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

Optimal. Leaf size=117 \[ \frac{1}{2} d e (2 d-3 e x) \sqrt{d^2-e^2 x^2}-\frac{(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac{3}{2} d^3 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^3 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0916725, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {813, 815, 844, 217, 203, 266, 63, 208} \[ \frac{1}{2} d e (2 d-3 e x) \sqrt{d^2-e^2 x^2}-\frac{(3 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{3 x}-\frac{3}{2} d^3 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^3 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.189624, size = 124, normalized size = 1.06 \[ -\frac{d^5 \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x \sqrt{d^2-e^2 x^2}}-\frac{1}{3} e \left (\sqrt{d^2-e^2 x^2} \left (e^2 x^2-4 d^2\right )+3 d^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*(Sqrt[d^2 - e^2*x^2]*(-4*d^2 + e^2*x^2) + 3*d^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]))/3 - (d^5*Sqrt[1 - (e^2*x^
2)/d^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, (e^2*x^2)/d^2])/(x*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.056, size = 182, normalized size = 1.6 \begin{align*}{\frac{e}{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+e{d}^{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}-{e{d}^{4}\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{1}{dx} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{2}x}{d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{3\,d{e}^{2}x}{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{3\,{e}^{2}{d}^{3}}{2}\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*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x)

[Out]

1/3*e*(-e^2*x^2+d^2)^(3/2)+e*d^2*(-e^2*x^2+d^2)^(1/2)-e*d^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)
^(1/2))/x)-1/d/x*(-e^2*x^2+d^2)^(5/2)-e^2/d*x*(-e^2*x^2+d^2)^(3/2)-3/2*d*e^2*x*(-e^2*x^2+d^2)^(1/2)-3/2*e^2*d^
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*x+d)*(-e^2*x^2+d^2)^(3/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79018, size = 257, normalized size = 2.2 \begin{align*} \frac{18 \, d^{3} e x \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + 6 \, d^{3} e x \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + 8 \, d^{3} e x -{\left (2 \, e^{3} x^{3} + 3 \, d e^{2} x^{2} - 8 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(18*d^3*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 6*d^3*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + 8*d
^3*e*x - (2*e^3*x^3 + 3*d*e^2*x^2 - 8*d^2*e*x + 6*d^3)*sqrt(-e^2*x^2 + d^2))/x

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Sympy [C]  time = 9.18669, size = 393, normalized size = 3.36 \begin{align*} d^{3} \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 ) + d^{2} 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}\: \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 ) - d 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}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{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^{3} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*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*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**
2/d**2)), True)) + d**2*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)/(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)) - d*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)/Abs(d**2) > 1), (d**
2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) - e**3*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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Giac [A]  time = 1.26675, size = 212, normalized size = 1.81 \begin{align*} -\frac{3}{2} \, d^{3} \arcsin \left (\frac{x e}{d}\right ) e \mathrm{sgn}\left (d\right ) - d^{3} e \log \left (\frac{{\left | -2 \, d e - 2 \, \sqrt{-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \,{\left | x \right |}}\right ) + \frac{d^{3} x e^{3}}{2 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{3} e^{\left (-1\right )}}{2 \, x} + \frac{1}{6} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left (8 \, d^{2} e -{\left (2 \, x e^{3} + 3 \, d e^{2}\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*d^3*arcsin(x*e/d)*e*sgn(d) - d^3*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/2*d^
3*x*e^3/(d*e + sqrt(-x^2*e^2 + d^2)*e) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^3*e^(-1)/x + 1/6*sqrt(-x^2*e^2 +
d^2)*(8*d^2*e - (2*x*e^3 + 3*d*e^2)*x)