\(\int \frac {x^5 (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\) [21]

Optimal result

Integrand size = 25, antiderivative size = 122 \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {833, 792, 223, 209} \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}+\frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

Int[(x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(x^4*(d + e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (x^2*(4*d + 5*e*x))/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (8*d + 15
*e*x)/(15*e^6*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^6

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 792

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 833

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])

Rubi steps \begin{align*} \text {integral}& = \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x^3 \left (4 d^3+5 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2} \\ & = \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {x \left (8 d^5+15 d^4 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4 e^4} \\ & = \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^5} \\ & = \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \\ & = \frac {x^4 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d+5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.94 \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4+7 d^3 e x-27 d^2 e^2 x^2-8 d e^3 x^3+23 e^4 x^4\right )}{(d-e x)^3 (d+e x)^2}+30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^6} \]

[In]

Integrate[(x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(8*d^4 + 7*d^3*e*x - 27*d^2*e^2*x^2 - 8*d*e^3*x^3 + 23*e^4*x^4))/((d - e*x)^3*(d + e*x)^
2) + 30*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(15*e^6)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.55

method	result	size
default	\(e \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+d \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )\)	\(189\)

[In]

int(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (108) = 216\).

Time = 0.30 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.02 \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {8 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x - 8 \, d^{5} + 30 \, {\left (e^{5} x^{5} - d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{3} e^{2} x^{2} + d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (23 \, e^{4} x^{4} - 8 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} + 7 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{11} x^{5} - d e^{10} x^{4} - 2 \, d^{2} e^{9} x^{3} + 2 \, d^{3} e^{8} x^{2} + d^{4} e^{7} x - d^{5} e^{6}\right )}} \]

[In]

integrate(x^5*(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 + 30*(e^5*x^5 - d*e^4*x^4
- 2*d^2*e^3*x^3 + 2*d^3*e^2*x^2 + d^4*e*x - d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (23*e^4*x^4 - 8*d
*e^3*x^3 - 27*d^2*e^2*x^2 + 7*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(e^11*x^5 - d*e^10*x^4 - 2*d^2*e^9*x^3 +
2*d^3*e^8*x^2 + d^4*e^7*x - d^5*e^6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (104) = 208\).

Time = 10.20 (sec) , antiderivative size = 1739, normalized size of antiderivative = 14.25 \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate(x**5*(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*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*sqr
t(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)) + e*Piecewise((30*I*d**5*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/(30*d**5*e**7*sqrt(-1 + e**
2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 15*p
i*d**5*sqrt(-1 + e**2*x**2/d**2)/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x*
*2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 30*I*d**4*e*x/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2)
- 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 60*I*d**3*e**2*x*
*2*sqrt(-1 + e**2*x**2/d**2)*acosh(e*x/d)/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1
+ e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 30*pi*d**3*e**2*x**2*sqrt(-1 + e**2*x**2/d**2
)/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt
(-1 + e**2*x**2/d**2)) + 70*I*d**2*e**3*x**3/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(
-1 + e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) + 30*I*d*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)
*acosh(e*x/d)/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**
11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 15*pi*d*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(30*d**5*e**7*sqrt(-1 + e**2*
x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**11*x**4*sqrt(-1 + e**2*x**2/d**2)) - 46*I*e
**5*x**5/(30*d**5*e**7*sqrt(-1 + e**2*x**2/d**2) - 60*d**3*e**9*x**2*sqrt(-1 + e**2*x**2/d**2) + 30*d*e**11*x*
*4*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (-15*d**5*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d*
*5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*
x**2/d**2)) + 15*d**4*e*x/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2)
+ 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) + 30*d**3*e**2*x**2*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d**5*
e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**
2/d**2)) - 35*d**2*e**3*x**3/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**
2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)) - 15*d*e**4*x**4*sqrt(1 - e**2*x**2/d**2)*asin(e*x/d)/(15*d**5*
e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d*e**11*x**4*sqrt(1 - e**2*x**
2/d**2)) + 23*e**5*x**5/(15*d**5*e**7*sqrt(1 - e**2*x**2/d**2) - 30*d**3*e**9*x**2*sqrt(1 - e**2*x**2/d**2) +
15*d*e**11*x**4*sqrt(1 - e**2*x**2/d**2)), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (108) = 216\).

Time = 0.29 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.15 \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{15} \, e x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )}}{3 \, e} + \frac {d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {4 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}} + \frac {4 \, d^{2} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {7 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{5}} \]

[In]

integrate(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 + d
^2)^(5/2)*e^6)) - 1/3*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4))/e + d*x^4/((
-e^2*x^2 + d^2)^(5/2)*e^2) - 4/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^6)
+ 4/15*d^2*x/((-e^2*x^2 + d^2)^(3/2)*e^5) - 7/15*x/(sqrt(-e^2*x^2 + d^2)*e^5) - arcsin(e^2*x/(d*sqrt(e^2)))/(s
qrt(e^2)*e^5)

Giac [F]

\[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )} x^{5}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(x^5*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)*x^5/(-e^2*x^2 + d^2)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^5\,\left (d+e\,x\right )}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[In]

int((x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^5*(d + e*x))/(d^2 - e^2*x^2)^(7/2), x)