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

Optimal. Leaf size=183 \[ \frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}} \]

[Out]

2/5*e^2*(-e*x+d)/d^3/(-e^2*x^2+d^2)^(5/2)+1/5*e^2*(-6*e*x+5*d)/d^5/(-e^2*x^2+d^2)^(3/2)-9/2*e^2*arctanh((-e^2*
x^2+d^2)^(1/2)/d)/d^7+2/5*e^2*(-11*e*x+10*d)/d^7/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^6/x^2+2*e*(-e
^2*x^2+d^2)^(1/2)/d^7/x

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 183, normalized size of antiderivative = 1.00, 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} \[ \frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*(d - e*x))/(5*d^3*(d^2 - e^2*x^2)^(5/2)) + (e^2*(5*d - 6*e*x))/(5*d^5*(d^2 - e^2*x^2)^(3/2)) + (2*e^2*(
10*d - 11*e*x))/(5*d^7*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^6*x^2) + (2*e*Sqrt[d^2 - e^2*x^2])/(d^7
*x) - (9*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^7)

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 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])

Rubi steps

\begin {align*} \int \frac {1}{x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {(d-e x)^2}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2+10 d e x-10 e^2 x^2+\frac {8 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2-30 d e x+45 e^2 x^2-\frac {36 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2+30 d e x-60 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {\int \frac {-60 d^3 e+135 d^2 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 \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^6}\\ &=\frac {2 e^2 (d-e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d-6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d-11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 127, normalized size = 0.69 \[ \frac {-45 e^2 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (5 d^5-10 d^4 e x-94 d^3 e^2 x^2-58 d^2 e^3 x^3+83 d e^4 x^4+64 e^5 x^5\right )}{x^2 (e x-d) (d+e x)^3}+45 e^2 \log (x)}{10 d^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(5*d^5 - 10*d^4*e*x - 94*d^3*e^2*x^2 - 58*d^2*e^3*x^3 + 83*d*e^4*x^4 + 64*e^5*x^5))/(x^2
*(-d + e*x)*(d + e*x)^3) + 45*e^2*Log[x] - 45*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(10*d^7)

________________________________________________________________________________________

fricas [A]  time = 0.98, size = 215, normalized size = 1.17 \[ \frac {54 \, e^{6} x^{6} + 108 \, d e^{5} x^{5} - 108 \, d^{3} e^{3} x^{3} - 54 \, d^{4} e^{2} x^{2} + 45 \, {\left (e^{6} x^{6} + 2 \, d e^{5} x^{5} - 2 \, d^{3} e^{3} x^{3} - d^{4} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (64 \, e^{5} x^{5} + 83 \, d e^{4} x^{4} - 58 \, d^{2} e^{3} x^{3} - 94 \, d^{3} e^{2} x^{2} - 10 \, d^{4} e x + 5 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (d^{7} e^{4} x^{6} + 2 \, d^{8} e^{3} x^{5} - 2 \, d^{10} e x^{3} - d^{11} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(54*e^6*x^6 + 108*d*e^5*x^5 - 108*d^3*e^3*x^3 - 54*d^4*e^2*x^2 + 45*(e^6*x^6 + 2*d*e^5*x^5 - 2*d^3*e^3*x^
3 - d^4*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (64*e^5*x^5 + 83*d*e^4*x^4 - 58*d^2*e^3*x^3 - 94*d^3*e^2
*x^2 - 10*d^4*e*x + 5*d^5)*sqrt(-e^2*x^2 + d^2))/(d^7*e^4*x^6 + 2*d^8*e^3*x^5 - 2*d^10*e*x^3 - d^11*x^2)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.02, size = 259, normalized size = 1.42 \[ -\frac {9 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{6}}-\frac {4 e^{3} x}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{7}}-\frac {12 e^{3} x}{5 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{7}}+\frac {1}{5 \left (x +\frac {d}{e}\right )^{2} \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4}}+\frac {6 e}{5 \left (x +\frac {d}{e}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{5}}+\frac {9 e^{2}}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6}}+\frac {2 e}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} x}-\frac {1}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d^5*e/x/(-e^2*x^2+d^2)^(1/2)-4/(-e^2*x^2+d^2)^(1/2)/d^7*e^3*x-1/2/d^4/x^2/(-e^2*x^2+d^2)^(1/2)+9/2/(-e^2*x^2
+d^2)^(1/2)/d^6*e^2-9/2/(d^2)^(1/2)/d^6*e^2*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+6/5/d^5*e/(x+d/e)
/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-12/5/d^7*e^3/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+1/5/d^4/(x+d/e)^2/(2*(
x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________