\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^7 (d+e x)^2} \, dx\) [169]

Optimal result

Integrand size = 27, antiderivative size = 169 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=-\frac {3 e^4 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac {3 e^6 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {866, 1821, 849, 821, 272, 43, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=\frac {3 e^6 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}-\frac {3 e^4 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3} \]

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 849

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)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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 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}& = \int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^7} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}-\frac {\int \frac {\left (12 d^3 e-9 d^2 e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x^6} \, dx}{6 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}+\frac {\int \frac {\left (45 d^4 e^2-24 d^3 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x^5} \, dx}{30 d^4} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}-\frac {\int \frac {\left (96 d^5 e^3-45 d^4 e^4 x\right ) \sqrt {d^2-e^2 x^2}}{x^4} \, dx}{120 d^6} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac {\left (3 e^4\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{x^3} \, dx}{8 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac {\left (3 e^4\right ) \text {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{16 d^2} \\ & = -\frac {3 e^4 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}-\frac {\left (3 e^6\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{32 d^2} \\ & = -\frac {3 e^4 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac {\left (3 e^4\right ) \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 )}{16 d^2} \\ & = -\frac {3 e^4 \sqrt {d^2-e^2 x^2}}{16 d^2 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac {3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac {4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac {3 e^6 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{16 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.73 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (40 d^5-96 d^4 e x+50 d^3 e^2 x^2+32 d^2 e^3 x^3-45 d e^4 x^4+64 e^5 x^5\right )+90 e^6 x^6 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{240 d^3 x^6} \]

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2),x]

[Out]

-1/240*(Sqrt[d^2 - e^2*x^2]*(40*d^5 - 96*d^4*e*x + 50*d^3*e^2*x^2 + 32*d^2*e^3*x^3 - 45*d*e^4*x^4 + 64*e^5*x^5
) + 90*e^6*x^6*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(d^3*x^6)

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (64 e^{5} x^{5}-45 d \,e^{4} x^{4}+32 d^{2} e^{3} x^{3}+50 d^{3} e^{2} x^{2}-96 d^{4} e x +40 d^{5}\right )}{240 d^{3} x^{6}}+\frac {3 e^{6} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{16 d^{2} \sqrt {d^{2}}}\)	\(121\)
default	\(\text {Expression too large to display}\)	\(1550\)

[In]

int((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

-1/240*(-e^2*x^2+d^2)^(1/2)*(64*e^5*x^5-45*d*e^4*x^4+32*d^2*e^3*x^3+50*d^3*e^2*x^2-96*d^4*e*x+40*d^5)/d^3/x^6+
3/16/d^2*e^6/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=-\frac {45 \, e^{6} x^{6} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (64 \, e^{5} x^{5} - 45 \, d e^{4} x^{4} + 32 \, d^{2} e^{3} x^{3} + 50 \, d^{3} e^{2} x^{2} - 96 \, d^{4} e x + 40 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, d^{3} x^{6}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/240*(45*e^6*x^6*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (64*e^5*x^5 - 45*d*e^4*x^4 + 32*d^2*e^3*x^3 + 50*d^3*e
^2*x^2 - 96*d^4*e*x + 40*d^5)*sqrt(-e^2*x^2 + d^2))/(d^3*x^6)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.28 (sec) , antiderivative size = 808, normalized size of antiderivative = 4.78 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=d^{2} \left (\begin {cases} - \frac {d^{2}}{6 e x^{7} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {5 e}{24 x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{3}}{48 d^{2} x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{5}}{16 d^{4} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{6} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{16 d^{5}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{6 e x^{7} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {5 i e}{24 x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{3}}{48 d^{2} x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{5}}{16 d^{4} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{6} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{16 d^{5}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \frac {3 i d^{3} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 i d e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 i e^{6} x^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {i e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{3} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 d e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 e^{6} x^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**7/(e*x+d)**2,x)

[Out]

d**2*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e**3/
(48*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*x))/
(16*d**5), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x**5*sqrt(-d
**2/(e**2*x**2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-d**2/(e**2
*x**2) + 1)) - I*e**6*asin(d/(e*x))/(16*d**5), True)) - 2*d*e*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-
15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*I*
e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2
)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2/d**2) > 1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5
+ 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 -
e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3*x**5 + 15*
d*e**2*x**7), True)) + e**2*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e*
*2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**
2)) > 1), (I*d**2/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3
/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=\frac {3 \, e^{6} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{16 \, d^{3}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}}{16 \, d^{4}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}{16 \, d^{4} x^{2}} + \frac {4 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{15 \, d^{3} x^{3}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d^{2} x^{4}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{5 \, d x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{6 \, x^{6}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x, algorithm="maxima")

[Out]

3/16*e^6*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^3 - 3/16*sqrt(-e^2*x^2 + d^2)*e^6/d^4 - 3/16*(-
e^2*x^2 + d^2)^(3/2)*e^4/(d^4*x^2) + 4/15*(-e^2*x^2 + d^2)^(3/2)*e^3/(d^3*x^3) - 3/8*(-e^2*x^2 + d^2)^(3/2)*e^
2/(d^2*x^4) + 2/5*(-e^2*x^2 + d^2)^(3/2)*e/(d*x^5) - 1/6*(-e^2*x^2 + d^2)^(3/2)/x^6

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.80 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=\frac {1}{7680} \, {\left (\frac {1440 \, e^{5} \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{3}} - \frac {1440 \, e^{5} \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{3}} + \frac {16 \, {\left (45 \, e^{5} \log \left (2\right ) - 90 \, e^{5} \log \left (i + 1\right ) + 128 i \, e^{5}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{3}} - \frac {45 \, e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {11}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 1025 \, e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 174 \, e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 594 \, e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 255 \, e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 45 \, e^{5} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{3} {\left (\frac {d}{e x + d} - 1\right )}^{6}}\right )} {\left | e \right |} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x, algorithm="giac")

[Out]

1/7680*(1440*e^5*log(sqrt(2*d/(e*x + d) - 1) + 1)*sgn(1/(e*x + d))*sgn(e)/d^3 - 1440*e^5*log(abs(sqrt(2*d/(e*x
+ d) - 1) - 1))*sgn(1/(e*x + d))*sgn(e)/d^3 + 16*(45*e^5*log(2) - 90*e^5*log(I + 1) + 128*I*e^5)*sgn(1/(e*x +
d))*sgn(e)/d^3 - (45*e^5*(2*d/(e*x + d) - 1)^(11/2)*sgn(1/(e*x + d))*sgn(e) + 1025*e^5*(2*d/(e*x + d) - 1)^(9
/2)*sgn(1/(e*x + d))*sgn(e) - 174*e^5*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) + 594*e^5*(2*d/(e*x +
d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) - 255*e^5*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) + 45*e^5*sqr
t(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))/(d^3*(d/(e*x + d) - 1)^6))*abs(e)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^7\,{\left (d+e\,x\right )}^2} \,d x \]

[In]

int((d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2), x)