3.2.29 \(\int \frac {x^3}{(d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [129]

Optimal. Leaf size=89 \[ \frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {864, 833, 792, 223, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 833

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 + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 864

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + c*(x/e))*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 102, normalized size = 1.15 \begin {gather*} \frac {\left (2 d^2-d e x-4 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{3 e^4 (-d+e x) (d+e x)^2}+\frac {\log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3 \sqrt {-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(203\) vs. \(2(79)=158\).
time = 0.07, size = 204, normalized size = 2.29

method	result	size
default	\(\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e}-\frac {d}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {x}{e^{3} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d^{3} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{4}}\)	\(204\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 91, normalized size = 1.02 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} + \frac {4 \, x e^{\left (-3\right )}}{3 \, \sqrt {-x^{2} e^{2} + d^{2}}} - \frac {d e^{\left (-4\right )}}{\sqrt {-x^{2} e^{2} + d^{2}}} + \frac {d^{2}}{3 \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x e^{5} + \sqrt {-x^{2} e^{2} + d^{2}} d e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(x*e/d)*e^(-4) + 4/3*x*e^(-3)/sqrt(-x^2*e^2 + d^2) - d*e^(-4)/sqrt(-x^2*e^2 + d^2) + 1/3*d^2/(sqrt(-x^2
*e^2 + d^2)*x*e^5 + sqrt(-x^2*e^2 + d^2)*d*e^4)

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Fricas [A]
time = 1.63, size = 148, normalized size = 1.66 \begin {gather*} -\frac {2 \, x^{3} e^{3} + 2 \, d x^{2} e^{2} - 2 \, d^{2} x e - 2 \, d^{3} - 6 \, {\left (x^{3} e^{3} + d x^{2} e^{2} - d^{2} x e - d^{3}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (4 \, x^{2} e^{2} + d x e - 2 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x^{3} e^{7} + d x^{2} e^{6} - d^{2} x e^{5} - d^{3} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*x^3*e^3 + 2*d*x^2*e^2 - 2*d^2*x*e - 2*d^3 - 6*(x^3*e^3 + d*x^2*e^2 - d^2*x*e - d^3)*arctan(-(d - sqrt(
-x^2*e^2 + d^2))*e^(-1)/x) + (4*x^2*e^2 + d*x*e - 2*d^2)*sqrt(-x^2*e^2 + d^2))/(x^3*e^7 + d*x^2*e^6 - d^2*x*e^
5 - d^3*e^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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