3.3.4 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{x (d+e x)^4} \, dx\)

Optimal. Leaf size=89 \[ \frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {852, 1805, 1809, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^4),x]

[Out]

(8*d*(d - e*x))/Sqrt[d^2 - e^2*x^2] + Sqrt[d^2 - e^2*x^2] + 4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] - d*ArcTanh[
Sqrt[d^2 - e^2*x^2]/d]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 852

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 1805

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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-d^4-4 d^3 e x+d^2 e^2 x^2}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^2}\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+\frac {\int \frac {d^4 e^2+4 d^3 e^3 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+d^2 \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx+(4 d e) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+\frac {1}{2} d^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )+(4 d e) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {d^2 \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 )}{e^2}\\ &=\frac {8 d (d-e x)}{\sqrt {d^2-e^2 x^2}}+\sqrt {d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 79, normalized size = 0.89 \begin {gather*} \sqrt {d^2-e^2 x^2} \left (\frac {8 d}{d+e x}+1\right )-d \log \left (\sqrt {d^2-e^2 x^2}+d\right )+4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+d \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^4),x]

[Out]

Sqrt[d^2 - e^2*x^2]*(1 + (8*d)/(d + e*x)) + 4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + d*Log[x] - d*Log[d + Sqrt[
d^2 - e^2*x^2]]

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IntegrateAlgebraic [A]  time = 0.64, size = 117, normalized size = 1.31 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} (9 d+e x)}{d+e x}+\frac {4 d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e}+2 d \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^4),x]

[Out]

((9*d + e*x)*Sqrt[d^2 - e^2*x^2])/(d + e*x) + 2*d*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d] + (4*d*Sqr
t[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e

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fricas [A]  time = 0.41, size = 111, normalized size = 1.25 \begin {gather*} \frac {9 \, d e x + 9 \, d^{2} - 8 \, {\left (d e x + d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (d e x + d^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + 9 \, d\right )}}{e x + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x/(e*x+d)^4,x, algorithm="fricas")

[Out]

(9*d*e*x + 9*d^2 - 8*(d*e*x + d^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (d*e*x + d^2)*log(-(d - sqrt(-e
^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(e*x + 9*d))/(e*x + d)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-54*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^2-18*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2
))^5*exp(1)^10*exp(2)^3-144*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^12*exp(2)^2
-180*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^10*exp(2)^3-54*d*(-1/2*(-2*d*exp(1
)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^4-78*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*
exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^2-370*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp
(1)^10*exp(2)^3-321*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^8*exp(2)^4-75*d*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(2)^5-252*d*(-1/2*(-2*d*exp(1)-2*sqrt(d
^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^3-432*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/
x/exp(2))^3*exp(1)^8*exp(2)^4-216*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp
(2)^5-36*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^4*exp(2)^6-624*d*(-1/2*(-2*d*e
xp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^4-528*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp
(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^5-132*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4
*exp(1)^4*exp(2)^6-12*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^8-540*d*(-1/2*(-2
*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^5-324*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^3*exp(1)^4*exp(2)^6-11*d*exp(1)^8*exp(2)^4-72*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^4*exp(2)^8-312*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)
^4*exp(2)^6-36*d*exp(1)^6*exp(2)^5-108*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^
8-76*d*exp(1)^4*exp(2)^6-144*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8-72*d*exp
(2)^8-44*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^14*exp(2)+48*d*(-2*d*exp(1)-2*
sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)+144*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(
2)^6/x/exp(2)+381/2*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+81*d*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/x/exp(2)+24*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*ex
p(1)^10*exp(2)^3/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-
2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^3/(3*exp(1)^10+9*exp(1)^6*exp(2)^2+3*exp(1)^4*exp(2)^3+9*exp(1)^8*exp
(2))+4*d*sign(d)*asin(x*exp(2)/d/exp(1))+1/2*(-12*d*exp(1)^10*exp(2)^2+2*d*exp(1)^8*exp(2)^3-8*d*exp(1)^6*exp(
2)^4-40*d*exp(1)^4*exp(2)^5-64*d*exp(2)^7-4*d*exp(1)^12*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))
*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(-exp(1)^12-3*exp(1)^8*exp(2)^2-exp(1)^6
*exp(2)^3-3*exp(1)^10*exp(2))-d*exp(2)*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/ex
p(1)^2+sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.01, size = 378, normalized size = 4.25 \begin {gather*} -\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {4 d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}}+\frac {4 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e x}{d}+\frac {8 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e x}{3 d^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 d^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 d^{4}}+\frac {32 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{15 d^{4}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{4} d^{2} e^{4}}+\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{3} d^{3} e^{3}}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d^{4} e^{2}}+\sqrt {-e^{2} x^{2}+d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x/(e*x+d)^4,x)

[Out]

-1/(d^2)^(1/2)*d^2*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+1/d^2/e^4/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)
^2*e^2)^(7/2)+2/d^3/e^3/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+7/3/d^4/e^2/(x+d/e)^2*(2*(x+d/e)*d*e-(x+
d/e)^2*e^2)^(7/2)+8/3/d^3*e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+4/d*e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+
(-e^2*x^2+d^2)^(1/2)+4*d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)+1/5/d^4*(-e^2
*x^2+d^2)^(5/2)+1/3/d^2*(-e^2*x^2+d^2)^(3/2)+32/15/d^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{4} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^4*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x\,{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x*(d + e*x)^4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{x \left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x*(d + e*x)**4), x)

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