3.2.13 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^6 (d+e x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {850, 807, 266, 47, 63, 208} \begin {gather*} -\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)),x]

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx &=\int \frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {1}{2} e \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {1}{8} \left (3 e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac {1}{16} \left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )\\ &=-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {1}{8} \left (3 e^3\right ) \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 )\\ &=-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 106, normalized size = 0.98 \begin {gather*} \frac {15 e^5 x^5 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\sqrt {d^2-e^2 x^2} \left (-8 d^4+10 d^3 e x+16 d^2 e^2 x^2-25 d e^3 x^3-8 e^4 x^4\right )-15 e^5 x^5 \log (x)}{40 d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-8*d^4 + 10*d^3*e*x + 16*d^2*e^2*x^2 - 25*d*e^3*x^3 - 8*e^4*x^4) - 15*e^5*x^5*Log[x] + 1
5*e^5*x^5*Log[d + Sqrt[d^2 - e^2*x^2]])/(40*d*x^5)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.59, size = 155, normalized size = 1.44 \begin {gather*} \frac {3 e^5 \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{8 d}-\frac {3 e^5 \log \left (-d \sqrt {d^2-e^2 x^2}+d^2+d \sqrt {-e^2} x\right )}{8 d}+\frac {\sqrt {d^2-e^2 x^2} \left (-8 d^4+10 d^3 e x+16 d^2 e^2 x^2-25 d e^3 x^3-8 e^4 x^4\right )}{40 d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-8*d^4 + 10*d^3*e*x + 16*d^2*e^2*x^2 - 25*d*e^3*x^3 - 8*e^4*x^4))/(40*d*x^5) + (3*e^5*Lo
g[d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*x^2]])/(8*d) - (3*e^5*Log[d^2 + d*Sqrt[-e^2]*x - d*Sqrt[d^2 - e^2*x^2]])/(
8*d)

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 97, normalized size = 0.90 \begin {gather*} -\frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (8 \, e^{4} x^{4} + 25 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} - 10 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{40 \, d x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d),x, algorithm="fricas")

[Out]

-1/40*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (8*e^4*x^4 + 25*d*e^3*x^3 - 16*d^2*e^2*x^2 - 10*d^3*e*x
+ 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*x^5)

________________________________________________________________________________________

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^6/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/960*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*(480*exp(1)^8*exp(2)^2-1440*exp(1)^6*exp(2)^3+1800*exp(1)^4*exp(2)^4-780*exp(2)^6
)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*(-120*exp(1)^6*exp(2)^3+360*exp(1)^4*exp(2)^4-
120*exp(2)^6)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(40*exp(1)^4*exp(2)^4-70*exp(2)^6)
+6*exp(2)^6+15/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^6/x/exp(2))/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))/x/exp(2))^5/exp(1)^6/d/exp(1)+1/33554432*(4194304*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)^34*exp(2)^8-4194304/3*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x
/exp(2))^3*exp(1)^32*exp(2)^9+524288*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^
30*exp(2)^10-1048576/5*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^28*exp(2)^11-1
2582912*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^32*exp(2)^9+4194304*d^4*(-1/2
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^30*exp(2)^10+4194304*d^4*(-1/2*(-2*d*exp(1)-2*
sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^30*exp(2)^10-5242880/3*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^3*exp(1)^28*exp(2)^11+5242880*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^
28*exp(2)^11/x/exp(2)-18874368*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^30*exp(2)^10/x/exp(2)+31
457280*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^32*exp(2)^9/x/exp(2)-25165824*d^4*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^34*exp(2)^8/x/exp(2)+8388608*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(
1))*exp(1)^36*exp(2)^7/x/exp(2))/d^5/exp(1)^35/exp(2)^5+1/2*(12*exp(1)^4*exp(2)^2-8*exp(2)^4-4*exp(1)^6*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+ex
p(2)^2)/d/exp(1)+1/8*(-24*exp(1)^6*exp(2)^2+28*exp(1)^4*exp(2)^3-9*exp(2)^5+8*exp(1)^8*exp(2))*ln(1/2*abs(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/exp(1)^4/d/exp(1)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 493, normalized size = 4.56 \begin {gather*} \frac {3 e^{6} \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 )}{8 \sqrt {e^{2}}\, d}-\frac {3 e^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, d}+\frac {3 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}+\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{6} x}{8 d^{3}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{6} x}{8 d^{3}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}}{8 d^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{6} x}{4 d^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{6} x}{4 d^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5}}{8 d^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{6} x}{5 d^{7}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{5}}{5 d^{6}}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}{40 d^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{4}}{5 d^{7} x}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{8 d^{6} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{5 d^{5} x^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{4 d^{4} x^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{5 d^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d),x)

[Out]

-1/8*(-e^2*x^2+d^2)^(3/2)/d^4*e^5-3/8*(-e^2*x^2+d^2)^(1/2)/d^2*e^5+3/8/(d^2)^(1/2)*e^5*ln((2*d^2+2*(d^2)^(1/2)
*(-e^2*x^2+d^2)^(1/2))/x)+1/4/d^5*e^6*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+3/8/d^3*e^6*(2*(x+d/e)*d*e-(x+d/e)
^2*e^2)^(1/2)*x+3/8/d*e^6/(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^5*e^2/x^
3*(-e^2*x^2+d^2)^(7/2)-1/5/d^7*e^4/x*(-e^2*x^2+d^2)^(7/2)-1/5/d^7*e^6*x*(-e^2*x^2+d^2)^(5/2)-1/4/d^5*e^6*x*(-e
^2*x^2+d^2)^(3/2)-3/8/d^3*e^6*x*(-e^2*x^2+d^2)^(1/2)-3/8/d*e^6/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(
1/2)*x)+1/4*e/d^4/x^4*(-e^2*x^2+d^2)^(7/2)+1/8*e^3/d^6/x^2*(-e^2*x^2+d^2)^(7/2)+1/5/d^6*e^5*(2*(x+d/e)*d*e-(x+
d/e)^2*e^2)^(5/2)-1/5/d^3/x^5*(-e^2*x^2+d^2)^(7/2)-3/40*e^5/d^6*(-e^2*x^2+d^2)^(5/2)

________________________________________________________________________________________

maxima [A]  time = 1.00, size = 153, normalized size = 1.42 \begin {gather*} \frac {3 \, e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}}{8 \, d^{2}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{8 \, d^{2} x^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{5 \, d x^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{4 \, x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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^6\,\left (d+e\,x\right )} \,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^6*(d + e*x)),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)), x)

________________________________________________________________________________________

sympy [C]  time = 13.94, size = 774, normalized size = 7.17 \begin {gather*} d^{3} \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 ) - d^{2} e \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 ) - d e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**6/(e*x+d),x)

[Out]

d**3*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)) - d**2*e*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)) - d*e**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**
2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(
e**2*x**2) + 1)/(3*d**2), True)) + e**3*Piecewise((-d**2/(2*e*x**3*sqrt(d**2/(e**2*x**2) - 1)) + e/(2*x*sqrt(d
**2/(e**2*x**2) - 1)) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) +
1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), True))

________________________________________________________________________________________