3.207 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{x^4 (d+e x)^4} \, dx\)

Optimal. Leaf size=137 \[ -\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{10 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

[Out]

(-8*e^3*(d - e*x))/(d^2*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*x^3) + (2*e*Sqrt[d^2 - e^2*x^2])/(d*x^2)
- (23*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^2*x) + (10*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^2

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Rubi [A]  time = 0.297347, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {852, 1805, 1807, 807, 266, 63, 208} \[ -\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{10 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^4*(d + e*x)^4),x]

[Out]

(-8*e^3*(d - e*x))/(d^2*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*x^3) + (2*e*Sqrt[d^2 - e^2*x^2])/(d*x^2)
- (23*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^2*x) + (10*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 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^4 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-d^4+4 d^3 e x-7 d^2 e^2 x^2+8 d e^3 x^3}{x^4 \sqrt{d^2-e^2 x^2}} \, dx}{d^2}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{\int \frac{-12 d^5 e+23 d^4 e^2 x-24 d^3 e^3 x^2}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{3 d^4}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{\int \frac{-46 d^6 e^2+60 d^5 e^3 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{6 d^6}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{\left (10 e^3\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{\left (5 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{d}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{(10 e) \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 )}{d}\\ &=-\frac{8 e^3 (d-e x)}{d^2 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}+\frac{2 e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{23 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{10 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.274926, size = 94, normalized size = 0.69 \[ -\frac{\frac{\sqrt{d^2-e^2 x^2} \left (-5 d^2 e x+d^3+17 d e^2 x^2+47 e^3 x^3\right )}{x^3 (d+e x)}-30 e^3 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+30 e^3 \log (x)}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^4*(d + e*x)^4),x]

[Out]

-((Sqrt[d^2 - e^2*x^2]*(d^3 - 5*d^2*e*x + 17*d*e^2*x^2 + 47*e^3*x^3))/(x^3*(d + e*x)) + 30*e^3*Log[x] - 30*e^3
*Log[d + Sqrt[d^2 - e^2*x^2]])/(3*d^2)

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Maple [B]  time = 0.087, size = 575, normalized size = 4.2 \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^4/(e*x+d)^4,x)

[Out]

65/6/d^6*e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+65/4/d^4*e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x+65/4/d
^2*e^4/(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^6/x^3*(-e^2*x^2+d^2)^(7/2)
+10/d*e^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+26/3/d^7*e^3*(-(d/e+x)^2*e^2+2*d*e*(d/e
+x))^(5/2)+1/d^6/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-2/d^7*e^3*(-e^2*x^2+d^2)^(5/2)-10/3/d^5*e^3*(-
e^2*x^2+d^2)^(3/2)-10/d^3*e^3*(-e^2*x^2+d^2)^(1/2)-26/3/d^8*e^2/x*(-e^2*x^2+d^2)^(7/2)-26/3/d^8*e^4*x*(-e^2*x^
2+d^2)^(5/2)-65/6/d^6*e^4*x*(-e^2*x^2+d^2)^(3/2)-65/4/d^4*e^4*x*(-e^2*x^2+d^2)^(1/2)-65/4/d^2*e^4/(e^2)^(1/2)*
arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+2/d^7*e/x^2*(-e^2*x^2+d^2)^(7/2)-1/d^5/e/(d/e+x)^4*(-(d/e+x)^2*e^2+
2*d*e*(d/e+x))^(7/2)+14/3/d^7*e/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{4} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^4/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^4*x^4), x)

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Fricas [A]  time = 1.57278, size = 252, normalized size = 1.84 \begin{align*} -\frac{24 \, e^{4} x^{4} + 24 \, d e^{3} x^{3} + 30 \,{\left (e^{4} x^{4} + d e^{3} x^{3}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (47 \, e^{3} x^{3} + 17 \, d e^{2} x^{2} - 5 \, d^{2} e x + d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \,{\left (d^{2} e x^{4} + d^{3} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^4/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/3*(24*e^4*x^4 + 24*d*e^3*x^3 + 30*(e^4*x^4 + d*e^3*x^3)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (47*e^3*x^3 +
17*d*e^2*x^2 - 5*d^2*e*x + d^3)*sqrt(-e^2*x^2 + d^2))/(d^2*e*x^4 + d^3*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}}}{x^{4} \left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**4/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x**4*(d + e*x)**4), x)

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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^4/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError