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

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 = 0.06, 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 = {850, 819, 780, 217, 203} \[ \frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \]

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 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 780

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 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) \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 ) \operatorname {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 = 0.06, size = 80, normalized size = 0.88 \[ \frac {\sqrt {d^2-e^2 x^2} \left (4 d^2+d e x-e^2 x^2\right )+3 d^2 (d+e x) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 101, normalized size = 1.11 \[ \frac {4 \, d^{2} e x + 4 \, d^{3} - 6 \, {\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (e^{2} x^{2} - d e x - 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: 3/2*d^2*sign(d)*asin(x*exp(2)/d/exp(1))/
exp(1)^4+2*d^2*exp(2)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2
))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^3/exp(1)+2*(-4*exp(1)^7*1/16/exp(1)^10*x+8*exp(1)^6*d*1/16/exp(1)^10)*sqrt(
-exp(2)*x^2+d^2)

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maple [A]  time = 0.01, size = 120, normalized size = 1.32 \[ \frac {3 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x}{2 e^{3}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2}}{\left (x +\frac {d}{e}\right ) e^{5}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.00, size = 86, normalized size = 0.95 \[ \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{5} x + d e^{4}} + \frac {3 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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