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

Optimal. Leaf size=146 \[ -\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6} \]

[Out]

(-2*e*(d - e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) - (e*(10*d - 13*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) - (e*(30*d
- 41*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^6*x) + (2*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/
d^6

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Rubi [A]  time = 0.300216, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {852, 1805, 807, 266, 63, 208} \[ -\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(d - e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) - (e*(10*d - 13*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) - (e*(30*d
- 41*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^6*x) + (2*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/
d^6

Rule 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac{(d-e x)^2}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^2+10 d e x-8 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^2-30 d e x+26 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^2+30 d e x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{(2 e) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^5}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{e \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{d^5}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}+\frac{2 \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 )}{d^5 e}\\ &=-\frac{2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{e (30 d-41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}+\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6}\\ \end{align*}

Mathematica [A]  time = 0.118418, size = 112, normalized size = 0.77 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (32 d^2 e^2 x^2+76 d^3 e x+15 d^4-82 d e^3 x^3-56 e^4 x^4\right )}{x (e x-d) (d+e x)^3}+30 e \log \left (\sqrt{d^2-e^2 x^2}+d\right )-30 e \log (x)}{15 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(15*d^4 + 76*d^3*e*x + 32*d^2*e^2*x^2 - 82*d*e^3*x^3 - 56*e^4*x^4))/(x*(-d + e*x)*(d + e
*x)^3) - 30*e*Log[x] + 30*e*Log[d + Sqrt[d^2 - e^2*x^2]])/(15*d^6)

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Maple [A]  time = 0.067, size = 234, normalized size = 1.6 \begin{align*} -2\,{\frac{e}{{d}^{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}+2\,{\frac{e}{{d}^{5}\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }-{\frac{13}{15\,{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{26\,{e}^{2}x}{15\,{d}^{6}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{1}{5\,{d}^{3}e} \left ({\frac{d}{e}}+x \right ) ^{-2}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{1}{{d}^{4}x}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+2\,{\frac{{e}^{2}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^2/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x)

[Out]

-2/d^5*e/(-e^2*x^2+d^2)^(1/2)+2/d^5*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-13/15/d^4/(
d/e+x)/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+26/15/d^6*e^2/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-1/5/d^3/e/(d/
e+x)^2/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-1/d^4/x/(-e^2*x^2+d^2)^(1/2)+2/d^6*e^2*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 )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.69005, size = 402, normalized size = 2.75 \begin{align*} -\frac{46 \, e^{5} x^{5} + 92 \, d e^{4} x^{4} - 92 \, d^{3} e^{2} x^{2} - 46 \, d^{4} e x + 30 \,{\left (e^{5} x^{5} + 2 \, d e^{4} x^{4} - 2 \, d^{3} e^{2} x^{2} - d^{4} e x\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (56 \, e^{4} x^{4} + 82 \, d e^{3} x^{3} - 32 \, d^{2} e^{2} x^{2} - 76 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{6} e^{4} x^{5} + 2 \, d^{7} e^{3} x^{4} - 2 \, d^{9} e x^{2} - d^{10} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(46*e^5*x^5 + 92*d*e^4*x^4 - 92*d^3*e^2*x^2 - 46*d^4*e*x + 30*(e^5*x^5 + 2*d*e^4*x^4 - 2*d^3*e^2*x^2 - d
^4*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (56*e^4*x^4 + 82*d*e^3*x^3 - 32*d^2*e^2*x^2 - 76*d^3*e*x - 15*d^4
)*sqrt(-e^2*x^2 + d^2))/(d^6*e^4*x^5 + 2*d^7*e^3*x^4 - 2*d^9*e*x^2 - d^10*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError