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

Optimal. Leaf size=94 \[ -\frac{8 e (d-e x)}{\sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{x}-e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+4 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[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]

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Rubi [A]  time = 0.216331, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {852, 1805, 1807, 844, 217, 203, 266, 63, 208} \[ -\frac{8 e (d-e x)}{\sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{x}-e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+4 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[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 852

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 1805

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 1807

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])

Rule 844

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 217

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 203

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

Rule 266

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 63

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 208

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

Rubi steps

\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^{5/2}}{x^2 (d+e x)^4} \, dx &=\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) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )-e^2 \operatorname{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) \operatorname{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]  time = 0.213147, size = 84, normalized size = 0.89 \[ \sqrt{d^2-e^2 x^2} \left (-\frac{8 e}{d+e x}-\frac{1}{x}\right )+4 e \log \left (\sqrt{d^2-e^2 x^2}+d\right )-e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-4 e \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.072, size = 515, normalized size = 5.5 \begin{align*} -{\frac{{e}^{2}x}{{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{e}^{2}x}{4\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{e}^{2}x}{8\,{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{7\,{e}^{2}x}{12\,{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{e}^{2}x}{8\,{d}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{1}{{e}^{3}{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}-{\frac{1}{{e}^{2}{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}-{\frac{1}{3\,e{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-2}}+4\,{\frac{de}{\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }-{\frac{1}{{d}^{6}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{15\,{e}^{2}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{4\,e}{5\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{4\,e}{3\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-4\,{\frac{e\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{d}}+{\frac{7\,e}{15\,{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{7\,{e}^{2}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^6*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/4/d^4*e^2*x*(-e^2*x^2+d^2)^(3/2)-15/8/d^2*e^2*x*(-e^2*x^2+d^2)^(1/2)+7/12/
d^4*e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+7/8/d^2*e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-1/e^3/d^3/(d
/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-1/e^2/d^4/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-1/3/e/d^
5/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+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)-1/d^6/x*(-e^2*x^2+d^2)^(7/2)-15/8*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-4/5/d^5*e*(
-e^2*x^2+d^2)^(5/2)-4/3/d^3*e*(-e^2*x^2+d^2)^(3/2)-4/d*e*(-e^2*x^2+d^2)^(1/2)+7/15/d^5*e*(-(d/e+x)^2*e^2+2*d*e
*(d/e+x))^(5/2)+7/8*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{4} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.59943, size = 262, normalized size = 2.79 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError