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

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

[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)

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Rubi [A]  time = 0.146877, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {852, 1807, 807, 266, 47, 63, 208} \[ -\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{8 x^2}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac{5 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^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 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 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^5 (d+e x)^2} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \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{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]  time = 0.188358, size = 95, normalized size = 0.88 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (-16 d^2 e x+6 d^3+9 d e^2 x^2+16 e^3 x^3\right )-15 e^4 x^4 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+15 e^4 x^4 \log (x)}{24 d x^4} \]

Antiderivative was successfully verified.

[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) + 15*e^4*x^4*Log[x] - 15*e^4*x^4*Log[d +
Sqrt[d^2 - e^2*x^2]])/(24*d*x^4)

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Maple [B]  time = 0.115, size = 513, normalized size = 4.8 \begin{align*} -{\frac{9\,{e}^{2}}{8\,{d}^{6}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{5\,{e}^{5}x}{3\,{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{5\,{e}^{5}x}{2\,{d}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{5\,{e}^{5}}{2\,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{1}{4\,{d}^{4}{x}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{4\,{e}^{4}}{3\,{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}^{4}}{8\,{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{e}^{4}}{24\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{e}^{4}}{8\,{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{5\,{e}^{4}}{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}}}}}+{\frac{2\,e}{3\,{d}^{5}{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{e}^{3}}{3\,{d}^{7}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{e}^{5}x}{3\,{d}^{7}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{e}^{5}x}{3\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{e}^{5}x}{2\,{d}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{5\,{e}^{5}}{2\,d}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{{e}^{2}}{3\,{d}^{6}} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/8/d^6*e^2/x^2*(-e^2*x^2+d^2)^(7/2)-5/3/d^5*e^5*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-5/2/d^3*e^5*(-(d/e+x)
^2*e^2+2*d*e*(d/e+x))^(1/2)*x-5/2/d*e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))
-1/4/d^4/x^4*(-e^2*x^2+d^2)^(7/2)-4/3/d^6*e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-1/8/d^6*e^4*(-e^2*x^2+d^2)^
(5/2)-5/24/d^4*e^4*(-e^2*x^2+d^2)^(3/2)-5/8/d^2*e^4*(-e^2*x^2+d^2)^(1/2)+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)+2/3/d^5*e/x^3*(-e^2*x^2+d^2)^(7/2)+4/3/d^7*e^3/x*(-e^2*x^2+d^2)^(7/2)+4/3/d^7*
e^5*x*(-e^2*x^2+d^2)^(5/2)+5/3/d^5*e^5*x*(-e^2*x^2+d^2)^(3/2)+5/2/d^3*e^5*x*(-e^2*x^2+d^2)^(1/2)+5/2/d*e^5/(e^
2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/3/d^6*e^2/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)

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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^5/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60306, size = 181, normalized size = 1.68 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 11.8741, size = 432, normalized size = 4. \begin{align*} 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}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{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}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{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{d^{2}}{2 e x^{3} \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} + \frac{e}{2 x \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} + \frac{e^{2} \operatorname{acosh}{\left (\frac{d}{e x} \right )}}{2 d} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\- \frac{i e \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac{i e^{2} \operatorname{asin}{\left (\frac{d}{e x} \right )}}{2 d} & \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**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)/(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*sq
rt(-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)/(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**2*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: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError