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

Optimal. Leaf size=152 \[ \frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

[Out]

(5*d - 4*e*x)/(3*d^4*x^2*Sqrt[d^2 - e^2*x^2]) + 1/(3*d^2*x^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^
2*x^2])/(2*d^5*x^2) + (8*e*Sqrt[d^2 - e^2*x^2])/(3*d^6*x) - (5*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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Rubi [A]  time = 0.129039, antiderivative size = 152, normalized size of antiderivative = 1., 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 = {857, 823, 835, 807, 266, 63, 208} \[ \frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*d - 4*e*x)/(3*d^4*x^2*Sqrt[d^2 - e^2*x^2]) + 1/(3*d^2*x^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^
2*x^2])/(2*d^5*x^2) + (8*e*Sqrt[d^2 - e^2*x^2])/(3*d^6*x) - (5*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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]

Rule 823

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] + Di
st[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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-5 d e^2+4 e^3 x}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d^2 e^2}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^3 e^4+8 d^2 e^5 x}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{3 d^6 e^4}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{\int \frac{-16 d^4 e^5+15 d^3 e^6 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{6 d^8 e^4}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}+\frac{\left (5 e^2\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{2 d^5}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}+\frac{\left (5 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^5}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}-\frac{5 \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^5}\\ &=\frac{5 d-4 e x}{3 d^4 x^2 \sqrt{d^2-e^2 x^2}}+\frac{1}{3 d^2 x^2 (d+e x) \sqrt{d^2-e^2 x^2}}-\frac{5 \sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{8 e \sqrt{d^2-e^2 x^2}}{3 d^6 x}-\frac{5 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6}\\ \end{align*}

Mathematica [A]  time = 0.115383, size = 115, normalized size = 0.76 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (-23 d^2 e^2 x^2-3 d^3 e x+3 d^4+d e^3 x^3+16 e^4 x^4\right )}{x^2 (e x-d) (d+e x)^2}-15 e^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+15 e^2 \log (x)}{6 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(3*d^4 - 3*d^3*e*x - 23*d^2*e^2*x^2 + d*e^3*x^3 + 16*e^4*x^4))/(x^2*(-d + e*x)*(d + e*x)
^2) + 15*e^2*Log[x] - 15*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(6*d^6)

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Maple [A]  time = 0.063, size = 216, normalized size = 1.4 \begin{align*}{\frac{5\,{e}^{2}}{2\,{d}^{5}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{5\,{e}^{2}}{2\,{d}^{5}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}+{\frac{e}{3\,{d}^{4}} \left ({\frac{d}{e}}+x \right ) ^{-1}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{2\,{e}^{3}x}{3\,{d}^{6}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{1}{2\,{d}^{3}{x}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{e}{{d}^{4}x}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-2\,{\frac{{e}^{3}x}{{d}^{6}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70155, size = 394, normalized size = 2.59 \begin{align*} \frac{14 \, e^{5} x^{5} + 14 \, d e^{4} x^{4} - 14 \, d^{2} e^{3} x^{3} - 14 \, d^{3} e^{2} x^{2} + 15 \,{\left (e^{5} x^{5} + d e^{4} x^{4} - d^{2} e^{3} x^{3} - d^{3} e^{2} x^{2}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (16 \, e^{4} x^{4} + d e^{3} x^{3} - 23 \, d^{2} e^{2} x^{2} - 3 \, d^{3} e x + 3 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \,{\left (d^{6} e^{3} x^{5} + d^{7} e^{2} x^{4} - d^{8} e x^{3} - d^{9} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(14*e^5*x^5 + 14*d*e^4*x^4 - 14*d^2*e^3*x^3 - 14*d^3*e^2*x^2 + 15*(e^5*x^5 + d*e^4*x^4 - d^2*e^3*x^3 - d^3
*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (16*e^4*x^4 + d*e^3*x^3 - 23*d^2*e^2*x^2 - 3*d^3*e*x + 3*d^4)*s
qrt(-e^2*x^2 + d^2))/(d^6*e^3*x^5 + d^7*e^2*x^4 - d^8*e*x^3 - d^9*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, 1]