3.140 \(\int \frac{x^3}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=91 \[ \frac{x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{x}{5 d^2 e^3 \sqrt{d^2-e^2 x^2}}-\frac{2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

(x^2*(d - e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*d - 3*e*x)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) - x/(5*d^2*e^3*Sq
rt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.0694935, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {850, 819, 778, 191} \[ \frac{x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{x}{5 d^2 e^3 \sqrt{d^2-e^2 x^2}}-\frac{2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(x^2*(d - e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*d - 3*e*x)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) - x/(5*d^2*e^3*Sq
rt[d^2 - e^2*x^2])

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]))

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 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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\int \frac{x^3 (d-e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{x \left (2 d^3-3 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac{x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\int \frac{1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{5 e^3}\\ &=\frac{x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x}{5 d^2 e^3 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0784474, size = 82, normalized size = 0.9 \[ \frac{\sqrt{d^2-e^2 x^2} \left (3 d^2 e^2 x^2-2 d^3 e x-2 d^4+3 d e^3 x^3+3 e^4 x^4\right )}{15 d^2 e^4 (d-e x)^2 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.048, size = 70, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( -3\,{x}^{4}{e}^{4}-3\,{x}^{3}d{e}^{3}-3\,{x}^{2}{d}^{2}{e}^{2}+2\,{d}^{3}xe+2\,{d}^{4} \right ) }{15\,{d}^{2}{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.65836, size = 340, normalized size = 3.74 \begin{align*} -\frac{2 \, e^{5} x^{5} + 2 \, d e^{4} x^{4} - 4 \, d^{2} e^{3} x^{3} - 4 \, d^{3} e^{2} x^{2} + 2 \, d^{4} e x + 2 \, d^{5} -{\left (3 \, e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} - 2 \, d^{3} e x - 2 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{2} e^{9} x^{5} + d^{3} e^{8} x^{4} - 2 \, d^{4} e^{7} x^{3} - 2 \, d^{5} e^{6} x^{2} + d^{6} e^{5} x + d^{7} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(2*e^5*x^5 + 2*d*e^4*x^4 - 4*d^2*e^3*x^3 - 4*d^3*e^2*x^2 + 2*d^4*e*x + 2*d^5 - (3*e^4*x^4 + 3*d*e^3*x^3
+ 3*d^2*e^2*x^2 - 2*d^3*e*x - 2*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2*e^9*x^5 + d^3*e^8*x^4 - 2*d^4*e^7*x^3 - 2*d^5*
e^6*x^2 + d^6*e^5*x + d^7*e^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**3/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

[undef, undef, undef, undef, undef, 1]