3.139 \(\int \frac{x^4}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4}{5 e^5 \sqrt{d^2-e^2 x^2}}+\frac{4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0742674, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {850, 805, 266, 43} \[ -\frac{x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4}{5 e^5 \sqrt{d^2-e^2 x^2}}+\frac{4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

-(x^4*(d - e*x))/(5*d*e*(d^2 - e^2*x^2)^(5/2)) + (4*d^2)/(15*e^5*(d^2 - e^2*x^2)^(3/2)) - 4/(5*e^5*Sqrt[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 805

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0948815, size = 82, normalized size = 0.96 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (-12 d^2 e^2 x^2+8 d^3 e x+8 d^4-12 d e^3 x^3+3 e^4 x^4\right )}{15 d e^5 (d-e x)^2 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

-(Sqrt[d^2 - e^2*x^2]*(8*d^4 + 8*d^3*e*x - 12*d^2*e^2*x^2 - 12*d*e^3*x^3 + 3*e^4*x^4))/(15*d*e^5*(d - e*x)^2*(
d + e*x)^3)

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Maple [A]  time = 0.055, size = 70, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( 3\,{x}^{4}{e}^{4}-12\,{x}^{3}d{e}^{3}-12\,{d}^{2}{x}^{2}{e}^{2}+8\,{d}^{3}xe+8\,{d}^{4} \right ) }{15\,d{e}^{5}} \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^4/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-1/15*(-e*x+d)*(3*e^4*x^4-12*d*e^3*x^3-12*d^2*e^2*x^2+8*d^3*e*x+8*d^4)/d/e^5/(-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^4/(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 = 1.61291, size = 344, normalized size = 4.05 \begin{align*} -\frac{8 \, e^{5} x^{5} + 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} - 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x + 8 \, d^{5} +{\left (3 \, e^{4} x^{4} - 12 \, d e^{3} x^{3} - 12 \, d^{2} e^{2} x^{2} + 8 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d e^{10} x^{5} + d^{2} e^{9} x^{4} - 2 \, d^{3} e^{8} x^{3} - 2 \, d^{4} e^{7} x^{2} + d^{5} e^{6} x + d^{6} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(8*e^5*x^5 + 8*d*e^4*x^4 - 16*d^2*e^3*x^3 - 16*d^3*e^2*x^2 + 8*d^4*e*x + 8*d^5 + (3*e^4*x^4 - 12*d*e^3*x
^3 - 12*d^2*e^2*x^2 + 8*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*e^10*x^5 + d^2*e^9*x^4 - 2*d^3*e^8*x^3 - 2*d
^4*e^7*x^2 + d^5*e^6*x + d^6*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**4/((-(-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^4/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

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