3.2.20 \(\int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [120]

Optimal. Leaf size=91 \[ \frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 91, 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, 794, 223, 209} \begin {gather*} \frac {3 d^2 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}+\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) + ((4*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*e^4) + (3*d^2*ArcTan[(e*x)/
Sqrt[d^2 - e^2*x^2]])/(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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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) \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^3 (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x \left (2 d^3-3 d^2 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {\left (3 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^3}\\ &=\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {\left (3 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}\\ &=\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 95, normalized size = 1.04 \begin {gather*} \frac {\frac {e \sqrt {d^2-e^2 x^2} \left (4 d^2+d e x-e^2 x^2\right )}{d+e x}+3 d^2 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[d^2 - e^2*x^2]*(4*d^2 + d*e*x - e^2*x^2))/(d + e*x) + 3*d^2*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2
- e^2*x^2]])/(2*e^5)

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Maple [A]
time = 0.09, size = 159, normalized size = 1.75

method	result	size
risch	\(\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{4}}+\frac {3 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{3} \sqrt {e^{2}}}+\frac {d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\)	\(108\)
default	\(\frac {-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}}{e}+\frac {d \sqrt {-e^{2} x^{2}+d^{2}}}{e^{4}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 79, normalized size = 0.87 \begin {gather*} \frac {3}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} x e^{\left (-3\right )} + \sqrt {-x^{2} e^{2} + d^{2}} d e^{\left (-4\right )} + \frac {\sqrt {-x^{2} e^{2} + d^{2}} d^{2}}{x e^{5} + d e^{4}} \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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.24, size = 98, normalized size = 1.08 \begin {gather*} \frac {4 \, d^{2} x e + 4 \, d^{3} - 6 \, {\left (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 (x^{2} e^{2} - d x e - 4 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{2 \, {\left (x e^{5} + 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \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)**(1/2),x)

[Out]

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

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Giac [A]
time = 1.05, size = 81, normalized size = 0.89 \begin {gather*} \frac {3}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} \mathrm {sgn}\left (d\right ) - \frac {2 \, d^{2} e^{\left (-4\right )}}{\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e^{\left (-3\right )} - 2 \, d e^{\left (-4\right )}\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)^(1/2),x, algorithm="giac")

[Out]

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

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

[Out]

int(x^3/((d^2 - e^2*x^2)^(1/2)*(d + e*x)), x)

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