3.1.11 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=118 \[ \frac {e^2 (3 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {(3 d+4 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 x^4}+e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {3}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {811, 844, 217, 203, 266, 63, 208} \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^5,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]

Rubi steps

\begin {align*} \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*}

________________________________________________________________________________________

Mathematica [C]  time = 0.09, size = 133, normalized size = 1.13 \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2} \left (3 d^2 \left (2 d^2-5 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}+9 e^4 x^4 \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )+8 d^3 e x \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )\right )}{24 d x^4 \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^5,x]

[Out]

-1/24*(Sqrt[d^2 - e^2*x^2]*(3*d^2*(2*d^2 - 5*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] + 9*e^4*x^4*ArcTanh[Sqrt[1 - (e^
2*x^2)/d^2]] + 8*d^3*e*x*Hypergeometric2F1[-3/2, -3/2, -1/2, (e^2*x^2)/d^2]))/(d*x^4*Sqrt[1 - (e^2*x^2)/d^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.59, size = 141, normalized size = 1.19 \begin {gather*} \frac {3}{4} e^4 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\sqrt {-e^2} e^3 \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\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 )}{24 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^5,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))/(24*x^4) + (3*e^4*ArcTanh[(Sqrt[-e^2]*x
)/d - Sqrt[d^2 - e^2*x^2]/d])/4 + e^3*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 119, normalized size = 1.01 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^5,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) - (
32*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

________________________________________________________________________________________

giac [B]  time = 0.23, size = 297, normalized size = 2.52 \begin {gather*} \arcsin \left (\frac {x e}{d}\right ) e^{4} \mathrm {sgn}\relax (d) + \frac {x^{4} {\left (\frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{8}}{x} - \frac {24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{6}}{x^{2}} - \frac {120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{4}}{x^{3}} + 3 \, e^{10}\right )} e^{2}}{192 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4}} + \frac {1}{192} \, {\left (\frac {120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{26}}{x} + \frac {24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{24}}{x^{2}} - \frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{22}}{x^{3}} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{20}}{x^{4}}\right )} e^{\left (-24\right )} - \frac {3}{8} \, e^{4} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^5,x, algorithm="giac")

[Out]

arcsin(x*e/d)*e^4*sgn(d) + 1/192*x^4*(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^8/x - 24*(d*e + sqrt(-x^2*e^2 + d^2)*
e)^2*e^6/x^2 - 120*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^4/x^3 + 3*e^10)*e^2/(d*e + sqrt(-x^2*e^2 + d^2)*e)^4 + 1
/192*(120*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^26/x + 24*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^24/x^2 - 8*(d*e + sqrt
(-x^2*e^2 + d^2)*e)^3*e^22/x^3 - 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^20/x^4)*e^(-24) - 3/8*e^4*log(1/2*abs(-2
*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))

________________________________________________________________________________________

maple [B]  time = 0.02, size = 260, normalized size = 2.20 \begin {gather*} -\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 {e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{d^{2}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5} x}{3 d^{4}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}{8 d^{3}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{3 d^{4} x}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{8 d^{3} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{3 d^{2} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{4 d \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^5,x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.98, size = 210, normalized size = 1.78 \begin {gather*} 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 {\sqrt {-e^{2} x^{2} + d^{2}} e^{5} x}{d^{2}} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}{8 \, d^{3}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{3 \, d^{2} x} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}}{8 \, d^{3} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{3 \, d^{2} x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, d x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^5,x)

[Out]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^5, x)

________________________________________________________________________________________

sympy [C]  time = 11.06, size = 541, normalized size = 4.58 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**5,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))

________________________________________________________________________________________