\(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^4} \, dx\) [10]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=\frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d e^3 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-1/6*(3*e*x+2*d)*(-e^2*x^2+d^2)^(3/2)/x^3+d*e^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))+3/2*d*e^3*arctanh((-e^2*x^2+d
^2)^(1/2)/d)+1/2*e^2*(-3*e*x+2*d)*(-e^2*x^2+d^2)^(1/2)/x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {825, 827, 858, 223, 209, 272, 65, 214} \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=d e^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} d e^3 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {e^2 (2 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3} \]

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

(e^2*(2*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*x) - ((2*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/(6*x^3) + d*e^3*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]] + (3*d*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/2

Rule 65

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 209

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

Rule 214

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

Rule 223

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 272

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 825

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/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)))*((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), 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 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 858

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^3-3 d^2 e x+8 d e^2 x^2-6 e^3 x^3\right )}{6 x^3}-2 d e^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )+\frac {3}{2} \sqrt {d^2} e^3 \log (x)-\frac {3}{2} \sqrt {d^2} e^3 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right ) \]

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-2*d^3 - 3*d^2*e*x + 8*d*e^2*x^2 - 6*e^3*x^3))/(6*x^3) - 2*d*e^3*ArcTan[(e*x)/(Sqrt[d^2]
- Sqrt[d^2 - e^2*x^2])] + (3*Sqrt[d^2]*e^3*Log[x])/2 - (3*Sqrt[d^2]*e^3*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])
/2

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.14

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d \left (-8 e^{2} x^{2}+3 d e x +2 d^{2}\right )}{6 x^{3}}+\frac {e^{4} d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-e^{3} \sqrt {-e^{2} x^{2}+d^{2}}+\frac {3 e^{3} d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}\)	\(137\)
default	\(e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{2 d^{2} x^{2}}-\frac {3 e^{2} \left (\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3}+d^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )\right )}{2 d^{2}}\right )+d \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{3 d^{2} x^{3}}-\frac {2 e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{d^{2} x}-\frac {4 e^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{d^{2}}\right )}{3 d^{2}}\right )\)	\(250\)

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=-\frac {12 \, d e^{3} x^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 9 \, d e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + 6 \, d e^{3} x^{3} + {\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 3 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, x^{3}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

-1/6*(12*d*e^3*x^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 9*d*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d^2))/x)
+ 6*d*e^3*x^3 + (6*e^3*x^3 - 8*d*e^2*x^2 + 3*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.46 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.81 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=d^{3} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + d^{2} e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 x} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - e^{3} \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=\frac {d e^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} + \frac {3}{2} \, d e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{4} x}{d} - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{2 \, d^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{3 \, d x} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{2 \, d^{2} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{3 \, d x^{3}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (106) = 212\).

Time = 0.28 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.38 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=\frac {d e^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} + \frac {{\left (d e^{4} + \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d e^{2}}{x} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{x^{2}}\right )} e^{6} x^{3}}{24 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} {\left | e \right |}} + \frac {3 \, d e^{4} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, {\left | e \right |}} - \sqrt {-e^{2} x^{2} + d^{2}} e^{3} + \frac {\frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d e^{4}}{x} - \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d e^{2}}{x^{2}} - \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d}{x^{3}}}{24 \, e^{2} {\left | e \right |}} \]

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

d*e^4*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/24*(d*e^4 + 3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d*e^2/x - 15*(d
*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d/x^2)*e^6*x^3/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*abs(e)) + 3/2*d*e^4*
log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/abs(e) - sqrt(-e^2*x^2 + d^2)*e^3 + 1/24*(15
*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d*e^4/x - 3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d*e^2/x^2 - (d*e + sqrt
(-e^2*x^2 + d^2)*abs(e))^3*d/x^3)/(e^2*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )}{x^4} \,d x \]

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^4,x)

[Out]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^4, x)