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

Optimal result

Integrand size = 27, antiderivative size = 146 \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (30 d-41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {2 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \]

[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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {866, 1819, 821, 272, 65, 214} \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}-\frac {2 e (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (30 d-41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {e (10 d-13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (15 d^4+76 d^3 e x+32 d^2 e^2 x^2-82 d e^3 x^3-56 e^4 x^4\right )}{x (-d+e x) (d+e x)^3}+30 \sqrt {d^2} e \log (x)-30 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^7} \]

[In]

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

[Out]

((d*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*Sqrt[d^2]*e*Log[x] - 30*Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(15*d^7)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.67

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

[In]

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

[Out]

-(-e^2*x^2+d^2)^(1/2)/d^6/x+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)-29/60/d^5/e/(
x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-313/120/d^6/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/10/d^
4/e^2/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/8/d^6/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {46 \, e^{5} x^{5} + 92 \, d e^{4} x^{4} - 92 \, d^{3} e^{2} x^{2} - 46 \, d^{4} e x + 30 \, {\left (e^{5} x^{5} + 2 \, d e^{4} x^{4} - 2 \, d^{3} e^{2} x^{2} - d^{4} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (56 \, e^{4} x^{4} + 82 \, d e^{3} x^{3} - 32 \, d^{2} e^{2} x^{2} - 76 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{6} e^{4} x^{5} + 2 \, d^{7} e^{3} x^{4} - 2 \, d^{9} e x^{2} - d^{10} x\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.22 \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {e^{7} {\left (\frac {240 \, \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right )}{d^{6} e^{5} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )} - \frac {240 \, \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right )}{d^{6} e^{5} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )} + \frac {30 \, {\left (\frac {17 \, d}{e x + d} - 9\right )}}{{\left ({\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {2 \, d}{e x + d} - 1}\right )} d^{6} e^{5} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )} - \frac {3 \, d^{24} e^{20} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} + 35 \, d^{24} e^{20} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} + 345 \, d^{24} e^{20} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4}}{d^{30} e^{25} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{5} \mathrm {sgn}\left (e\right )^{5}}\right )} + \frac {8 \, {\left (15 \, e^{2} \log \left (2\right ) - 30 \, e^{2} \log \left (i + 1\right ) + 56 i \, e^{2}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{6}}}{120 \, {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{x^2\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,d x \]

[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)