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

Optimal. Leaf size=183 \[ \frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}-\frac{13 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

[Out]

(4*e^2*(d - e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e^2*(25*d - 31*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(90*d - 107*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^5*x^2) + (3*e*Sqrt[d^2 - e^2*x^2])/(
d^6*x) - (13*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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Rubi [A]  time = 0.380112, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {852, 1805, 1807, 807, 266, 63, 208} \[ \frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}-\frac{13 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^2*(d - e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e^2*(25*d - 31*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(90*d - 107*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^5*x^2) + (3*e*Sqrt[d^2 - e^2*x^2])/(
d^6*x) - (13*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*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 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])

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)^3 \sqrt{d^2-e^2 x^2}} \, dx &=\int \frac{(d-e x)^3}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^3+15 d^2 e x-20 d e^2 x^2+16 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^3-45 d^2 e x+75 d e^2 x^2-62 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^3+45 d^2 e x-90 d e^2 x^2}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{\int \frac{-90 d^4 e+195 d^3 e^2 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}+\frac{\left (13 e^2\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{2 d^5}\\ &=\frac{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}+\frac{\left (13 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{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{13 \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{4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (90 d-107 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^5 x^2}+\frac{3 e \sqrt{d^2-e^2 x^2}}{d^6 x}-\frac{13 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^6}\\ \end{align*}

Mathematica [A]  time = 0.191481, size = 107, normalized size = 0.58 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (479 d^2 e^2 x^2+45 d^3 e x-15 d^4+717 d e^3 x^3+304 e^4 x^4\right )}{x^2 (d+e x)^3}-195 e^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+195 e^2 \log (x)}{30 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-15*d^4 + 45*d^3*e*x + 479*d^2*e^2*x^2 + 717*d*e^3*x^3 + 304*e^4*x^4))/(x^2*(d + e*x)^3
) + 195*e^2*Log[x] - 195*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(30*d^6)

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Maple [A]  time = 0.066, size = 222, normalized size = 1.2 \begin{align*} -{\frac{13\,{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{107\,e}{15\,{d}^{6}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}}+{\frac{17}{15\,{d}^{5}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{1}{2\,{d}^{5}{x}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{1}{5\,{d}^{4}e}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-3}}+3\,{\frac{e\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{{d}^{6}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/2/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)+107/15*e/d^6/(d/e+x)*(-(d/e+x)^2*e^
2+2*d*e*(d/e+x))^(1/2)+17/15/d^5/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^5/x
^2+1/5/d^4/e/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+3*e*(-e^2*x^2+d^2)^(1/2)/d^6/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82406, size = 424, normalized size = 2.32 \begin{align*} \frac{254 \, e^{5} x^{5} + 762 \, d e^{4} x^{4} + 762 \, d^{2} e^{3} x^{3} + 254 \, d^{3} e^{2} x^{2} + 195 \,{\left (e^{5} x^{5} + 3 \, d e^{4} x^{4} + 3 \, 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 (304 \, e^{4} x^{4} + 717 \, d e^{3} x^{3} + 479 \, d^{2} e^{2} x^{2} + 45 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{30 \,{\left (d^{6} e^{3} x^{5} + 3 \, d^{7} e^{2} x^{4} + 3 \, 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)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/30*(254*e^5*x^5 + 762*d*e^4*x^4 + 762*d^2*e^3*x^3 + 254*d^3*e^2*x^2 + 195*(e^5*x^5 + 3*d*e^4*x^4 + 3*d^2*e^3
*x^3 + d^3*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (304*e^4*x^4 + 717*d*e^3*x^3 + 479*d^2*e^2*x^2 + 45*d
^3*e*x - 15*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^3*x^5 + 3*d^7*e^2*x^4 + 3*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} \sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16422, size = 1, normalized size = 0.01 \begin{align*} +\infty \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

+Infinity