3.113 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{x^6 (d+e x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac{3 e^3 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.0893033, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {850, 807, 266, 47, 63, 208} \[ -\frac{3 e^3 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)),x]

[Out]

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

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 807

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 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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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^6 (d+e x)} \, dx &=\int \frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx\\ &=-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-e \int \frac{\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx\\ &=-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac{1}{2} e \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac{1}{8} \left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d^2-e^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{3 e^3 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}-\frac{1}{16} \left (3 e^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )\\ &=-\frac{3 e^3 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac{1}{8} \left (3 e^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 )\\ &=-\frac{3 e^3 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.159132, size = 106, normalized size = 0.98 \[ \frac{\sqrt{d^2-e^2 x^2} \left (16 d^2 e^2 x^2+10 d^3 e x-8 d^4-25 d e^3 x^3-8 e^4 x^4\right )+15 e^5 x^5 \log \left (\sqrt{d^2-e^2 x^2}+d\right )-15 e^5 x^5 \log (x)}{40 d x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^6*(d + e*x)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-8*d^4 + 10*d^3*e*x + 16*d^2*e^2*x^2 - 25*d*e^3*x^3 - 8*e^4*x^4) - 15*e^5*x^5*Log[x] + 1
5*e^5*x^5*Log[d + Sqrt[d^2 - e^2*x^2]])/(40*d*x^5)

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Maple [B]  time = 0.097, size = 493, normalized size = 4.6 \begin{align*}{\frac{e}{4\,{d}^{4}{x}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{{e}^{3}}{8\,{d}^{6}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{1}{5\,{d}^{3}{x}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{{e}^{5}}{5\,{d}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{2}}{5\,{d}^{5}{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{{e}^{4}}{5\,{d}^{7}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{{e}^{6}x}{5\,{d}^{7}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{6}x}{4\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{5}}{40\,{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{e}^{6}x}{4\,{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{e}^{6}x}{8\,{d}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{3\,{e}^{6}}{8\,d}\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{3\,{e}^{6}x}{8\,{d}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{3\,{e}^{6}}{8\,d}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{{e}^{5}}{8\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{5}}{8\,{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{3\,{e}^{5}}{8}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d),x)

[Out]

1/4*e/d^4/x^4*(-e^2*x^2+d^2)^(7/2)+1/8*e^3/d^6/x^2*(-e^2*x^2+d^2)^(7/2)-1/5/d^3/x^5*(-e^2*x^2+d^2)^(7/2)+1/5*e
^5/d^6*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-1/5*e^2/d^5/x^3*(-e^2*x^2+d^2)^(7/2)-1/5*e^4/d^7/x*(-e^2*x^2+d^2)^
(7/2)-1/5*e^6/d^7*x*(-e^2*x^2+d^2)^(5/2)-1/4*e^6/d^5*x*(-e^2*x^2+d^2)^(3/2)-3/40*e^5/d^6*(-e^2*x^2+d^2)^(5/2)+
1/4*e^6/d^5*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+3/8*e^6/d^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x+3/8*e^6/
d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))-3/8*e^6/d^3*x*(-e^2*x^2+d^2)^(1/2)-3/
8*e^6/d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/8*e^5/d^4*(-e^2*x^2+d^2)^(3/2)-3/8*e^5/d^2*(-
e^2*x^2+d^2)^(1/2)+3/8*e^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

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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^6/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63698, size = 204, normalized size = 1.89 \begin{align*} -\frac{15 \, e^{5} x^{5} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (8 \, e^{4} x^{4} + 25 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} - 10 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{40 \, d x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^6/(e*x+d),x, algorithm="fricas")

[Out]

-1/40*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (8*e^4*x^4 + 25*d*e^3*x^3 - 16*d^2*e^2*x^2 - 10*d^3*e*x
+ 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*x^5)

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Sympy [C]  time = 14.1072, size = 785, normalized size = 7.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**6/(e*x+d),x)

[Out]

d**3*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 +
e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d
**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2)/Abs(d*
*2) > 1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d
**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7)
- e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), True)) - d**2*e*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)/(Abs(e**2)*Abs(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)) - d*e**2*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)/(Abs(e**2)*Abs(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**3*Piecewise((-d**2/(2*e*x**3*sqrt(d**2/(e**2*x*
*2) - 1)) + e/(2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2)/(Abs(e**2)*Abs(x**2)) >
1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), True))

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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^6/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError