3.3.7 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{x^4 (d+e x)^4} \, dx\) [207]

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

[Out]

10*e^3*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^2-8*e^3*(-e*x+d)/d^2/(-e^2*x^2+d^2)^(1/2)-1/3*(-e^2*x^2+d^2)^(1/2)/x^
3+2*e*(-e^2*x^2+d^2)^(1/2)/d/x^2-23/3*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x

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Rubi [A]
time = 0.40, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {866, 1819, 1821, 821, 272, 65, 214} \begin {gather*} -\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 {8 e^3 (d-e x)}{d^2 \sqrt {d^2-e^2 x^2}}+\frac {10 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^2} \end {gather*}

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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 821

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 866

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 1819

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 1821

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])

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 ) \text {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) \text {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*}

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Mathematica [A]
time = 0.46, size = 103, normalized size = 0.75 \begin {gather*} -\frac {\frac {\sqrt {d^2-e^2 x^2} \left (d^3-5 d^2 e x+17 d e^2 x^2+47 e^3 x^3\right )}{x^3 (d+e x)}+60 e^3 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((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)) + 60*e^3*ArcTanh[(Sq
rt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/d^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1625\) vs. \(2(123)=246\).
time = 0.10, size = 1626, normalized size = 11.87

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (23 e^{2} x^{2}-6 d e x +d^{2}\right )}{3 x^{3} d^{2}}-\frac {8 e^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{2} \left (x +\frac {d}{e}\right )}+\frac {10 e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d \sqrt {d^{2}}}\)	\(131\)
default	\(\text {Expression too large to display}\)	\(1626\)

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,method=_RETURNVERBOSE)

[Out]

20*e^3/d^7*(1/5*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*
e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^2)^(
1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)))))+4*e/d^5*(1/d/e/(x+d/e)^3*(-(x+d/e)^2*e^2+2*
d*e*(x+d/e))^(7/2)+4*e/d*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+5/3*e/d*(1/5*(-(x+d/e)^2*e^2+
2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(
-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d
/e)^2*e^2+2*d*e*(x+d/e))^(1/2)))))))-4/d^5*e*(-1/2/d^2/x^2*(-e^2*x^2+d^2)^(7/2)-5/2*e^2/d^2*(1/5*(-e^2*x^2+d^2
)^(5/2)+d^2*(1/3*(-e^2*x^2+d^2)^(3/2)+d^2*((-e^2*x^2+d^2)^(1/2)-d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*
x^2+d^2)^(1/2))/x)))))+1/d^4*(-1/d/e/(x+d/e)^4*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)-3*e/d*(1/d/e/(x+d/e)^3*(-(
x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+4*e/d*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+5/3*e/d*(1/5*(
-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)+
3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2)
^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))))))+10/d^6*e^2*(-1/d^2/x*(-e^2*x^2+d^2)^(7/2)-6*e^2/d^2*(1/6*
x*(-e^2*x^2+d^2)^(5/2)+5/6*d^2*(1/4*x*(-e^2*x^2+d^2)^(3/2)+3/4*d^2*(1/2*x*(-e^2*x^2+d^2)^(1/2)+1/2*d^2/(e^2)^(
1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))))))+10*e^2/d^6*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e
))^(7/2)+5/3*e/d*(1/5*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(5/2)+d*e*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^
2+2*d*e*(x+d/e))^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(
e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))))+1/d^4*(-1/3/d^2/x^3*(-e^2*x^2+d^2)^(
7/2)-4/3*e^2/d^2*(-1/d^2/x*(-e^2*x^2+d^2)^(7/2)-6*e^2/d^2*(1/6*x*(-e^2*x^2+d^2)^(5/2)+5/6*d^2*(1/4*x*(-e^2*x^2
+d^2)^(3/2)+3/4*d^2*(1/2*x*(-e^2*x^2+d^2)^(1/2)+1/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
)))))-20/d^7*e^3*(1/5*(-e^2*x^2+d^2)^(5/2)+d^2*(1/3*(-e^2*x^2+d^2)^(3/2)+d^2*((-e^2*x^2+d^2)^(1/2)-d^2/(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 3.46, size = 117, normalized size = 0.85 \begin {gather*} -\frac {24 \, x^{4} e^{4} + 24 \, d x^{3} e^{3} + 30 \, {\left (x^{4} e^{4} + d x^{3} e^{3}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (47 \, x^{3} e^{3} + 17 \, d x^{2} e^{2} - 5 \, d^{2} x e + d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (d^{2} x^{4} e + d^{3} x^{3}\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (118) = 236\).
time = 1.34, size = 300, normalized size = 2.19 \begin {gather*} \frac {10 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{d^{2}} - \frac {x^{3} {\left (\frac {11 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e}{x} - \frac {81 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-1\right )}}{x^{2}} - \frac {477 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-3\right )}}{x^{3}} - e^{3}\right )} e^{6}}{24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{2} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}} - \frac {\frac {93 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{4} e}{x} - \frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{\left (-1\right )}}{x^{2}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{\left (-3\right )}}{x^{3}}}{24 \, d^{6}} \end {gather*}

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]

10*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^2 - 1/24*x^3*(11*(d*e + sqrt(-x^2*e^2 +
d^2)*e)*e/x - 81*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^(-1)/x^2 - 477*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-3)/x^
3 - e^3)*e^6/((d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^2*((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)) - 1/24*(93*(
d*e + sqrt(-x^2*e^2 + d^2)*e)*d^4*e/x - 12*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^(-1)/x^2 + (d*e + sqrt(-x^2*
e^2 + d^2)*e)^3*d^4*e^(-3)/x^3)/d^6

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^4\,{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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