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

Optimal. Leaf size=119 \[ \frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+e^4 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-\frac {3}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-1/12*(-4*e*x+3*d)*(-e^2*x^2+d^2)^(3/2)/x^4-e^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))-3/8*e^4*arctanh((-e^2*x^2+d^2
)^(1/2)/d)+1/8*e^2*(-8*e*x+3*d)*(-e^2*x^2+d^2)^(1/2)/x^2

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {850, 811, 844, 217, 203, 266, 63, 208} \[ \frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+e^4 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-\frac {3}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(3*d - 8*e*x)*Sqrt[d^2 - e^2*x^2])/(8*x^2) - ((3*d - 4*e*x)*(d^2 - e^2*x^2)^(3/2))/(12*x^4) - e^4*ArcTan[
(e*x)/Sqrt[d^2 - e^2*x^2]] - (3*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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 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^5 (d+e x)} \, dx &=\int \frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}-\frac {\int \frac {\left (6 d^3 e^2-8 d^2 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x^3} \, dx}{8 d^2}\\ &=\frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+\frac {\int \frac {12 d^5 e^4-32 d^4 e^5 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{32 d^4}\\ &=\frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+\frac {1}{8} \left (3 d e^4\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-e^5 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+\frac {1}{16} \left (3 d e^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-e^5 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}-e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{8} \left (3 d e^2\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 {e^2 (3 d-8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d-4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}-e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {3}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 111, normalized size = 0.93 \[ \frac {1}{24} \left (-9 e^4 \log \left (\sqrt {d^2-e^2 x^2}+d\right )-24 e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^3+8 d^2 e x+15 d e^2 x^2-32 e^3 x^3\right )}{x^4}+9 e^4 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 8*d^2*e*x + 15*d*e^2*x^2 - 32*e^3*x^3))/x^4 - 24*e^4*ArcTan[(e*x)/Sqrt[d^2 - e
^2*x^2]] + 9*e^4*Log[x] - 9*e^4*Log[d + Sqrt[d^2 - e^2*x^2]])/24

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fricas [A]  time = 0.95, size = 119, normalized size = 1.00 \[ \frac {48 \, e^{4} x^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 9 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (32 \, e^{3} x^{3} - 15 \, d e^{2} x^{2} - 8 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(48*e^4*x^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 9*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (3
2*e^3*x^3 - 15*d*e^2*x^2 - 8*d^2*e*x + 6*d^3)*sqrt(-e^2*x^2 + d^2))/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/192*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^3*(-96*exp(1)^6*exp(2)^2+288*exp(1)^4*exp(2)^3-72*exp(2)^5)+(-1/2*(-2*d*exp(1)-2*sq
rt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(24*exp(1)^4*exp(2)^3-48*exp(2)^5)+3*exp(2)^5+4*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))*exp(2)^5/x/exp(2))/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4/exp(1)^6
+1/65536*(-8192*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^22*exp(2)^7+8192/3*(-1/2*
(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^20*exp(2)^8-1024*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^18*exp(2)^9+24576*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/e
xp(2))^2*exp(1)^20*exp(2)^8-8192*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^18*exp(2
)^9-12288*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^20*exp(2)^8/x/exp(2)+49152*(-2*d*exp(1)-2*sqrt(d^
2-x^2*exp(2))*exp(1))*exp(1)^22*exp(2)^7/x/exp(2)-16384*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^24*
exp(2)^6/x/exp(2))/exp(1)^24/exp(2)^4+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+exp(2)^2)/exp(1)^2+
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)^5/exp(1)-sign(d)*asin(x*exp(2)/d/exp(1))*exp(1)^4

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maple [B]  time = 0.01, size = 463, normalized size = 3.89 \[ -\frac {3 d \,e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}-\frac {3 e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {5 e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{5} x}{8 d^{2}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{8 d^{2}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{5} x}{4 d^{4}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5} x}{12 d^{4}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}{8 d^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5} x}{3 d^{6}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{4}}{5 d^{5}}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}{40 d^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{3 d^{6} x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{8 d^{5} x^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{3 d^{4} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{4 d^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8/(e^2)^(1/2)*e^5*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/8*(-e^2*x^2+d^2)^(3/2)/d^3*e^4+3/8*(-e^2*x^2
+d^2)^(1/2)/d*e^4-1/4/d^4*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-3/8/d^2*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^
(1/2)*x+1/3*e/d^4/x^3*(-e^2*x^2+d^2)^(7/2)-1/3/d^6*e^3/x*(-e^2*x^2+d^2)^(7/2)-1/3/d^6*e^5*x*(-e^2*x^2+d^2)^(5/
2)-1/8/d^5*e^2/x^2*(-e^2*x^2+d^2)^(7/2)-5/12*(-e^2*x^2+d^2)^(3/2)/d^4*e^5*x-5/8*(-e^2*x^2+d^2)^(1/2)/d^2*e^5*x
-3/8/(d^2)^(1/2)*d*e^4*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/5/d^5*e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e
^2)^(5/2)-3/8*e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)-1/4/d^3/x^4*(-e^2*x^2+
d^2)^(7/2)+3/40/d^5*e^4*(-e^2*x^2+d^2)^(5/2)

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maxima [A]  time = 1.01, size = 159, normalized size = 1.34 \[ -e^{4} \arcsin \left (\frac {e x}{d}\right ) - \frac {3}{8} \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{x} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d x^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{3 \, x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.37, size = 541, normalized size = 4.55 \[ d^{3} \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^{2} e \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 ) - d e^{2} \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 ) + e^{3} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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