3.190 \(\int \frac{x^3 \sqrt{d^2-e^2 x^2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=148 \[ \frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac{14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac{8 d \sqrt{d^2-e^2 x^2}}{e^4 (d+e x)}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac{4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

(8*d*Sqrt[d^2 - e^2*x^2])/(e^4*(d + e*x)) + (d^2*(d^2 - e^2*x^2)^(3/2))/(5*e^4*(d + e*x)^4) - (14*d*(d^2 - e^2
*x^2)^(3/2))/(15*e^4*(d + e*x)^3) - (d^2 - e^2*x^2)^(3/2)/(e^4*(d + e*x)^2) + (4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2
*x^2]])/e^4

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Rubi [A]  time = 0.246061, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {1639, 1637, 659, 651, 663, 217, 203} \[ \frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac{14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac{8 d \sqrt{d^2-e^2 x^2}}{e^4 (d+e x)}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac{4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*d*Sqrt[d^2 - e^2*x^2])/(e^4*(d + e*x)) + (d^2*(d^2 - e^2*x^2)^(3/2))/(5*e^4*(d + e*x)^4) - (14*d*(d^2 - e^2
*x^2)^(3/2))/(15*e^4*(d + e*x)^3) - (d^2 - e^2*x^2)^(3/2)/(e^4*(d + e*x)^2) + (4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2
*x^2]])/e^4

Rule 1639

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

Rule 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 663

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

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 \sqrt{d^2-e^2 x^2}}{(d+e x)^4} \, dx &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac{\int \frac{\sqrt{d^2-e^2 x^2} \left (2 d^3 e^2+5 d^2 e^3 x+4 d e^4 x^2\right )}{(d+e x)^4} \, dx}{e^5}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac{\int \left (\frac{d^3 e^2 \sqrt{d^2-e^2 x^2}}{(d+e x)^4}-\frac{3 d^2 e^2 \sqrt{d^2-e^2 x^2}}{(d+e x)^3}+\frac{4 d e^2 \sqrt{d^2-e^2 x^2}}{(d+e x)^2}\right ) \, dx}{e^5}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac{(4 d) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^3}+\frac{\left (3 d^2\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^3}-\frac{d^3 \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^4} \, dx}{e^3}\\ &=\frac{8 d \sqrt{d^2-e^2 x^2}}{e^4 (d+e x)}+\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac{(4 d) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^3}-\frac{d^2 \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e^3}\\ &=\frac{8 d \sqrt{d^2-e^2 x^2}}{e^4 (d+e x)}+\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac{14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac{(4 d) \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{8 d \sqrt{d^2-e^2 x^2}}{e^4 (d+e x)}+\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac{14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac{4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.144645, size = 85, normalized size = 0.57 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (222 d^2 e x+94 d^3+149 d e^2 x^2+15 e^3 x^3\right )}{(d+e x)^3}+60 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{15 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(94*d^3 + 222*d^2*e*x + 149*d*e^2*x^2 + 15*e^3*x^3))/(d + e*x)^3 + 60*d*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/(15*e^4)

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Maple [A]  time = 0.063, size = 212, normalized size = 1.4 \begin{align*}{\frac{{d}^{2}}{5\,{e}^{8}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}-{\frac{14\,d}{15\,{e}^{7}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}+4\,{\frac{1}{{e}^{4}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+4\,{\frac{d}{{e}^{3}\sqrt{{e}^{2}}}\arctan \left ({\sqrt{{e}^{2}}x{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ) }+3\,{\frac{1}{{e}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2} \left ({\frac{d}{e}}+x \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d^2/e^8/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)-14/15*d/e^7/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x)
)^(3/2)+4/e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+4/e^3*d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*
d*e*(d/e+x))^(1/2))+3/e^6/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)

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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^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78005, size = 377, normalized size = 2.55 \begin{align*} \frac{94 \, d e^{3} x^{3} + 282 \, d^{2} e^{2} x^{2} + 282 \, d^{3} e x + 94 \, d^{4} - 120 \,{\left (d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (15 \, e^{3} x^{3} + 149 \, d e^{2} x^{2} + 222 \, d^{2} e x + 94 \, d^{3}\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^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/15*(94*d*e^3*x^3 + 282*d^2*e^2*x^2 + 282*d^3*e*x + 94*d^4 - 120*(d*e^3*x^3 + 3*d^2*e^2*x^2 + 3*d^3*e*x + d^4
)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^3*x^3 + 149*d*e^2*x^2 + 222*d^2*e*x + 94*d^3)*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 )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(x**3*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, 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^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError