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

Optimal. Leaf size=120 \[ \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/5*d^2*(-e*x+d)^3/e^4/(-e^2*x^2+d^2)^(5/2)-13/15*d*(-e*x+d)^2/e^4/(-e^2*x^2+d^2)^(3/2)+arctan(e*x/(-e^2*x^2+d
^2)^(1/2))/e^4+32/15*(-e*x+d)/e^4/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {852, 1635, 778, 217, 203} \[ \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(d - e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d - e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d -
e*x))/(15*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 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^3 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^2 \left (-\frac {3 d^3}{e^3}+\frac {5 d^2 x}{e^2}-\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\left (-\frac {17 d^3}{e^3}+\frac {15 d^2 x}{e^2}\right ) (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3}\\ &=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 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 {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 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 = 0.11, size = 73, normalized size = 0.61 \[ \frac {\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2+51 d e x+32 e^2 x^2\right )}{(d+e x)^3}+15 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{15 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(22*d^2 + 51*d*e*x + 32*e^2*x^2))/(d + e*x)^3 + 15*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(1
5*e^4)

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fricas [A]  time = 0.82, size = 157, normalized size = 1.31 \[ \frac {22 \, e^{3} x^{3} + 66 \, d e^{2} x^{2} + 66 \, d^{2} e x + 22 \, d^{3} - 30 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (32 \, e^{2} x^{2} + 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(22*e^3*x^3 + 66*d*e^2*x^2 + 66*d^2*e*x + 22*d^3 - 30*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (32*e^2*x^2 + 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 + 3*d*e^6*x
^2 + 3*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 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(
2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^3-10*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp
(1)^8*exp(2)-4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^2+(-1/2*(-2*d*exp
(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^5-5*exp(1)^4*exp(2)^3+2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^5+2*exp(2)^5-7/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^5/x/e
xp(2)+8*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^2/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^2/(-exp(1)^10+2*
exp(1)^6*exp(2)^2-exp(1)^2*exp(2)^4)+1/2*(-10*exp(1)^4*exp(2)^3+4*exp(2)^5+12*exp(1)^8*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)^1
2-2*exp(1)^8*exp(2)^2+exp(1)^4*exp(2)^4)+sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)^4

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maple [A]  time = 0.01, size = 163, normalized size = 1.36 \[ \frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{3}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2}}{5 \left (x +\frac {d}{e}\right )^{3} e^{7}}-\frac {13 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d}{15 \left (x +\frac {d}{e}\right )^{2} e^{6}}+\frac {32 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{15 \left (x +\frac {d}{e}\right ) e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(e^2)^(1/2)/e^3*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/5*d^2/e^7/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^
2)^(1/2)-13/15*d/e^6/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+32/15/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 = 0.99, size = 136, normalized size = 1.13 \[ \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {13 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{15 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {32 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(-e^2*x^2 + d^2)*d^2/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) - 13/15*sqrt(-e^2*x^2 + d^2)*d/(e
^6*x^2 + 2*d*e^5*x + d^2*e^4) + 32/15*sqrt(-e^2*x^2 + d^2)/(e^5*x + d*e^4) + arcsin(e*x/d)/e^4

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

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)^3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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