3.39 \(\int \frac {(d+e x)^2}{x^2 \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {\sqrt {d^2-e^2 x^2}}{x}+e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1807, 844, 217, 203, 266, 63, 208} \[ -\frac {\sqrt {d^2-e^2 x^2}}{x}+e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 1807

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 68, normalized size = 1.00 \[ -\frac {\sqrt {d^2-e^2 x^2}}{x}+e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 79, normalized size = 1.16 \[ -\frac {2 \, e x \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 2 \, e x \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 2*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 +
d^2))/x

________________________________________________________________________________________

giac [A]  time = 0.26, size = 107, normalized size = 1.57 \[ \arcsin \left (\frac {x e}{d}\right ) e \mathrm {sgn}\relax (d) - 2 \, e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 93, normalized size = 1.37 \[ -\frac {2 d e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-(-e^2*x^2+d^2)^(1/2)/x-2*d*e/(d^2)^(1/2)*ln((2*d^2+
2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 64, normalized size = 0.94 \[ e \arcsin \left (\frac {e x}{d}\right ) - 2 \, e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \left \{\begin {array}{cl} \frac {e^2\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{\sqrt {-e^2}}-\frac {\sqrt {d^2-e^2\,x^2}}{x}-\frac {2\,d\,e\,\ln \left (\frac {\sqrt {d^2}+\sqrt {d^2-e^2\,x^2}}{x}\right )}{\sqrt {d^2}} & \text {\ if\ \ }e^2<0\\ \int \frac {e^2}{\sqrt {d^2-e^2\,x^2}}+\frac {d^2}{x^2\,\sqrt {d^2-e^2\,x^2}}+\frac {2\,d\,e}{x\,\sqrt {d^2-e^2\,x^2}} \,d x & \text {\ if\ \ }\neg e^2<0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(e^2 < 0, - (d^2 - e^2*x^2)^(1/2)/x + (e^2*log(x*(-e^2)^(1/2) + (d^2 - e^2*x^2)^(1/2)))/(-e^2)^(1/2)
- (2*d*e*log(((d^2)^(1/2) + (d^2 - e^2*x^2)^(1/2))/x))/(d^2)^(1/2), ~e^2 < 0, int(e^2/(d^2 - e^2*x^2)^(1/2) +
d^2/(x^2*(d^2 - e^2*x^2)^(1/2)) + (2*d*e)/(x*(d^2 - e^2*x^2)^(1/2)), x))

________________________________________________________________________________________

sympy [C]  time = 4.30, size = 207, normalized size = 3.04 \[ d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{d^{2}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {d}{e x} \right )}}{d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i \operatorname {asin}{\left (\frac {d}{e x} \right )}}{d} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {asin}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac {\sqrt {- \frac {d^{2}}{e^{2}}} \operatorname {asinh}{\left (x \sqrt {- \frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {acosh}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: d^{2} < 0 \wedge e^{2} < 0 \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*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)) + 2*d*e*Piecewise((-acosh(d/(e*x))/d, Abs(d**2/(e**2*x**2)) > 1), (I*asin(d/(e*x))/d, True))
+ e**2*Piecewise((sqrt(d**2/e**2)*asin(x*sqrt(e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 > 0)), (sqrt(-d**2/e*
*2)*asinh(x*sqrt(-e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 < 0)), (sqrt(d**2/e**2)*acosh(x*sqrt(e**2/d**2))/
sqrt(-d**2), (d**2 < 0) & (e**2 < 0)))

________________________________________________________________________________________