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

Optimal. Leaf size=102 \[ \frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}+e^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-1/3*(-e^2*x^2+d^2)^(3/2)/x^3-e^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))-e^3*arctanh((-e^2*x^2+d^2)^(1/2)/d)+e*(-e*x
+d)*(-e^2*x^2+d^2)^(1/2)/x^2

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Rubi [A]  time = 0.16, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {852, 1807, 811, 844, 217, 203, 266, 63, 208} \[ \frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}+e^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(d - e*x)*Sqrt[d^2 - e^2*x^2])/x^2 - (d^2 - e^2*x^2)^(3/2)/(3*x^3) - e^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]
- e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

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])

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^4 (d+e x)^2} \, dx &=\int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^4} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}-\frac {\int \frac {\left (6 d^3 e-3 d^2 e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x^3} \, dx}{3 d^2}\\ &=\frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}+\frac {\int \frac {12 d^5 e^3-12 d^4 e^4 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{12 d^4}\\ &=\frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}+\left (d e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-e^4 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}+\frac {1}{2} \left (d e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-e^4 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}-e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-(d 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 )\\ &=\frac {e (d-e x) \sqrt {d^2-e^2 x^2}}{x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 x^3}-e^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 96, normalized size = 0.94 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (d^2-3 d e x+2 e^2 x^2\right )}{3 x^3}-e^3 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+e^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )+e^3 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[d^2 - e^2*x^2]*(d^2 - 3*d*e*x + 2*e^2*x^2))/x^3 - e^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + e^3*Log[x
] - e^3*Log[d + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.93, size = 106, normalized size = 1.04 \[ \frac {6 \, e^{3} x^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 3 \, e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (2 \, e^{2} x^{2} - 3 \, d e x + d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, x^{3}} \]

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)^2,x, algorithm="fricas")

[Out]

1/3*(6*e^3*x^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 3*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (2*e
^2*x^2 - 3*d*e*x + d^2)*sqrt(-e^2*x^2 + d^2))/x^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.71index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [B]  time = 0.02, size = 479, normalized size = 4.70 \[ -\frac {d \,e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {17 e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {25 e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}+\frac {17 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{4} x}{8 d^{2}}-\frac {25 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4} x}{8 d^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{d}+\frac {17 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{4} x}{12 d^{4}}-\frac {25 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4} x}{12 d^{4}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}{3 d^{3}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4} x}{3 d^{6}}+\frac {17 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{3}}{15 d^{5}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{5 d^{5}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} e}{3 \left (x +\frac {d}{e}\right )^{2} d^{5}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{3 d^{6} x}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{d^{5} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{3 d^{4} x^{3}} \]

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)^2,x)

[Out]

17/12/d^4*e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+17/8/d^2*e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+1/3/d^5
*e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+1/d^5*e/x^2*(-e^2*x^2+d^2)^(7/2)-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)-5/3/d^6*e^2/x*(-e^2*x^2+d^2)^(7/2)-5/3/d^6*e^4*x*(-e^2*x^2+d^2)^(5/2)-2
5/12/d^4*e^4*x*(-e^2*x^2+d^2)^(3/2)-25/8/d^2*e^4*x*(-e^2*x^2+d^2)^(1/2)+17/15/d^5*e^3*(2*(x+d/e)*d*e-(x+d/e)^2
*e^2)^(5/2)+17/8*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)-1/3/d^4/x^3*(-e^2*x
^2+d^2)^(7/2)+1/5/d^5*e^3*(-e^2*x^2+d^2)^(5/2)+1/3/d^3*e^3*(-e^2*x^2+d^2)^(3/2)+1/d*e^3*(-e^2*x^2+d^2)^(1/2)-2
5/8*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)

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maxima [A]  time = 0.99, size = 134, normalized size = 1.31 \[ -e^{3} \arcsin \left (\frac {e x}{d}\right ) - e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{d} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{2}}{x} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{d x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

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)^2,x, algorithm="maxima")

[Out]

-e^3*arcsin(e*x/d) - e^3*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) + sqrt(-e^2*x^2 + d^2)*e^3/d - sq
rt(-e^2*x^2 + d^2)*e^2/x + (-e^2*x^2 + d^2)^(3/2)*e/(d*x^2) - 1/3*(-e^2*x^2 + d^2)^(3/2)/x^3

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

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)^2),x)

[Out]

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

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sympy [C]  time = 9.73, size = 338, normalized size = 3.31 \[ d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

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)**2,x)

[Out]

d**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/(e
**2*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), Tru
e)) - 2*d*e*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/(e**2*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)) + e**2*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1
+ e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sq
rt(1 - e**2*x**2/d**2)), True))

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