/(15*d^9)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.29

method result size
risch \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (14 e^{2} x^{2}+3 d e x +d^{2}\right )}{3 d^{8} x^{3}}-\frac {7 e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{7} \sqrt {d^{2}}}-\frac {e^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{8 d^{8} \left (x +\frac {d}{e}\right )}-\frac {833 e^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{120 d^{8} \left (x -\frac {d}{e}\right )}+\frac {49 e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{60 d^{7} \left (x -\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{10 d^{6} \left (x -\frac {d}{e}\right )^{3}}\) \(270\)
default \(e^{2} \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )+d^{2} \left (-\frac {1}{3 d^{2} x^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {8 e^{2} \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )}{3 d^{2}}\right )+2 d e \left (-\frac {1}{2 d^{2} x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {7 e^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )}{2 d^{2}}\right )\) \(381\)

[In]

int((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(-e^2*x^2+d^2)^(1/2)*(14*e^2*x^2+3*d*e*x+d^2)/d^8/x^3-7*e^3/d^7/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2
*x^2+d^2)^(1/2))/x)-1/8*e^2/d^8/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-833/120*e^2/d^8/(x-d/e)*(-(x-d/e)
^2*e^2-2*d*e*(x-d/e))^(1/2)+49/60/d^7*e/(x-d/e)^2*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)-1/10/d^6/(x-d/e)^3*(-(x
-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {116 \, e^{7} x^{7} - 232 \, d e^{6} x^{6} + 232 \, d^{3} e^{4} x^{4} - 116 \, d^{4} e^{3} x^{3} + 105 \, {\left (e^{7} x^{7} - 2 \, d e^{6} x^{6} + 2 \, d^{3} e^{4} x^{4} - d^{4} e^{3} x^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (176 \, e^{6} x^{6} - 247 \, d e^{5} x^{5} - 122 \, d^{2} e^{4} x^{4} + 246 \, d^{3} e^{3} x^{3} - 40 \, d^{4} e^{2} x^{2} - 5 \, d^{5} e x - 5 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{8} e^{4} x^{7} - 2 \, d^{9} e^{3} x^{6} + 2 \, d^{11} e x^{4} - d^{12} x^{3}\right )}} \]

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(116*e^7*x^7 - 232*d*e^6*x^6 + 232*d^3*e^4*x^4 - 116*d^4*e^3*x^3 + 105*(e^7*x^7 - 2*d*e^6*x^6 + 2*d^3*e^4
*x^4 - d^4*e^3*x^3)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (176*e^6*x^6 - 247*d*e^5*x^5 - 122*d^2*e^4*x^4 + 246*
d^3*e^3*x^3 - 40*d^4*e^2*x^2 - 5*d^5*e*x - 5*d^6)*sqrt(-e^2*x^2 + d^2))/(d^8*e^4*x^7 - 2*d^9*e^3*x^6 + 2*d^11*
e*x^4 - d^12*x^3)

Sympy [F]

\[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((e*x+d)**2/x**4/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**2/(x**4*(-(-d + e*x)*(d + e*x))**(7/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {22 \, e^{4} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4}} + \frac {7 \, e^{3}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {88 \, e^{4} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{6}} + \frac {7 \, e^{3}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} - \frac {11 \, e^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x} + \frac {176 \, e^{4} x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{8}} - \frac {7 \, e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{8}} + \frac {7 \, e^{3}}{\sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x^{2}} - \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{3}} \]

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

22/5*e^4*x/((-e^2*x^2 + d^2)^(5/2)*d^4) + 7/5*e^3/((-e^2*x^2 + d^2)^(5/2)*d^3) + 88/15*e^4*x/((-e^2*x^2 + d^2)
^(3/2)*d^6) + 7/3*e^3/((-e^2*x^2 + d^2)^(3/2)*d^5) - 11/3*e^2/((-e^2*x^2 + d^2)^(5/2)*d^2*x) + 176/15*e^4*x/(s
qrt(-e^2*x^2 + d^2)*d^8) - 7*e^3*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^8 + 7*e^3/(sqrt(-e^2*x^
2 + d^2)*d^7) - e/((-e^2*x^2 + d^2)^(5/2)*d*x^2) - 1/3/((-e^2*x^2 + d^2)^(5/2)*x^3)

Giac [F]

\[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x^{4}} \,d x } \]

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^2/((-e^2*x^2 + d^2)^(7/2)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{x^4\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[In]

int((d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

int((d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)), x)