3.22 \(\int \frac {x^4 (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=84 \[ \frac {x^4 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {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)^(7/2),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]

Rubi steps

\begin {align*} \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 = 0.03, size = 82, normalized size = 0.98 \[ \frac {8 d^4-8 d^3 e x-12 d^2 e^2 x^2+12 d e^3 x^3+3 e^4 x^4}{15 d e^5 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(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)*Sqrt[d^2 - e^2
*x^2])

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fricas [B]  time = 0.92, size = 171, normalized size = 2.04 \[ \frac {8 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x - 8 \, d^{5} - {\left (3 \, e^{4} x^{4} + 12 \, d e^{3} x^{3} - 12 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{10} x^{5} - d^{2} e^{9} x^{4} - 2 \, d^{3} e^{8} x^{3} + 2 \, d^{4} e^{7} x^{2} + d^{5} e^{6} x - d^{6} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-e^2*x^2+d^2)^(7/2),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 [A]  time = 0.27, size = 64, normalized size = 0.76 \[ -\frac {{\left (8 \, d^{4} e^{\left (-5\right )} + {\left (3 \, x^{2} {\left (\frac {x}{d} + 5 \, e^{\left (-1\right )}\right )} - 20 \, d^{2} e^{\left (-3\right )}\right )} x^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*(8*d^4*e^(-5) + (3*x^2*(x/d + 5*e^(-1)) - 20*d^2*e^(-3))*x^2)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3

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maple [A]  time = 0.01, size = 77, normalized size = 0.92 \[ \frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (3 x^{4} e^{4}+12 x^{3} d \,e^{3}-12 d^{2} x^{2} e^{2}-8 d^{3} x e +8 d^{4}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d \,e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/15*(-e*x+d)*(e*x+d)^2*(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)^(7/2)

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maxima [B]  time = 0.45, size = 159, normalized size = 1.89 \[ \frac {x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {4 \, d^{2} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {3 \, d^{3} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {d x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} + \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

x^4/((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 4/3*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*
e^3) - 3/10*d^3*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8/15*d^4/((-e^2*x^2 + d^2)^(5/2)*e^5) + 1/10*d*x/((-e^2*x^2 +
d^2)^(3/2)*e^4) + 1/5*x/(sqrt(-e^2*x^2 + d^2)*d*e^4)

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mupad [B]  time = 2.70, size = 78, normalized size = 0.93 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (8\,d^4-8\,d^3\,e\,x-12\,d^2\,e^2\,x^2+12\,d\,e^3\,x^3+3\,e^4\,x^4\right )}{15\,d\,e^5\,{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x))/(d^2 - e^2*x^2)^(7/2),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)^2*
(d - e*x)^3)

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sympy [C]  time = 63.87, size = 418, normalized size = 4.98 \[ d \left (\begin {cases} - \frac {i x^{5}}{5 d^{7} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 10 d^{5} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 5 d^{3} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {x^{5}}{5 d^{7} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 10 d^{5} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 5 d^{3} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} \frac {8 d^{4}}{15 d^{4} e^{6} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{8} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{10} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} - \frac {20 d^{2} e^{2} x^{2}}{15 d^{4} e^{6} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{8} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{10} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} + \frac {15 e^{4} x^{4}}{15 d^{4} e^{6} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{8} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{10} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \left (d^{2}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((-I*x**5/(5*d**7*sqrt(-1 + e**2*x**2/d**2) - 10*d**5*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 5*d**3*
e**4*x**4*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (x**5/(5*d**7*sqrt(1 - e**2*x**2/d**2) - 10*d*
*5*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 5*d**3*e**4*x**4*sqrt(1 - e**2*x**2/d**2)), True)) + e*Piecewise((8*d*
*4/(15*d**4*e**6*sqrt(d**2 - e**2*x**2) - 30*d**2*e**8*x**2*sqrt(d**2 - e**2*x**2) + 15*e**10*x**4*sqrt(d**2 -
e**2*x**2)) - 20*d**2*e**2*x**2/(15*d**4*e**6*sqrt(d**2 - e**2*x**2) - 30*d**2*e**8*x**2*sqrt(d**2 - e**2*x**
2) + 15*e**10*x**4*sqrt(d**2 - e**2*x**2)) + 15*e**4*x**4/(15*d**4*e**6*sqrt(d**2 - e**2*x**2) - 30*d**2*e**8*
x**2*sqrt(d**2 - e**2*x**2) + 15*e**10*x**4*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**6/(6*(d**2)**(7/2)), True)
)

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