3.2.26 \(\int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4}-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {857, 835, 807, 266, 63, 208} \begin {gather*} \frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 807

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 857

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

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 127, normalized size = 1.12 \begin {gather*} -\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}-\frac {d^3+d e^2 x^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )-2 d^2 e x-3 d e^2 x^2+4 e^3 x^3}{2 d^4 x^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 97, normalized size = 0.86 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-d^2+d e x+4 e^2 x^2\right )}{2 d^4 x^2 (d+e x)}+\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-d^2 + d*e*x + 4*e^2*x^2))/(2*d^4*x^2*(d + e*x)) + (3*e^2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqr
t[d^2 - e^2*x^2]/d])/d^4

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fricas [A]  time = 0.41, size = 113, normalized size = 1.00 \begin {gather*} \frac {2 \, e^{3} x^{3} + 2 \, d e^{2} x^{2} + 3 \, {\left (e^{3} x^{3} + d e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (4 \, e^{2} x^{2} + d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (d^{4} e x^{3} + d^{5} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/8*(exp(2)^3+2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))*exp(2)^3/x/exp(2))/d^4/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2/exp(1
)^4+1/16*(-2*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^5-4*d^4*(-2*d*e
xp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^4/x/exp(2))/d^8/exp(1)^6/exp(2)^3+1/2*(-exp(2)^3-2*exp(1)
^4*exp(2))*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^4/exp(1)^3/exp(1)+2*exp(1)^3
*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))/d^4/sqrt(-e
xp(1)^4+exp(2)^2)/exp(1)

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maple [A]  time = 0.01, size = 133, normalized size = 1.18 \begin {gather*} -\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}}\, d^{3}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e}{\left (x +\frac {d}{e}\right ) d^{4}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e}{d^{4} x}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)), x)

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