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

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 = 0.07, 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 = {850, 819, 778, 217, 203} \[ \frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

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 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 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 778

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 819

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 850

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 {\operatorname {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 = 0.12, size = 80, normalized size = 0.90 \[ \frac {\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^2+d e x+4 e^2 x^2\right )}{(d-e x) (d+e x)^2}-3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{3 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 157, normalized size = 1.76 \[ -\frac {2 \, e^{3} x^{3} + 2 \, d e^{2} x^{2} - 2 \, d^{2} e x - 2 \, d^{3} - 6 \, {\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (4 \, e^{2} x^{2} + d e x - 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{7} x^{3} + d e^{6} x^{2} - d^{2} e^{5} x - d^{3} e^{4}\right )}} \]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 153, normalized size = 1.72 \[ \frac {2 x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}-\frac {2 x}{3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{3}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{3}}+\frac {d^{2}}{3 \left (x +\frac {d}{e}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{5}}-\frac {d}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}} \]

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)

[Out]

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

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maxima [A]  time = 1.00, size = 99, normalized size = 1.11 \[ \frac {d^{2}}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{5} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}\right )}} + \frac {4 \, x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} - \frac {d}{\sqrt {-e^{2} x^{2} + d^{2}} e^{4}} \]

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]

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

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

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

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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