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

Optimal. Leaf size=96 \[ d (d-e x) \sqrt {d^2-e^2 x^2}-\frac {1}{3} \left (d^2-e^2 x^2\right )^{3/2}+d^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-d^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 96, 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, 1809, 815, 844, 217, 203, 266, 63, 208} \begin {gather*} d (d-e x) \sqrt {d^2-e^2 x^2}-\frac {1}{3} \left (d^2-e^2 x^2\right )^{3/2}+d^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )-d^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

d*(d - e*x)*Sqrt[d^2 - e^2*x^2] - (d^2 - e^2*x^2)^(3/2)/3 - d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] - d^3*ArcTan
h[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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 96, normalized size = 1.00 \begin {gather*} d^3 \log (x)+\sqrt {d^2-e^2 x^2} \left (\frac {2 d^2}{3}-d e x+\frac {e^2 x^2}{3}\right )-d^3 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+d^3 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.48, size = 128, normalized size = 1.33 \begin {gather*} \frac {1}{3} \sqrt {d^2-e^2 x^2} \left (2 d^2-3 d e x+e^2 x^2\right )-\frac {d^3 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e}+2 d^3 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 95, normalized size = 0.99 \begin {gather*} 2 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \frac {1}{3} \, {\left (e^{2} x^{2} - 3 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.01, size = 290, normalized size = 3.02 \begin {gather*} -\frac {d^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}-\frac {d^{3} e \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 )}{\sqrt {e^{2}}}-\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d e x +\sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}-\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e x}{3 d}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 d^{2}}-\frac {8 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{15 d^{2}}-\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^{2} e^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/d^2*(-e^2*x^2+d^2)^(5/2)+1/3*(-e^2*x^2+d^2)^(3/2)+d^2*(-e^2*x^2+d^2)^(1/2)-d^4/(d^2)^(1/2)*ln((2*d^2+2*(d^
2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-8/15/d^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-2/3/d*e*(2*(x+d/e)*d*e-(x+d/e)^
2*e^2)^(3/2)*x-d*e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-d^3*e/(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^2/e^2/(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 = 0.99, size = 103, normalized size = 1.07 \begin {gather*} -d^{3} \arcsin \left (\frac {e x}{d}\right ) - d^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \sqrt {-e^{2} x^{2} + d^{2}} d e x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} - \frac {1}{3} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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\,{\left (d+e\,x\right )}^2} \,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*(d + e*x)^2),x)

[Out]

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

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sympy [C]  time = 14.83, size = 267, normalized size = 2.78 \begin {gather*} d^{2} \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 ) - 2 d e \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 ) + e^{2} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*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)) - 2*d*e*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)) + e**2*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2
), True))

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