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

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

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Rubi [A]  time = 0.07, antiderivative size = 85, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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 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]))

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*}

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Mathematica [A]  time = 0.09, size = 82, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2} \left (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\right )}{15 d e^5 (d-e x)^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.45, size = 82, normalized size = 0.96 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-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\right )}{15 d e^5 (d-e x)^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [B]  time = 0.41, size = 168, normalized size = 1.98 \begin {gather*} -\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 {gather*}

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 70, normalized size = 0.82 \begin {gather*} -\frac {\left (-e x +d \right ) \left (3 x^{4} e^{4}-12 x^{3} d \,e^{3}-12 d^{2} x^{2} e^{2}+8 d^{3} x e +8 d^{4}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,e^{5}} \end {gather*}

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 [A]  time = 0.50, size = 134, normalized size = 1.58 \begin {gather*} -\frac {d^{3}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{5}\right )}} + \frac {x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {2 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {d^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} + \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}} \end {gather*}

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]

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

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mupad [B]  time = 2.95, size = 78, normalized size = 0.92 \begin {gather*} -\frac {\sqrt {d^2-e^2\,x^2}\,\left (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\right )}{15\,d\,e^5\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)

[Out]

-((d^2 - e^2*x^2)^(1/2)*(8*d^4 + 3*e^4*x^4 - 12*d*e^3*x^3 - 12*d^2*e^2*x^2 + 8*d^3*e*x))/(15*d*e^5*(d + e*x)^3
*(d - e*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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