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

Optimal. Leaf size=113 \[ \frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{3}{8} d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

(d^2*(8*d + 3*e*x)*Sqrt[d^2 - e^2*x^2])/8 + ((4*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/12 + (3*d^4*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/8 - d^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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Rubi [A]  time = 0.0983893, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {815, 844, 217, 203, 266, 63, 208} \[ \frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{3}{8} d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(8*d + 3*e*x)*Sqrt[d^2 - e^2*x^2])/8 + ((4*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/12 + (3*d^4*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/8 - d^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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} \, dx &=\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac{\int \frac{\left (-4 d^3 e^2-3 d^2 e^3 x\right ) \sqrt{d^2-e^2 x^2}}{x} \, dx}{4 e^2}\\ &=\frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{\int \frac{8 d^5 e^4+3 d^4 e^5 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{8 e^4}\\ &=\frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+d^5 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+\frac{1}{8} \left (3 d^4 e\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{2} d^5 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+\frac{1}{8} \left (3 d^4 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 )\\ &=\frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{3}{8} d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{d^5 \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}\\ &=\frac{1}{8} d^2 (8 d+3 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{3}{8} d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [A]  time = 0.202308, size = 124, normalized size = 1.1 \[ \frac{1}{24} \sqrt{d^2-e^2 x^2} \left (15 d^2 e x+32 d^3-8 d e^2 x^2-6 e^3 x^3\right )+\frac{3 d^3 \sqrt{d^2-e^2 x^2} \sin ^{-1}\left (\frac{e x}{d}\right )}{8 \sqrt{1-\frac{e^2 x^2}{d^2}}}+d^4 \left (-\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(32*d^3 + 15*d^2*e*x - 8*d*e^2*x^2 - 6*e^3*x^3))/24 + (3*d^3*Sqrt[d^2 - e^2*x^2]*ArcSin[(
e*x)/d])/(8*Sqrt[1 - (e^2*x^2)/d^2]) - d^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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Maple [A]  time = 0.054, size = 151, normalized size = 1.3 \begin{align*}{\frac{ex}{4} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,e{d}^{2}x}{8}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{3\,e{d}^{4}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{d}{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{d}^{3}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}-{{d}^{5}\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,x)

[Out]

1/4*e*x*(-e^2*x^2+d^2)^(3/2)+3/8*e*d^2*x*(-e^2*x^2+d^2)^(1/2)+3/8*e*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2
*x^2+d^2)^(1/2))+1/3*d*(-e^2*x^2+d^2)^(3/2)+d^3*(-e^2*x^2+d^2)^(1/2)-d^5/(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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82572, size = 228, normalized size = 2.02 \begin{align*} -\frac{3}{4} \, d^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{4} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) - \frac{1}{24} \,{\left (6 \, e^{3} x^{3} + 8 \, d e^{2} x^{2} - 15 \, d^{2} e x - 32 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}} \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,x, algorithm="fricas")

[Out]

-3/4*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + d^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - 1/24*(6*e^3*x^3
+ 8*d*e^2*x^2 - 15*d^2*e*x - 32*d^3)*sqrt(-e^2*x^2 + d^2)

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Sympy [C]  time = 23.338, size = 476, normalized size = 4.21 \begin{align*} d^{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 ) + d^{2} 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}\: \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 ) - d 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 ) - e^{3} \left (\begin{cases} - \frac{i d^{4} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{8 e^{3}} + \frac{i d^{3} x}{8 e^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{3 i d x^{3}}{8 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{5}}{4 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^{4} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{8 e^{3}} - \frac{d^{3} x}{8 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{3 d x^{3}}{8 \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} - \frac{e^{2} x^{5}}{4 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*x+d)*(-e**2*x**2+d**2)**(3/2)/x,x)

[Out]

d**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)) + d**2*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)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e) +
d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) - d*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)) - e**3*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*
x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2
*x**2)/Abs(d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqr
t(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.31211, size = 134, normalized size = 1.19 \begin{align*} \frac{3}{8} \, d^{4} \arcsin \left (\frac{x e}{d}\right ) \mathrm{sgn}\left (d\right ) - d^{4} \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{1}{24} \,{\left (32 \, d^{3} +{\left (15 \, d^{2} e - 2 \,{\left (3 \, x e^{3} + 4 \, d e^{2}\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \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,x, algorithm="giac")

[Out]

3/8*d^4*arcsin(x*e/d)*sgn(d) - d^4*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/24*(32*d^
3 + (15*d^2*e - 2*(3*x*e^3 + 4*d*e^2)*x)*x)*sqrt(-x^2*e^2 + d^2)