\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^5 (d+e x)^2} \, dx\) [167]

Optimal result

Integrand size = 27, antiderivative size = 108 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=-\frac {5 e^2 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}+\frac {5 e^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 108, 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 = {866, 1821, 821, 272, 43, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {5 e^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d}-\frac {5 e^2 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3} \]

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^3+16 d^2 e x-9 d e^2 x^2-16 e^3 x^3\right )}{24 d x^4}+\frac {5 e^4 \log (x)}{8 \sqrt {d^2}}-\frac {5 e^4 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{8 \sqrt {d^2}} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 16*d^2*e*x - 9*d*e^2*x^2 - 16*e^3*x^3))/(24*d*x^4) + (5*e^4*Log[x])/(8*Sqrt[d^2
]) - (5*e^4*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(8*Sqrt[d^2])

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (16 e^{3} x^{3}+9 d \,e^{2} x^{2}-16 d^{2} e x +6 d^{3}\right )}{24 x^{4} d}+\frac {5 e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}\)	\(96\)
default	\(\text {Expression too large to display}\)	\(1153\)

[In]

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

[Out]

-1/24*(-e^2*x^2+d^2)^(1/2)*(16*e^3*x^3+9*d*e^2*x^2-16*d^2*e*x+6*d^3)/x^4/d+5/8*e^4/(d^2)^(1/2)*ln((2*d^2+2*(d^
2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=-\frac {15 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (16 \, e^{3} x^{3} + 9 \, d e^{2} x^{2} - 16 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, d x^{4}} \]

[In]

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

[Out]

-1/24*(15*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (16*e^3*x^3 + 9*d*e^2*x^2 - 16*d^2*e*x + 6*d^3)*sqrt(-e
^2*x^2 + d^2))/(d*x^4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.18 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.91 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=d^{2} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 x} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**2*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(
8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*
x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(
e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) - 2*d*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x
**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)
/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), True)) + e**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1
)/(2*x) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(2*e*x**3*sqrt(-d**2/(e**2*x**2) + 1)
) - I*e/(2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**2*asin(d/(e*x))/(2*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.20 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {5 \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d^{2}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d^{2} x^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{3 \, d x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{4 \, x^{4}} \]

[In]

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

[Out]

5/8*e^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 5/8*sqrt(-e^2*x^2 + d^2)*e^4/d^2 - 5/8*(-e^2*x
^2 + d^2)^(3/2)*e^2/(d^2*x^2) + 2/3*(-e^2*x^2 + d^2)^(3/2)*e/(d*x^3) - 1/4*(-e^2*x^2 + d^2)^(3/2)/x^4

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.28 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {1}{192} \, {\left (\frac {120 \, e^{3} \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} - \frac {120 \, e^{3} \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} + \frac {4 \, {\left (15 \, e^{3} \log \left (2\right ) - 30 \, e^{3} \log \left (i + 1\right ) + 32 i \, e^{3}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} - \frac {15 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 73 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 55 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 15 \, e^{3} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d {\left (\frac {d}{e x + d} - 1\right )}^{4}}\right )} {\left | e \right |} \]

[In]

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

[Out]

1/192*(120*e^3*log(sqrt(2*d/(e*x + d) - 1) + 1)*sgn(1/(e*x + d))*sgn(e)/d - 120*e^3*log(abs(sqrt(2*d/(e*x + d)
- 1) - 1))*sgn(1/(e*x + d))*sgn(e)/d + 4*(15*e^3*log(2) - 30*e^3*log(I + 1) + 32*I*e^3)*sgn(1/(e*x + d))*sgn(
e)/d - (15*e^3*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) + 73*e^3*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x
+ d))*sgn(e) - 55*e^3*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) + 15*e^3*sqrt(2*d/(e*x + d) - 1)*sgn(
1/(e*x + d))*sgn(e))/(d*(d/(e*x + d) - 1)^4))*abs(e)

Mupad [F(-1)]

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

[In]

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

[Out]

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