3.2.33 \(\int \frac {1}{x (d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [133]

Optimal. Leaf size=88 \[ \frac {3 d-2 e x}{3 d^4 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 88, 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 = {871, 837, 12, 272, 65, 214} \begin {gather*} \frac {1}{3 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}+\frac {3 d-2 e x}{3 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 871

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

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Mathematica [A]
time = 0.35, size = 95, normalized size = 1.08 \begin {gather*} \frac {\frac {\left (4 d^2+d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{(d-e x) (d+e x)^2}+6 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{3 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*d^2 + d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/((d - e*x)*(d + e*x)^2) + 6*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^
2 - e^2*x^2])/d])/(3*d^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(170\) vs. \(2(78)=156\).
time = 0.06, size = 171, normalized size = 1.94

method	result	size
default	\(-\frac {-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{d}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.03, size = 150, normalized size = 1.70 \begin {gather*} \frac {4 \, x^{3} e^{3} + 4 \, d x^{2} e^{2} - 4 \, d^{2} x e - 4 \, d^{3} + 3 \, {\left (x^{3} e^{3} + d x^{2} e^{2} - d^{2} x e - d^{3}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (2 \, x^{2} e^{2} - d x e - 4 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (d^{4} x^{3} e^{3} + d^{5} x^{2} e^{2} - d^{6} x e - d^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(4*x^3*e^3 + 4*d*x^2*e^2 - 4*d^2*x*e - 4*d^3 + 3*(x^3*e^3 + d*x^2*e^2 - d^2*x*e - d^3)*log(-(d - sqrt(-x^2
*e^2 + d^2))/x) + (2*x^2*e^2 - d*x*e - 4*d^2)*sqrt(-x^2*e^2 + d^2))/(d^4*x^3*e^3 + d^5*x^2*e^2 - d^6*x*e - d^7
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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