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

Optimal. Leaf size=105 \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2}}{x}-\frac{1}{2} e (4 d+e x) \sqrt{d^2-e^2 x^2}-\frac{1}{2} d^2 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+2 d^2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

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Rubi [A]  time = 0.16069, antiderivative size = 105, 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, 1807, 815, 844, 217, 203, 266, 63, 208} \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2}}{x}-\frac{1}{2} e (4 d+e x) \sqrt{d^2-e^2 x^2}-\frac{1}{2} d^2 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+2 d^2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*(4*d + e*x)*Sqrt[d^2 - e^2*x^2])/2 - (d^2 - e^2*x^2)^(3/2)/x - (d^2*e*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/2
+ 2*d^2*e*ArcTanh[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 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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

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

Mathematica [A]  time = 0.142649, size = 100, normalized size = 0.95 \[ \left (-\frac{d^2}{x}-2 d e+\frac{e^2 x}{2}\right ) \sqrt{d^2-e^2 x^2}+2 d^2 e \log \left (\sqrt{d^2-e^2 x^2}+d\right )-\frac{1}{2} d^2 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-2 d^2 e \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.073, size = 425, normalized size = 4.1 \begin{align*} -{\frac{2\,e}{5\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{2\,e}{3\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-2\,de\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}+2\,{\frac{{d}^{3}e}{\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }+{\frac{11\,e}{15\,{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{11\,{e}^{2}x}{12\,{d}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{11\,{e}^{2}x}{8}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{11\,{d}^{2}{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}}}}}+{\frac{1}{3\,{d}^{3}e} \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}}-{\frac{1}{{d}^{4}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{{e}^{2}x}{{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{e}^{2}x}{4\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{e}^{2}x}{8}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{15\,{d}^{2}{e}^{2}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \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)^2,x)

[Out]

-2/5/d^3*e*(-e^2*x^2+d^2)^(5/2)-2/3/d*e*(-e^2*x^2+d^2)^(3/2)-2*d*e*(-e^2*x^2+d^2)^(1/2)+2*d^3*e/(d^2)^(1/2)*ln
((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+11/15/d^3*e*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+11/12/d^2*e^2*
(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+11/8*e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x+11/8*d^2*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))+1/3/d^3/e/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x)
)^(7/2)-1/d^4/x*(-e^2*x^2+d^2)^(7/2)-1/d^4*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/4/d^2*e^2*x*(-e^2*x^2+d^2)^(3/2)-15/8*
e^2*x*(-e^2*x^2+d^2)^(1/2)-15/8*d^2*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59977, size = 231, normalized size = 2.2 \begin{align*} \frac{2 \, d^{2} e x \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) - 4 \, d^{2} e x \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) - 4 \, d^{2} e x +{\left (e^{2} x^{2} - 4 \, d e x - 2 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{2 \, 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)^2,x, algorithm="fricas")

[Out]

1/2*(2*d^2*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 4*d^2*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - 4*d^
2*e*x + (e^2*x^2 - 4*d*e*x - 2*d^2)*sqrt(-e^2*x^2 + d^2))/x

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Sympy [C]  time = 9.24709, size = 354, normalized size = 3.37 \begin{align*} d^{2} \left (\begin{cases} \frac{i d}{x \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + i e \operatorname{acosh}{\left (\frac{e x}{d} \right )} - \frac{i e^{2} x}{d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\- \frac{d}{x \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} - e \operatorname{asin}{\left (\frac{e x}{d} \right )} + \frac{e^{2} x}{d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) - 2 d e \left (\begin{cases} \frac{d^{2}}{e x \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname{acosh}{\left (\frac{d}{e x} \right )} - \frac{e x}{\sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\- \frac{i d^{2}}{e x \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname{asin}{\left (\frac{d}{e x} \right )} + \frac{i e x}{\sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e} - \frac{i d x}{2 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{3}}{2 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e} + \frac{d x \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}}{2} & \text{otherwise} \end{cases}\right ) \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)**2,x)

[Out]

d**2*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d**2)),
Abs(e**2*x**2)/Abs(d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**
2/d**2)), True)) - 2*d*e*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**2/(
e**2*x**2) - 1), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asin(d
/(e*x)) + I*e*x/sqrt(-d**2/(e**2*x**2) + 1), True)) + e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sq
rt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*a
sin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} 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)^2,x, algorithm="giac")

[Out]

sage0*x