\(\int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [186]

Optimal result

Integrand size = 27, antiderivative size = 146 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]

[Out]

-4/5*e*(-e*x+d)/d/(-e^2*x^2+d^2)^(5/2)-1/5*e*(-7*e*x+5*d)/d^3/(-e^2*x^2+d^2)^(3/2)+3*e*arctanh((-e^2*x^2+d^2)^
(1/2)/d)/d^5-1/5*e*(-19*e*x+15*d)/d^5/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^5/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} \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(-4*e*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) - (e*(5*d - 7*e*x))/(5*d^3*(d^2 - e^2*x^2)^(3/2)) - (e*(15*d - 19
*e*x))/(5*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^5*x) + (3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^5

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)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3+15 d^2 e x-16 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3-45 d^2 e x+42 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3+45 d^2 e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {(3 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {(3 e) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 \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^4 e} \\ & = -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (5 d^3+39 d^2 e x+57 d e^2 x^2+24 e^3 x^3\right )}{x (d+e x)^3}-15 \sqrt {d^2} e \log (x)+15 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{5 d^6} \]

[In]

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

[Out]

-1/5*((d*Sqrt[d^2 - e^2*x^2]*(5*d^3 + 39*d^2*e*x + 57*d*e^2*x^2 + 24*e^3*x^3))/(x*(d + e*x)^3) - 15*Sqrt[d^2]*
e*Log[x] + 15*Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/d^6

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x}+\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{4} \sqrt {d^{2}}}-\frac {4 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{4} e \left (x +\frac {d}{e}\right )^{2}}-\frac {19 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{5} \left (x +\frac {d}{e}\right )}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{3} e^{2} \left (x +\frac {d}{e}\right )^{3}}\)	\(199\)
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x}+\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{4} \sqrt {d^{2}}}+\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}}{e \,d^{2}}+\frac {-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{d^{3}}-\frac {3 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{5} \left (x +\frac {d}{e}\right )}\)	\(349\)

[In]

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

[Out]

-(-e^2*x^2+d^2)^(1/2)/d^5/x+3/d^4*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-4/5/d^4/e/(x+
d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-19/5/d^5/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/5/d^3/e^2/
(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {24 \, e^{4} x^{4} + 72 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 24 \, d^{3} e x + 15 \, {\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (24 \, e^{3} x^{3} + 57 \, d e^{2} x^{2} + 39 \, d^{2} e x + 5 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{5} e^{3} x^{4} + 3 \, d^{6} e^{2} x^{3} + 3 \, d^{7} e x^{2} + d^{8} x\right )}} \]

[In]

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

[Out]

-1/5*(24*e^4*x^4 + 72*d*e^3*x^3 + 72*d^2*e^2*x^2 + 24*d^3*e*x + 15*(e^4*x^4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2 + d^
3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (24*e^3*x^3 + 57*d*e^2*x^2 + 39*d^2*e*x + 5*d^3)*sqrt(-e^2*x^2 + d
^2))/(d^5*e^3*x^4 + 3*d^6*e^2*x^3 + 3*d^7*e*x^2 + d^8*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/x^2/(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^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (131) = 262\).

Time = 0.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.08 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{5} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, d^{5} x {\left | e \right |}} + \frac {{\left (5 \, e^{2} + \frac {121 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x} + \frac {410 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{2} x^{2}} + \frac {610 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{4} x^{3}} + \frac {425 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{6} x^{4}} + \frac {125 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{8} x^{5}}\right )} e^{2} x}{10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{5} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

3*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^5*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^
2 + d^2)*abs(e))/(d^5*x*abs(e)) + 1/10*(5*e^2 + 121*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/x + 410*(d*e + sqrt(-e
^2*x^2 + d^2)*abs(e))^2/(e^2*x^2) + 610*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^4*x^3) + 425*(d*e + sqrt(-e^2
*x^2 + d^2)*abs(e))^4/(e^6*x^4) + 125*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^8*x^5))*e^2*x/((d*e + sqrt(-e^2
*x^2 + d^2)*abs(e))*d^5*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [F(-1)]

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

[In]

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

[Out]

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