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

Optimal. Leaf size=105 \[ -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{x}-\frac {1}{2} e (4 d+e x) \sqrt {d^2-e^2 x^2}-\frac {1}{2} d^2 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+2 d^2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

-(-e^2*x^2+d^2)^(3/2)/x-1/2*d^2*e*arctan(e*x/(-e^2*x^2+d^2)^(1/2))+2*d^2*e*arctanh((-e^2*x^2+d^2)^(1/2)/d)-1/2
*e*(e*x+4*d)*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 105, 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, 815, 844, 217, 203, 266, 63, 208} \[ -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{x}-\frac {1}{2} e (4 d+e x) \sqrt {d^2-e^2 x^2}-\frac {1}{2} d^2 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+2 d^2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*(4*d + e*x)*Sqrt[d^2 - e^2*x^2])/2 - (d^2 - e^2*x^2)^(3/2)/x - (d^2*e*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/2
+ 2*d^2*e*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 815

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

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

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Mathematica [A]  time = 0.13, size = 100, normalized size = 0.95 \[ \left (-\frac {d^2}{x}-2 d e+\frac {e^2 x}{2}\right ) \sqrt {d^2-e^2 x^2}+2 d^2 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )-\frac {1}{2} d^2 e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-2 d^2 e \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 111, normalized size = 1.06 \[ \frac {2 \, d^{2} e x \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 4 \, d^{2} e x \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 4 \, d^{2} e x + {\left (e^{2} x^{2} - 4 \, d e x - 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d^2*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 4*d^2*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - 4*d^
2*e*x + (e^2*x^2 - 4*d*e*x - 2*d^2)*sqrt(-e^2*x^2 + d^2))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.01, size = 425, normalized size = 4.05 \[ \frac {2 d^{3} e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {11 d^{2} e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {15 d^{2} e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x}{8}+\frac {11 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{2} x}{8}-2 \sqrt {-e^{2} x^{2}+d^{2}}\, d e -\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2} x}{4 d^{2}}+\frac {11 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{2} x}{12 d^{2}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{3 d}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2} x}{d^{4}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{5 d^{3}}+\frac {11 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e}{15 d^{3}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d^{3} e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{d^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^4/x*(-e^2*x^2+d^2)^(7/2)-1/d^4*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/4/d^2*e^2*x*(-e^2*x^2+d^2)^(3/2)-15/8*e^2*x*(
-e^2*x^2+d^2)^(1/2)-15/8*d^2*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-2/5/d^3*e*(-e^2*x^2+d^
2)^(5/2)-2/3/d*e*(-e^2*x^2+d^2)^(3/2)-2*d*e*(-e^2*x^2+d^2)^(1/2)+2*d^3*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(
-e^2*x^2+d^2)^(1/2))/x)+11/15/d^3*e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+11/12/d^2*e^2*(2*(x+d/e)*d*e-(x+d/e)^2
*e^2)^(3/2)*x+11/8*e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+11/8*d^2*e^2/(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^3/e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)

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maxima [A]  time = 1.12, size = 112, normalized size = 1.07 \[ -\frac {1}{2} \, d^{2} e \arcsin \left (\frac {e x}{d}\right ) + 2 \, d^{2} e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2} x - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.85, size = 347, normalized size = 3.30 \[ d^{2} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**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*sqrt(1 - e**2*x**2/d**
2)), True)) - 2*d*e*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**2/(e**2*
x**2) - 1), Abs(d**2/(e**2*x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asin(d/(e*x)) + I*e*x
/sqrt(-d**2/(e**2*x**2) + 1), True)) + e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x*
*2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d
*x*sqrt(1 - e**2*x**2/d**2)/2, True))

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