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

Optimal. Leaf size=140 \[ \frac{e^3 \sqrt{d^2-e^2 x^2}}{4 d x^2}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}-\frac{e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{4 d^2} \]

[Out]

(e^3*Sqrt[d^2 - e^2*x^2])/(4*d*x^2) - (d^2 - e^2*x^2)^(3/2)/(5*x^5) + (e*(d^2 - e^2*x^2)^(3/2))/(2*d*x^4) - (7
*e^2*(d^2 - e^2*x^2)^(3/2))/(15*d^2*x^3) - (e^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(4*d^2)

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Rubi [A]  time = 0.176687, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {852, 1807, 835, 807, 266, 47, 63, 208} \[ \frac{e^3 \sqrt{d^2-e^2 x^2}}{4 d x^2}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}-\frac{e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[d^2 - e^2*x^2])/(4*d*x^2) - (d^2 - e^2*x^2)^(3/2)/(5*x^5) + (e*(d^2 - e^2*x^2)^(3/2))/(2*d*x^4) - (7
*e^2*(d^2 - e^2*x^2)^(3/2))/(15*d^2*x^3) - (e^5*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(4*d^2)

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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)^2} \, dx &=\int \frac{(d-e x)^2 \sqrt{d^2-e^2 x^2}}{x^6} \, dx\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}-\frac{\int \frac{\left (10 d^3 e-7 d^2 e^2 x\right ) \sqrt{d^2-e^2 x^2}}{x^5} \, dx}{5 d^2}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}+\frac{\int \frac{\left (28 d^4 e^2-10 d^3 e^3 x\right ) \sqrt{d^2-e^2 x^2}}{x^4} \, dx}{20 d^4}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac{e^3 \int \frac{\sqrt{d^2-e^2 x^2}}{x^3} \, dx}{2 d}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac{e^3 \operatorname{Subst}\left (\int \frac{\sqrt{d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{4 d}\\ &=\frac{e^3 \sqrt{d^2-e^2 x^2}}{4 d x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}+\frac{e^5 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{8 d}\\ &=\frac{e^3 \sqrt{d^2-e^2 x^2}}{4 d x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac{e^3 \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 )}{4 d}\\ &=\frac{e^3 \sqrt{d^2-e^2 x^2}}{4 d x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac{7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac{e^5 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.190551, size = 106, normalized size = 0.76 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-16 d^2 e^2 x^2+30 d^3 e x-12 d^4-15 d e^3 x^3+28 e^4 x^4\right )-15 e^5 x^5 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+15 e^5 x^5 \log (x)}{60 d^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-12*d^4 + 30*d^3*e*x - 16*d^2*e^2*x^2 - 15*d*e^3*x^3 + 28*e^4*x^4) + 15*e^5*x^5*Log[x] -
15*e^5*x^5*Log[d + Sqrt[d^2 - e^2*x^2]])/(60*d^2*x^5)

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Maple [B]  time = 0.1, size = 541, normalized size = 3.9 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

1/2/d^5*e/x^4*(-e^2*x^2+d^2)^(7/2)+5/4/d^7*e^3/x^2*(-e^2*x^2+d^2)^(7/2)-1/4/d*e^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2
)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+23/15/d^7*e^5*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-1/5/d^4/x^5*(-e^2*x^2+d^2)
^(7/2)+1/20/d^7*e^5*(-e^2*x^2+d^2)^(5/2)+1/12/d^5*e^5*(-e^2*x^2+d^2)^(3/2)+1/4/d^3*e^5*(-e^2*x^2+d^2)^(1/2)-13
/15/d^6*e^2/x^3*(-e^2*x^2+d^2)^(7/2)-23/15/d^8*e^4/x*(-e^2*x^2+d^2)^(7/2)-23/15/d^8*e^6*x*(-e^2*x^2+d^2)^(5/2)
-23/12/d^6*e^6*x*(-e^2*x^2+d^2)^(3/2)-23/8/d^4*e^6*x*(-e^2*x^2+d^2)^(1/2)-23/8/d^2*e^6/(e^2)^(1/2)*arctan((e^2
)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+23/12/d^6*e^6*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+23/8/d^4*e^6*(-(d/e+x)^2*
e^2+2*d*e*(d/e+x))^(1/2)*x+23/8/d^2*e^6/(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^7*e^3/(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^6/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65197, size = 208, normalized size = 1.49 \begin{align*} \frac{15 \, e^{5} x^{5} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (28 \, e^{4} x^{4} - 15 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 30 \, d^{3} e x - 12 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{60 \, d^{2} 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)^2,x, algorithm="fricas")

[Out]

1/60*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (28*e^4*x^4 - 15*d*e^3*x^3 - 16*d^2*e^2*x^2 + 30*d^3*e*x
- 12*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2*x^5)

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Sympy [C]  time = 13.0823, size = 668, normalized size = 4.77 \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)**2,x)

[Out]

d**2*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)) - 2*d*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)) + 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))

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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^6/(e*x+d)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError