3.10 \(\int \frac{(d+e x) (d^2-e^2 x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=120 \[ \frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{3}{2} d e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

(e^2*(2*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*x) - ((2*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/(6*x^3) + d*e^3*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]] + (3*d*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/2

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Rubi [A]  time = 0.0921883, antiderivative size = 120, 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 = {811, 813, 844, 217, 203, 266, 63, 208} \[ \frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{3}{2} d e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

(e^2*(2*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*x) - ((2*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/(6*x^3) + d*e^3*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]] + (3*d*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/2

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 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{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx &=-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac{\int \frac{\left (4 d^3 e^2+6 d^2 e^3 x\right ) \sqrt{d^2-e^2 x^2}}{x^2} \, dx}{4 d^2}\\ &=\frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+\frac{\int \frac{-12 d^4 e^3+8 d^3 e^4 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{8 d^2}\\ &=\frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac{1}{2} \left (3 d^2 e^3\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+\left (d e^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}-\frac{1}{4} \left (3 d^2 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+\left (d e^4\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^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{1}{2} \left (3 d^2 e\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 )\\ &=\frac{e^2 (2 d-3 e x) \sqrt{d^2-e^2 x^2}}{2 x}-\frac{(2 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{6 x^3}+d e^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{3}{2} d e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [C]  time = 0.0685181, size = 111, normalized size = 0.92 \[ -\frac{d^3 \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 x^3 \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{e^3 \left (d^2-e^2 x^2\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};1-\frac{e^2 x^2}{d^2}\right )}{5 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^4,x]

[Out]

-(d^3*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (e^2*x^2)/d^2])/(3*x^3*Sqrt[1 - (e^2*x^2)/d^2])
- (e^3*(d^2 - e^2*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 - (e^2*x^2)/d^2])/(5*d^4)

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Maple [B]  time = 0.064, size = 235, normalized size = 2. \begin{align*} -{\frac{1}{3\,d{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{e}^{2}}{3\,{d}^{3}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{e}^{4}x}{3\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{4}x}{d}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{d{e}^{4}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{e}{2\,{d}^{2}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{3}}{2\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{3}}{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{3\,{e}^{3}{d}^{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}}}}} \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^4,x)

[Out]

-1/3/d/x^3*(-e^2*x^2+d^2)^(5/2)+2/3*e^2/d^3/x*(-e^2*x^2+d^2)^(5/2)+2/3*e^4/d^3*x*(-e^2*x^2+d^2)^(3/2)+e^4/d*x*
(-e^2*x^2+d^2)^(1/2)+d*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/2*e/d^2/x^2*(-e^2*x^2+d^2)
^(5/2)-1/2*e^3/d^2*(-e^2*x^2+d^2)^(3/2)-3/2*e^3*(-e^2*x^2+d^2)^(1/2)+3/2*e^3*d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)
^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00324, size = 269, normalized size = 2.24 \begin{align*} -\frac{12 \, d e^{3} x^{3} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + 9 \, d e^{3} x^{3} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + 6 \, d e^{3} x^{3} +{\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 3 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

-1/6*(12*d*e^3*x^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 9*d*e^3*x^3*log(-(d - sqrt(-e^2*x^2 + d^2))/x)
+ 6*d*e^3*x^3 + (6*e^3*x^3 - 8*d*e^2*x^2 + 3*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/x^3

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Sympy [C]  time = 7.67882, size = 469, normalized size = 3.91 \begin{align*} d^{3} \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 ) + d^{2} 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 ) - d 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 ) - e^{3} \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*x+d)*(-e**2*x**2+d**2)**(3/2)/x**4,x)

[Out]

d**3*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)) + d**2*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)) - d*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*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**2)), True)) - e**3*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 [B]  time = 1.25321, size = 352, normalized size = 2.93 \begin{align*} d \arcsin \left (\frac{x e}{d}\right ) e^{3} \mathrm{sgn}\left (d\right ) + \frac{3}{2} \, d e^{3} \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{{\left (d e^{8} + \frac{3 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d e^{6}}{x} - \frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{4}}{x^{2}}\right )} x^{3} e}{24 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3}} + \frac{1}{24} \,{\left (\frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d e^{16}}{x} - \frac{3 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{14}}{x^{2}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{12}}{x^{3}}\right )} e^{\left (-15\right )} - \sqrt{-x^{2} e^{2} + d^{2}} e^{3} \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^4,x, algorithm="giac")

[Out]

d*arcsin(x*e/d)*e^3*sgn(d) + 3/2*d*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/24*(d
*e^8 + 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^6/x - 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^4/x^2)*x^3*e/(d*e +
sqrt(-x^2*e^2 + d^2)*e)^3 + 1/24*(15*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^16/x - 3*(d*e + sqrt(-x^2*e^2 + d^2)*e
)^2*d*e^14/x^2 - (d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^12/x^3)*e^(-15) - sqrt(-x^2*e^2 + d^2)*e^3