3.165 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{x^3 (d+e x)^2} \, dx\)

Optimal. Leaf size=110 \[ \frac{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{1}{2} d e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

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Rubi [A]  time = 0.162917, antiderivative size = 110, 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, 813, 844, 217, 203, 266, 63, 208} \[ \frac{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{1}{2} d e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^2),x]

[Out]

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

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 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 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^3 (d+e x)^2} \, dx &=\int \frac{(d-e x)^2 \sqrt{d^2-e^2 x^2}}{x^3} \, dx\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac{\int \frac{\left (4 d^3 e-d^2 e^2 x\right ) \sqrt{d^2-e^2 x^2}}{x^2} \, dx}{2 d^2}\\ &=\frac{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac{\int \frac{2 d^4 e^2+8 d^3 e^3 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{4 d^2}\\ &=\frac{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac{1}{2} \left (d^2 e^2\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+\left (2 d e^3\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac{1}{4} \left (d^2 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+\left (2 d e^3\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{e (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{1}{2} d^2 \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 (4 d+e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{1}{2} d e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [A]  time = 0.152905, size = 102, normalized size = 0.93 \[ \left (-\frac{d^2}{2 x^2}+\frac{2 d e}{x}+e^2\right ) \sqrt{d^2-e^2 x^2}-\frac{1}{2} d e^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+2 d e^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{1}{2} d e^2 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^2),x]

[Out]

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

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Maple [B]  time = 0.07, size = 456, normalized size = 4.2 \begin{align*}{\frac{{e}^{2}}{10\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{e}^{2}}{6\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{2}}{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{d}^{2}{e}^{2}}{2}\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{14\,{e}^{2}}{15\,{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{7\,{e}^{3}x}{6\,{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{e}^{3}x}{4\,d}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{7\,d{e}^{3}}{4}\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}} \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{1}{2\,{d}^{4}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+2\,{\frac{e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{7/2}}{{d}^{5}x}}+2\,{\frac{{e}^{3}x \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}{{d}^{5}}}+{\frac{5\,{e}^{3}x}{2\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{15\,{e}^{3}x}{4\,d}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{15\,d{e}^{3}}{4}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^2,x)

[Out]

1/10/d^4*e^2*(-e^2*x^2+d^2)^(5/2)+1/6/d^2*e^2*(-e^2*x^2+d^2)^(3/2)+1/2*e^2*(-e^2*x^2+d^2)^(1/2)-1/2*d^2*e^2/(d
^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-14/15/d^4*e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-
7/6/d^3*e^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-7/4/d*e^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-7/4*d*e^3/
(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/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e
*(d/e+x))^(7/2)-1/2/d^4/x^2*(-e^2*x^2+d^2)^(7/2)+2/d^5*e/x*(-e^2*x^2+d^2)^(7/2)+2/d^5*e^3*x*(-e^2*x^2+d^2)^(5/
2)+5/2/d^3*e^3*x*(-e^2*x^2+d^2)^(3/2)+15/4/d*e^3*x*(-e^2*x^2+d^2)^(1/2)+15/4*d*e^3/(e^2)^(1/2)*arctan((e^2)^(1
/2)*x/(-e^2*x^2+d^2)^(1/2))

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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^3/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62689, size = 240, normalized size = 2.18 \begin{align*} -\frac{8 \, d e^{2} x^{2} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) - d e^{2} x^{2} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) - 2 \, d e^{2} x^{2} -{\left (2 \, e^{2} x^{2} + 4 \, d e x - d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/2*(8*d*e^2*x^2*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - d*e^2*x^2*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - 2
*d*e^2*x^2 - (2*e^2*x^2 + 4*d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/x^2

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Sympy [C]  time = 8.92536, size = 355, normalized size = 3.23 \begin{align*} d^{2} \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 ) - 2 d e \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 ) + e^{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}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/x**3/(e*x+d)**2,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^2,x, algorithm="giac")

[Out]

sage0*x