\(\int \frac {1}{x^3 (d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [135]

Optimal result

Integrand size = 27, antiderivative size = 152 \[ \int \frac {1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}-\frac {5 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6} \]

[Out]

-5/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^6+1/3*(-4*e*x+5*d)/d^4/x^2/(-e^2*x^2+d^2)^(1/2)+1/3/d^2/x^2/(e*x+d)
/(-e^2*x^2+d^2)^(1/2)-5/2*(-e^2*x^2+d^2)^(1/2)/d^5/x^2+8/3*e*(-e^2*x^2+d^2)^(1/2)/d^6/x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {871, 837, 849, 821, 272, 65, 214} \[ \int \frac {1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {5 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(5*d - 4*e*x)/(3*d^4*x^2*Sqrt[d^2 - e^2*x^2]) + 1/(3*d^2*x^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^
2*x^2])/(2*d^5*x^2) + (8*e*Sqrt[d^2 - e^2*x^2])/(3*d^6*x) - (5*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*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 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 849

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)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 871

Int[(((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[d*(f + g*x)^
(n + 1)*((a + c*x^2)^(p + 1)/(2*a*p*(e*f - d*g)*(d + e*x))), x] + Dist[1/(p*(2*c*d)*(e*f - d*g)), Int[(f + g*x
)^n*(a + c*x^2)^p*(c*e*f*(2*p + 1) - c*d*g*(n + 2*p + 1) + c*e*g*(n + 2*p + 2)*x), x], x] /; FreeQ[{a, c, d, e
, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[n, 0] && ILtQ[n + 2*p, 0] &
&  !IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-5 d e^2+4 e^3 x}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d^2 e^2} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3 e^4+8 d^2 e^5 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{3 d^6 e^4} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {\int \frac {-16 d^4 e^5+15 d^3 e^6 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{6 d^8 e^4} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}+\frac {\left (5 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^5} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}+\frac {\left (5 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^5} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}-\frac {5 \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 )}{2 d^5} \\ & = \frac {5 d-4 e x}{3 d^4 x^2 \sqrt {d^2-e^2 x^2}}+\frac {1}{3 d^2 x^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {8 e \sqrt {d^2-e^2 x^2}}{3 d^6 x}-\frac {5 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (3 d^4-3 d^3 e x-23 d^2 e^2 x^2+d e^3 x^3+16 e^4 x^4\right )}{x^2 (-d+e x) (d+e x)^2}-15 \sqrt {d^2} e^2 \log (x)+15 \sqrt {d^2} e^2 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{6 d^7} \]

[In]

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

[Out]

((d*Sqrt[d^2 - e^2*x^2]*(3*d^4 - 3*d^3*e*x - 23*d^2*e^2*x^2 + d*e^3*x^3 + 16*e^4*x^4))/(x^2*(-d + e*x)*(d + e*
x)^2) - 15*Sqrt[d^2]*e^2*Log[x] + 15*Sqrt[d^2]*e^2*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(6*d^7)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-2 e x +d \right )}{2 d^{6} x^{2}}-\frac {5 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{5} \sqrt {d^{2}}}+\frac {23 e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{12 d^{6} \left (x +\frac {d}{e}\right )}-\frac {e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{4 d^{6} \left (x -\frac {d}{e}\right )}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{6 d^{5} \left (x +\frac {d}{e}\right )^{2}}\)	\(206\)
default	\(\frac {-\frac {1}{2 d^{2} x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 e^{2} \left (\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}\right )}{2 d^{2}}}{d}+\frac {e^{2} \left (\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}\right )}{d^{3}}-\frac {e \left (-\frac {1}{d^{2} x \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 e^{2} x}{d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{d^{2}}-\frac {e^{2} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{d^{3}}\)	\(325\)

[In]

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

[Out]

-1/2*(-e^2*x^2+d^2)^(1/2)*(-2*e*x+d)/d^6/x^2-5/2/d^5*e^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1
/2))/x)+23/12/d^6*e/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/4/d^6*e/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/
e))^(1/2)+1/6/d^5/(x+d/e)^2*(-(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 = 201, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {14 \, e^{5} x^{5} + 14 \, d e^{4} x^{4} - 14 \, d^{2} e^{3} x^{3} - 14 \, d^{3} e^{2} x^{2} + 15 \, {\left (e^{5} x^{5} + d e^{4} x^{4} - d^{2} e^{3} x^{3} - d^{3} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (16 \, e^{4} x^{4} + d e^{3} x^{3} - 23 \, d^{2} e^{2} x^{2} - 3 \, d^{3} e x + 3 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (d^{6} e^{3} x^{5} + d^{7} e^{2} x^{4} - d^{8} e x^{3} - d^{9} x^{2}\right )}} \]

[In]

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

[Out]

1/6*(14*e^5*x^5 + 14*d*e^4*x^4 - 14*d^2*e^3*x^3 - 14*d^3*e^2*x^2 + 15*(e^5*x^5 + d*e^4*x^4 - d^2*e^3*x^3 - d^3
*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (16*e^4*x^4 + d*e^3*x^3 - 23*d^2*e^2*x^2 - 3*d^3*e*x + 3*d^4)*s
qrt(-e^2*x^2 + d^2))/(d^6*e^3*x^5 + d^7*e^2*x^4 - d^8*e*x^3 - d^9*x^2)

Sympy [F]

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

[In]

integrate(1/x**3/(e*x+d)/(-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)), x)

Maxima [F]

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

[In]

integrate(1/x^3/(e*x+d)/(-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)*x^3), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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