\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^3 (d+e x)^4} \, dx\) [206]

Optimal result

Integrand size = 27, antiderivative size = 110 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {15 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d} \]

[Out]

-15/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d+8*e^2*(-e*x+d)/d/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/x^2
+4*e*(-e^2*x^2+d^2)^(1/2)/d/x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 110, 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, 1819, 1821, 821, 272, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=-\frac {15 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d}+\frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2} \]

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4),x]

[Out]

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

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^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx \\ & = \frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-d^4+4 d^3 e x-7 d^2 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{d^2} \\ & = \frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {\int \frac {-8 d^5 e+15 d^4 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{2 d^4} \\ & = \frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}+\frac {1}{2} \left (15 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}+\frac {1}{4} \left (15 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right ) \\ & = \frac {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {15}{2} \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 {8 e^2 (d-e x)}{d \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {15 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\frac {1}{2} \left (\frac {\sqrt {d^2-e^2 x^2} \left (-d^2+7 d e x+24 e^2 x^2\right )}{d x^2 (d+e x)}-\frac {15 e^2 \log (x)}{\sqrt {d^2}}+\frac {15 e^2 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{\sqrt {d^2}}\right ) \]

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(-d^2 + 7*d*e*x + 24*e^2*x^2))/(d*x^2*(d + e*x)) - (15*e^2*Log[x])/Sqrt[d^2] + (15*e^2*L
og[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/Sqrt[d^2])/2

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-8 e x +d \right )}{2 d \,x^{2}}-\frac {15 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}+\frac {8 e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d \left (x +\frac {d}{e}\right )}\)	\(115\)
default	\(\text {Expression too large to display}\)	\(1461\)

[In]

int((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/2*(-e^2*x^2+d^2)^(1/2)*(-8*e*x+d)/d/x^2-15/2*e^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/
x)+8*e/d/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\frac {16 \, e^{3} x^{3} + 16 \, d e^{2} x^{2} + 15 \, {\left (e^{3} x^{3} + d e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (24 \, e^{2} x^{2} + 7 \, d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (d e x^{3} + d^{2} x^{2}\right )}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/2*(16*e^3*x^3 + 16*d*e^2*x^2 + 15*(e^3*x^3 + d*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (24*e^2*x^2 + 7
*d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/(d*e*x^3 + d^2*x^2)

Sympy [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{x^{3} \left (d + e x\right )^{4}}\, dx \]

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**3/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x**3*(d + e*x)**4), x)

Maxima [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{4} x^{3}} \,d x } \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^4*x^3), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (99) = 198\).

Time = 0.32 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.29 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\frac {{\left (e^{3} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e}{x} - \frac {144 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e x^{2}}\right )} e^{4} x^{2}}{8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} - \frac {15 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, d {\left | e \right |}} + \frac {\frac {16 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d e {\left | e \right |}}{x} - \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left | e \right |}}{e x^{2}}}{8 \, d^{2} e^{2}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x, algorithm="giac")

[Out]

1/8*(e^3 - 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e/x - 144*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e*x^2))*e^4
*x^2/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e)) - 15/2
*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d*abs(e)) + 1/8*(16*(d*e + sqrt(-e^2*x
^2 + d^2)*abs(e))*d*e*abs(e)/x - (d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d*abs(e)/(e*x^2))/(d^2*e^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^3\,{\left (d+e\,x\right )}^4} \,d x \]

[In]

int((d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4), x)