3.2.77 \(\int \frac {1}{x^2 (d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [177]

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

[Out]

-2/5*e*(-e*x+d)/d^2/(-e^2*x^2+d^2)^(5/2)-1/15*e*(-13*e*x+10*d)/d^4/(-e^2*x^2+d^2)^(3/2)+2*e*arctanh((-e^2*x^2+
d^2)^(1/2)/d)/d^6-1/15*e*(-41*e*x+30*d)/d^6/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^6/x

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Rubi [A]
time = 0.19, antiderivative size = 146, normalized size of antiderivative = 1.00, 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 = {866, 1819, 821, 272, 65, 214} \begin {gather*} -\frac {2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\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}-\frac {e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

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 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 821

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 866

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 1819

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]

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 \text {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 \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^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*}

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Mathematica [A]
time = 0.46, size = 123, normalized size = 0.84 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (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\right )}{x (-d+e x) (d+e x)^3}-60 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^6} \end {gather*}

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) - 60*e*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(15*d^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(378\) vs. \(2(130)=260\).
time = 0.09, size = 379, normalized size = 2.60

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{6} x}-\frac {29 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{60 d^{5} e \left (x +\frac {d}{e}\right )^{2}}-\frac {313 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{120 d^{6} \left (x +\frac {d}{e}\right )}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{8 d^{6} \left (x -\frac {d}{e}\right )}+\frac {2 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{5} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{10 d^{4} e^{2} \left (x +\frac {d}{e}\right )^{3}}\)	\(244\)
default	\(\frac {2 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{d^{3}}+\frac {-\frac {1}{d^{2} x \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 e^{2} x}{d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}+\frac {-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}}{d^{2}}-\frac {2 e \left (\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}\right )}{d^{3}}\)	\(379\)

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,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 1.96, size = 185, normalized size = 1.27 \begin {gather*} -\frac {46 \, x^{5} e^{5} + 92 \, d x^{4} e^{4} - 92 \, d^{3} x^{2} e^{2} - 46 \, d^{4} x e + 30 \, {\left (x^{5} e^{5} + 2 \, d x^{4} e^{4} - 2 \, d^{3} x^{2} e^{2} - d^{4} x e\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (56 \, x^{4} e^{4} + 82 \, d x^{3} e^{3} - 32 \, d^{2} x^{2} e^{2} - 76 \, d^{3} x e - 15 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{6} x^{5} e^{4} + 2 \, d^{7} x^{4} e^{3} - 2 \, d^{9} x^{2} e - d^{10} x\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [C] Result contains complex when optimal does not.
time = 3.17, size = 297, normalized size = 2.03 \begin {gather*} \frac {1}{120} \, {\left ({\left (\frac {240 \, e^{\left (-5\right )} \log \left (\sqrt {\frac {2 \, d}{x e + d} - 1} + 1\right )}{d^{6} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {240 \, e^{\left (-5\right )} \log \left ({\left | \sqrt {\frac {2 \, d}{x e + d} - 1} - 1 \right |}\right )}{d^{6} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} + \frac {30 \, {\left (\frac {17 \, d}{x e + d} - 9\right )} e^{\left (-5\right )}}{{\left ({\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {2 \, d}{x e + d} - 1}\right )} d^{6} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {{\left (3 \, d^{24} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{20} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} + 35 \, d^{24} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{20} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} + 345 \, d^{24} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{20} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4}\right )} e^{\left (-25\right )}}{d^{30} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{5}}\right )} e^{7} + \frac {8 \, {\left (15 \, e^{2} \log \left (2\right ) - 30 \, e^{2} \log \left (i + 1\right ) + 56 i \, e^{2}\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{6}}\right )} e^{\left (-1\right )} \end {gather*}

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]

1/120*((240*e^(-5)*log(sqrt(2*d/(x*e + d) - 1) + 1)/(d^6*sgn(1/(x*e + d))) - 240*e^(-5)*log(abs(sqrt(2*d/(x*e
+ d) - 1) - 1))/(d^6*sgn(1/(x*e + d))) + 30*(17*d/(x*e + d) - 9)*e^(-5)/(((2*d/(x*e + d) - 1)^(3/2) - sqrt(2*d
/(x*e + d) - 1))*d^6*sgn(1/(x*e + d))) - (3*d^24*(2*d/(x*e + d) - 1)^(5/2)*e^20*sgn(1/(x*e + d))^4 + 35*d^24*(
2*d/(x*e + d) - 1)^(3/2)*e^20*sgn(1/(x*e + d))^4 + 345*d^24*sqrt(2*d/(x*e + d) - 1)*e^20*sgn(1/(x*e + d))^4)*e
^(-25)/(d^30*sgn(1/(x*e + d))^5))*e^7 + 8*(15*e^2*log(2) - 30*e^2*log(I + 1) + 56*I*e^2)*sgn(1/(x*e + d))/d^6)
*e^(-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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