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]

(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]

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Rubi [A]  time = 0.162587, antiderivative size = 102, normalized size of antiderivative = 1., 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 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])

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 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 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 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)^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*}

Mathematica [A]  time = 0.187797, 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]

-(Sqrt[d^2 - e^2*x^2]*(d^2 - 3*d*e*x + 2*e^2*x^2))/(3*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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Maple [B]  time = 0.076, size = 479, normalized size = 4.7 \begin{align*} -{\frac{5\,{e}^{2}}{3\,{d}^{6}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{5\,{e}^{4}x}{3\,{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{25\,{e}^{4}x}{12\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{25\,{e}^{4}x}{8\,{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{17\,{e}^{4}x}{12\,{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{17\,{e}^{4}x}{8\,{d}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{d{e}^{3}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}+{\frac{e}{{d}^{5}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{e}{3\,{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-2}}+{\frac{17\,{e}^{3}}{15\,{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{17\,{e}^{4}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{1}{3\,{d}^{4}{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{25\,{e}^{4}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{{e}^{3}}{3\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{3}}{d}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{e}^{3}}{5\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}} \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)^2,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60153, size = 217, normalized size = 2.13 \begin{align*} \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}} \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)^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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Sympy [C]  time = 8.45791, size = 347, normalized size = 3.4 \begin{align*} 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}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{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}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{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}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{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 ) \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)**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)/(
Abs(e**2)*Abs(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), True)) - 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)/(Abs(e**2)*Abs(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)/Abs(d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*a
sin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError