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

Optimal. Leaf size=137 \[ -\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}+\frac {10 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \]

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Rubi [A]  time = 0.30, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1805, 1807, 807, 266, 63, 208} \begin {gather*} -\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {10 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*e^3*(d - e*x))/(d^2*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*x^3) + (2*e*Sqrt[d^2 - e^2*x^2])/(d*x^2)
- (23*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^2*x) + (10*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^2

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 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 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 807

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 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 1807

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 {\left (d^2-e^2 x^2\right )^{5/2}}{x^4 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-d^4+4 d^3 e x-7 d^2 e^2 x^2+8 d e^3 x^3}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{d^2}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {\int \frac {-12 d^5 e+23 d^4 e^2 x-24 d^3 e^3 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{3 d^4}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {\int \frac {-46 d^6 e^2+60 d^5 e^3 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{6 d^6}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}-\frac {\left (10 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}-\frac {\left (5 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{d}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}+\frac {(10 e) \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 )}{d}\\ &=-\frac {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 x^3}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d x^2}-\frac {23 e^2 \sqrt {d^2-e^2 x^2}}{3 d^2 x}+\frac {10 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 94, normalized size = 0.69 \begin {gather*} -\frac {-30 e^3 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (d^3-5 d^2 e x+17 d e^2 x^2+47 e^3 x^3\right )}{x^3 (d+e x)}+30 e^3 \log (x)}{3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((Sqrt[d^2 - e^2*x^2]*(d^3 - 5*d^2*e*x + 17*d*e^2*x^2 + 47*e^3*x^3))/(x^3*(d + e*x)) + 30*e^3*Log[x] - 30
*e^3*Log[d + Sqrt[d^2 - e^2*x^2]])/d^2

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IntegrateAlgebraic [A]  time = 0.76, size = 109, normalized size = 0.80 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-d^3+5 d^2 e x-17 d e^2 x^2-47 e^3 x^3\right )}{3 d^2 x^3 (d+e x)}-\frac {20 e^3 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-d^3 + 5*d^2*e*x - 17*d*e^2*x^2 - 47*e^3*x^3))/(3*d^2*x^3*(d + e*x)) - (20*e^3*ArcTanh[(
Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^2

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fricas [A]  time = 0.40, size = 123, normalized size = 0.90 \begin {gather*} -\frac {24 \, e^{4} x^{4} + 24 \, d e^{3} x^{3} + 30 \, {\left (e^{4} x^{4} + d e^{3} x^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (47 \, e^{3} x^{3} + 17 \, d e^{2} x^{2} - 5 \, d^{2} e x + d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (d^{2} e x^{4} + d^{3} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(24*e^4*x^4 + 24*d*e^3*x^3 + 30*(e^4*x^4 + d*e^3*x^3)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (47*e^3*x^3 +
17*d*e^2*x^2 - 5*d^2*e*x + d^3)*sqrt(-e^2*x^2 + d^2))/(d^2*e*x^4 + d^3*x^3)

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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^4/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/512*(256*d^4*(-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)^16-64/3*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*ex
p(1))/x/exp(2))^3*exp(1)^8*exp(2)^17+96*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^17/x/e
xp(2)-384*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*exp(2)^16/x/exp(2)+1280*d^4*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^12*exp(2)^15/x/exp(2))/d^6/exp(1)^15/exp(2)^12+1/72*((-1/2*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*(16416*exp(1)^16*exp(2)^2+34560*exp(1)^14*exp(2)^3+20376*exp(1)^12*e
xp(2)^4+8712*exp(1)^10*exp(2)^5+9234*exp(1)^8*exp(2)^6-1674*exp(1)^6*exp(2)^7-594*exp(1)^4*exp(2)^8+234*exp(2)
^10)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^6*(13824*exp(1)^16*exp(2)^2+17952*exp(1)^14*e
xp(2)^3+27216*exp(1)^12*exp(2)^4+14724*exp(1)^10*exp(2)^5-4104*exp(1)^8*exp(2)^6+1380*exp(1)^6*exp(2)^7+468*ex
p(1)^4*exp(2)^8+12*exp(2)^10+7104*exp(1)^18*exp(2))+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2)
)^7*(7776*exp(1)^16*exp(2)^2+14688*exp(1)^14*exp(2)^3+6192*exp(1)^12*exp(2)^4+3240*exp(1)^10*exp(2)^5+5454*exp
(1)^8*exp(2)^6-702*exp(1)^6*exp(2)^7-270*exp(1)^4*exp(2)^8+126*exp(2)^10)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^8*(2160*exp(1)^14*exp(2)^3+3888*exp(1)^12*exp(2)^4+648*exp(1)^10*exp(2)^5-756*exp(1)^8
*exp(2)^6+855*exp(1)^6*exp(2)^7+117*exp(1)^4*exp(2)^8)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^4*(12528*exp(1)^14*exp(2)^3+27648*exp(1)^12*exp(2)^4+13392*exp(1)^10*exp(2)^5-3204*exp(1)^8*exp(2)^6+774*
exp(1)^6*exp(2)^7+306*exp(1)^4*exp(2)^8+36*exp(2)^10)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(
2))^3*(3528*exp(1)^12*exp(2)^4+8064*exp(1)^10*exp(2)^5+4146*exp(1)^8*exp(2)^6-1218*exp(1)^6*exp(2)^7-378*exp(1
)^4*exp(2)^8+90*exp(2)^10)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(180*exp(1)^10*exp(2)
^5+432*exp(1)^8*exp(2)^6+252*exp(1)^6*exp(2)^7-36*exp(1)^4*exp(2)^8+36*exp(2)^10)+3*exp(1)^6*exp(2)^7+9*exp(1)
^4*exp(2)^8+12*exp(2)^10-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*(-18*exp(1)^8*exp(2)^6-54*exp(1)^6*ex
p(2)^7-54*exp(1)^4*exp(2)^8-18*exp(2)^10)/x/exp(2))/d^2/(2*exp(2))^3/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))
*exp(1))/x/exp(2))^3/((-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/exp(1)^5-(10*exp(2)^3-20*exp(1)^4*exp(2))*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d
^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^2/exp(1)/exp(2)+1/2*(-192*exp(1)^8*exp(2)^2-64*exp(1)^6*exp(2)^3+136*e
xp(1)^4*exp(2)^4+74*exp(2)^6-80*exp(1)^10*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)/(d^2*exp(1)^7+3*d^2*exp(1)^5*exp(2)+4*d^2*exp(1)*exp(2
)^3)

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maple [B]  time = 0.02, size = 575, normalized size = 4.20 \begin {gather*} \frac {10 e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d}+\frac {65 e^{4} \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 )}{4 \sqrt {e^{2}}\, d^{2}}-\frac {65 e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{4 \sqrt {e^{2}}\, d^{2}}-\frac {65 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4} x}{4 d^{4}}+\frac {65 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{4} x}{4 d^{4}}-\frac {10 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{d^{3}}-\frac {65 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4} x}{6 d^{6}}+\frac {65 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{4} x}{6 d^{6}}-\frac {10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}{3 d^{5}}-\frac {26 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4} x}{3 d^{8}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{d^{7}}+\frac {26 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{3}}{3 d^{7}}-\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^{5} e}+\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 )^{3} d^{6}}+\frac {14 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} e}{3 \left (x +\frac {d}{e}\right )^{2} d^{7}}-\frac {26 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{3 d^{8} x}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{d^{7} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{3 d^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-26/3/d^8*e^4*x*(-e^2*x^2+d^2)^(5/2)-65/6/d^6*e^4*x*(-e^2*x^2+d^2)^(3/2)-65/4/d^4*e^4*x*(-e^2*x^2+d^2)^(1/2)-6
5/4/d^2*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+2/d^7*e/x^2*(-e^2*x^2+d^2)^(7/2)-1/d^5/e/(x
+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+14/3/d^7*e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+65/6/d^6*
e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+65/4/d^4*e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+65/4/d^2*e^4/(e^2
)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)-26/3/d^8*e^2/x*(-e^2*x^2+d^2)^(7/2)+10/(d^2)
^(1/2)/d*e^3*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/3/d^6/x^3*(-e^2*x^2+d^2)^(7/2)-2/d^7*e^3*(-e^2
*x^2+d^2)^(5/2)-10/3/d^5*e^3*(-e^2*x^2+d^2)^(3/2)-10/d^3*e^3*(-e^2*x^2+d^2)^(1/2)+26/3/d^7*e^3*(2*(x+d/e)*d*e-
(x+d/e)^2*e^2)^(5/2)+1/d^6/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)

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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^{4}}\,{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^4/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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