\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^2 (d+e x)^4} \, dx\) [205]

Optimal result

Integrand size = 27, antiderivative size = 94 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=-\frac {8 e (d-e x)}{\sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{x}-e \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+4 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-e*arctan(e*x/(-e^2*x^2+d^2)^(1/2))+4*e*arctanh((-e^2*x^2+d^2)^(1/2)/d)-8*e*(-e*x+d)/(-e^2*x^2+d^2)^(1/2)-(-e^
2*x^2+d^2)^(1/2)/x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {866, 1819, 1821, 858, 223, 209, 272, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=-e \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+4 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-\frac {8 e (d-e x)}{\sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{x} \]

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)^4),x]

[Out]

(-8*e*(d - e*x))/Sqrt[d^2 - e^2*x^2] - Sqrt[d^2 - e^2*x^2]/x - e*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + 4*e*ArcTa
nh[Sqrt[d^2 - e^2*x^2]/d]

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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]

Rule 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=\frac {(-d-9 e x) \sqrt {d^2-e^2 x^2}}{x (d+e x)}+2 e \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )+\frac {4 \sqrt {d^2} e \log (x)}{d}-\frac {4 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{d} \]

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^2*(d + e*x)^4),x]

[Out]

((-d - 9*e*x)*Sqrt[d^2 - e^2*x^2])/(x*(d + e*x)) + 2*e*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])] + (4*Sq
rt[d^2]*e*Log[x])/d - (4*Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/d

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.41

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{x}-\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {4 d e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}-\frac {8 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{x +\frac {d}{e}}\)	\(133\)
default	\(\text {Expression too large to display}\)	\(1322\)

[In]

int((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-(-e^2*x^2+d^2)^(1/2)/x-e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+4*d*e/(d^2)^(1/2)*ln((2*d^2
+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-8/(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 = 127, normalized size of antiderivative = 1.35 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=-\frac {8 \, e^{2} x^{2} + 8 \, d e x - 2 \, {\left (e^{2} x^{2} + d e x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 4 \, {\left (e^{2} x^{2} + d e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (9 \, e x + d\right )}}{e x^{2} + d x} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

-(8*e^2*x^2 + 8*d*e*x - 2*(e^2*x^2 + d*e*x)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 4*(e^2*x^2 + d*e*x)*lo
g(-(d - sqrt(-e^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(9*e*x + d))/(e*x^2 + d*x)

Sympy [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{x^{2} \left (d + e x\right )^{4}}\, dx \]

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**2/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x**2*(d + e*x)**4), x)

Maxima [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (87) = 174\).

Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.01 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=-\frac {e^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} + \frac {4 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{{\left | e \right |}} + \frac {{\left (e^{2} + \frac {33 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x}\right )} e^{2} x}{2 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, x {\left | e \right |}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^2/(e*x+d)^4,x, algorithm="giac")

[Out]

-e^2*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 4*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x
)))/abs(e) + 1/2*(e^2 + 33*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/x)*e^2*x/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*(
(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(x*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]

[In]

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

[Out]

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