\(\int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [185]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

4/5*(-e*x+d)/(-e^2*x^2+d^2)^(5/2)+1/15*(-11*e*x+5*d)/d^2/(-e^2*x^2+d^2)^(3/2)-arctanh((-e^2*x^2+d^2)^(1/2)/d)/
d^4+1/15*(-22*e*x+15*d)/d^4/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {866, 1819, 837, 12, 272, 65, 214} \[ \int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(4*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d - 11*e*x)/(15*d^2*(d^2 - e^2*x^2)^(3/2)) + (15*d - 22*e*x)/(15*
d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 \left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3+11 d^2 e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^5 e^2+22 d^4 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^7 e^4}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^4} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^3} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^3} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\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^3 e^2} \\ & = \frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (32 d^2+51 d e x+22 e^2 x^2\right )}{(d+e x)^3}-15 \sqrt {d^2} \log (x)+15 \sqrt {d^2} \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^5} \]

[In]

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

[Out]

((d*Sqrt[d^2 - e^2*x^2]*(32*d^2 + 51*d*e*x + 22*e^2*x^2))/(d + e*x)^3 - 15*Sqrt[d^2]*Log[x] + 15*Sqrt[d^2]*Log
[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(15*d^5)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(101)=202\).

Time = 0.41 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.89

method	result	size
default	\(-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}-\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{e \,d^{2}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e \,d^{4} \left (x +\frac {d}{e}\right )}-\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}}{e^{2} d}\)	\(332\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {32 \, e^{3} x^{3} + 96 \, d e^{2} x^{2} + 96 \, d^{2} e x + 32 \, d^{3} + 15 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (22 \, e^{2} x^{2} + 51 \, d e x + 32 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{3} x^{3} + 3 \, d^{5} e^{2} x^{2} + 3 \, d^{6} e x + d^{7}\right )}} \]

[In]

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

[Out]

1/15*(32*e^3*x^3 + 96*d*e^2*x^2 + 96*d^2*e*x + 32*d^3 + 15*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*log(-(d -
sqrt(-e^2*x^2 + d^2))/x) + (22*e^2*x^2 + 51*d*e*x + 32*d^2)*sqrt(-e^2*x^2 + d^2))/(d^4*e^3*x^3 + 3*d^5*e^2*x^
2 + 3*d^6*e*x + d^7)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/x/(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), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (102) = 204\).

Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.84 \[ \int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{4} {\left | e \right |}} - \frac {2 \, {\left (32 \, e + \frac {115 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e x} + \frac {185 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{3} x^{2}} + \frac {135 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{5} x^{3}} + \frac {45 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{7} x^{4}}\right )}}{15 \, d^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

-e*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^4*abs(e)) - 2/15*(32*e + 115*(d*e + sq
rt(-e^2*x^2 + d^2)*abs(e))/(e*x) + 185*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^3*x^2) + 135*(d*e + sqrt(-e^2*
x^2 + d^2)*abs(e))^3/(e^5*x^3) + 45*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^7*x^4))/(d^4*((d*e + sqrt(-e^2*x^
2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [F(-1)]

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

[In]

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

[Out]

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