\(\int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx\) [40]

Optimal result

Integrand size = 27, antiderivative size = 80 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1821, 821, 272, 65, 214} \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2} \]

[In]

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

[Out]

-1/2*Sqrt[d^2 - e^2*x^2]/x^2 - (2*e*Sqrt[d^2 - e^2*x^2])/(d*x) - (3*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d)

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 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 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 821

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 1821

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {(d+4 e x) \sqrt {d^2-e^2 x^2}-3 e^2 x^2 \log \left (d \left (-d-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )\right )+3 e^2 x^2 \log \left (d-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 d x^2} \]

[In]

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

[Out]

-1/2*((d + 4*e*x)*Sqrt[d^2 - e^2*x^2] - 3*e^2*x^2*Log[d*(-d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*x^2])] + 3*e^2*x^2
*Log[d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*x^2]])/(d*x^2)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (4 e x +d \right )}{2 d \,x^{2}}-\frac {3 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}\)	\(72\)
default	\(-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+d^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{2} \sqrt {d^{2}}}\right )-\frac {2 e \sqrt {-e^{2} x^{2}+d^{2}}}{d x}\)	\(139\)

[In]

int((e*x+d)^2/x^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(-e^2*x^2+d^2)^(1/2)*(4*e*x+d)/d/x^2-3/2*e^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 = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.79 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 \, e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - \sqrt {-e^{2} x^{2} + d^{2}} {\left (4 \, e x + d\right )}}{2 \, d x^{2}} \]

[In]

integrate((e*x+d)^2/x^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.68 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 d^{2} x} - \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{d^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {d}{e x} \right )}}{d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i \operatorname {asin}{\left (\frac {d}{e x} \right )}}{d} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((e*x+d)**2/x**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} e}{d x} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{2 \, x^{2}} \]

[In]

integrate((e*x+d)^2/x^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (70) = 140\).

Time = 0.28 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.35 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {{\left (e^{3} + \frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e}{x}\right )} e^{4} x^{2}}{8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left | e \right |}} - \frac {3 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, d {\left | e \right |}} - \frac {\frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d e {\left | e \right |}}{x} + \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left | e \right |}}{e x^{2}}}{8 \, d^{2} e^{2}} \]

[In]

integrate((e*x+d)^2/x^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((d + e*x)^2/(x^3*(d^2 - e^2*x^2)^(1/2)),x)

[Out]

int((d + e*x)^2/(x^3*(d^2 - e^2*x^2)^(1/2)), x)