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]

(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

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Rubi [A]  time = 0.261021, antiderivative size = 120, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.116699, 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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Maple [A]  time = 0.065, size = 163, normalized size = 1.4 \begin{align*}{\frac{1}{{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{32}{15\,{e}^{5}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{13\,d}{15\,{e}^{6}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-2}}+{\frac{{d}^{2}}{5\,{e}^{7}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-3}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.57298, size = 336, normalized size = 2.8 \begin{align*} \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 )}} \end{align*}

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

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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