3.169 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{x^7 (d+e x)^2} \, dx\)

Optimal. Leaf size=169 \[ -\frac{3 e^4 \sqrt{d^2-e^2 x^2}}{16 d^2 x^2}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{3 e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

[Out]

(-3*e^4*Sqrt[d^2 - e^2*x^2])/(16*d^2*x^2) - (d^2 - e^2*x^2)^(3/2)/(6*x^6) + (2*e*(d^2 - e^2*x^2)^(3/2))/(5*d*x
^5) - (3*e^2*(d^2 - e^2*x^2)^(3/2))/(8*d^2*x^4) + (4*e^3*(d^2 - e^2*x^2)^(3/2))/(15*d^3*x^3) + (3*e^6*ArcTanh[
Sqrt[d^2 - e^2*x^2]/d])/(16*d^3)

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Rubi [A]  time = 0.207751, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{3 e^4 \sqrt{d^2-e^2 x^2}}{16 d^2 x^2}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{3 e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2),x]

[Out]

(-3*e^4*Sqrt[d^2 - e^2*x^2])/(16*d^2*x^2) - (d^2 - e^2*x^2)^(3/2)/(6*x^6) + (2*e*(d^2 - e^2*x^2)^(3/2))/(5*d*x
^5) - (3*e^2*(d^2 - e^2*x^2)^(3/2))/(8*d^2*x^4) + (4*e^3*(d^2 - e^2*x^2)^(3/2))/(15*d^3*x^3) + (3*e^6*ArcTanh[
Sqrt[d^2 - e^2*x^2]/d])/(16*d^3)

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^7 (d+e x)^2} \, dx &=\int \frac{(d-e x)^2 \sqrt{d^2-e^2 x^2}}{x^7} \, dx\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}-\frac{\int \frac{\left (12 d^3 e-9 d^2 e^2 x\right ) \sqrt{d^2-e^2 x^2}}{x^6} \, dx}{6 d^2}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}+\frac{\int \frac{\left (45 d^4 e^2-24 d^3 e^3 x\right ) \sqrt{d^2-e^2 x^2}}{x^5} \, dx}{30 d^4}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}-\frac{\int \frac{\left (96 d^5 e^3-45 d^4 e^4 x\right ) \sqrt{d^2-e^2 x^2}}{x^4} \, dx}{120 d^6}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac{\left (3 e^4\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{x^3} \, dx}{8 d^2}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{16 d^2}\\ &=-\frac{3 e^4 \sqrt{d^2-e^2 x^2}}{16 d^2 x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}-\frac{\left (3 e^6\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{32 d^2}\\ &=-\frac{3 e^4 \sqrt{d^2-e^2 x^2}}{16 d^2 x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac{\left (3 e^4\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 )}{16 d^2}\\ &=-\frac{3 e^4 \sqrt{d^2-e^2 x^2}}{16 d^2 x^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{6 x^6}+\frac{2 e \left (d^2-e^2 x^2\right )^{3/2}}{5 d x^5}-\frac{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{8 d^2 x^4}+\frac{4 e^3 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^3 x^3}+\frac{3 e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^3}\\ \end{align*}

Mathematica [A]  time = 0.264799, size = 117, normalized size = 0.69 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (50 d^3 e^2 x^2+32 d^2 e^3 x^3-96 d^4 e x+40 d^5-45 d e^4 x^4+64 e^5 x^5\right )-45 e^6 x^6 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+45 e^6 x^6 \log (x)}{240 d^3 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^7*(d + e*x)^2),x]

[Out]

-(Sqrt[d^2 - e^2*x^2]*(40*d^5 - 96*d^4*e*x + 50*d^3*e^2*x^2 + 32*d^2*e^3*x^3 - 45*d*e^4*x^4 + 64*e^5*x^5) + 45
*e^6*x^6*Log[x] - 45*e^6*x^6*Log[d + Sqrt[d^2 - e^2*x^2]])/(240*d^3*x^6)

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Maple [B]  time = 0.12, size = 566, normalized size = 3.4 \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^7/(e*x+d)^2,x)

[Out]

16/15/d^7*e^3/x^3*(-e^2*x^2+d^2)^(7/2)+26/15/d^9*e^5/x*(-e^2*x^2+d^2)^(7/2)+26/15/d^9*e^7*x*(-e^2*x^2+d^2)^(5/
2)+13/6/d^7*e^7*x*(-e^2*x^2+d^2)^(3/2)+13/4/d^5*e^7*x*(-e^2*x^2+d^2)^(1/2)+13/4/d^3*e^7/(e^2)^(1/2)*arctan((e^
2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/6/d^4/x^6*(-e^2*x^2+d^2)^(7/2)-1/3/d^8*e^4/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*
(d/e+x))^(7/2)-3/80/d^8*e^6*(-e^2*x^2+d^2)^(5/2)-1/16/d^6*e^6*(-e^2*x^2+d^2)^(3/2)-3/16/d^4*e^6*(-e^2*x^2+d^2)
^(1/2)-26/15/d^8*e^6*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+3/16/d^2*e^6/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e
^2*x^2+d^2)^(1/2))/x)+2/5/d^5*e/x^5*(-e^2*x^2+d^2)^(7/2)-17/24/d^6*e^2/x^4*(-e^2*x^2+d^2)^(7/2)-23/16/d^8*e^4/
x^2*(-e^2*x^2+d^2)^(7/2)-13/6/d^7*e^7*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-13/4/d^5*e^7*(-(d/e+x)^2*e^2+2*d*
e*(d/e+x))^(1/2)*x-13/4/d^3*e^7/(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(-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^7/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70395, size = 234, normalized size = 1.38 \begin{align*} -\frac{45 \, e^{6} x^{6} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (64 \, e^{5} x^{5} - 45 \, d e^{4} x^{4} + 32 \, d^{2} e^{3} x^{3} + 50 \, d^{3} e^{2} x^{2} - 96 \, d^{4} e x + 40 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{240 \, d^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/240*(45*e^6*x^6*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (64*e^5*x^5 - 45*d*e^4*x^4 + 32*d^2*e^3*x^3 + 50*d^3*e
^2*x^2 - 96*d^4*e*x + 40*d^5)*sqrt(-e^2*x^2 + d^2))/(d^3*x^6)

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Sympy [C]  time = 18.4442, size = 816, normalized size = 4.83 \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**7/(e*x+d)**2,x)

[Out]

d**2*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e**3/
(48*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*x))/
(16*d**5), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x*
*5*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-
d**2/(e**2*x**2) + 1)) - I*e**6*asin(d/(e*x))/(16*d**5), True)) - 2*d*e*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)) + e**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*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), 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^7/(e*x+d)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError