\(\int \frac {x^5}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\) [138]

Optimal result

Integrand size = 27, antiderivative size = 122 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \]

[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 = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {864, 833, 792, 223, 209} \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}+\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

Rule 864

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*} \text {integral}& = \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 {\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 = 0.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4-7 d^3 e x-27 d^2 e^2 x^2+8 d e^3 x^3+23 e^4 x^4\right )}{(d-e x)^2 (d+e x)^3}-30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^6} \]

[In]

Integrate[x^5/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),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)^2*(d + e*x)^
3) - 30*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(15*e^6)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(108)=216\).

Time = 0.39 (sec) , antiderivative size = 450, normalized size of antiderivative = 3.69

method	result	size
default	\(\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e}+\frac {d^{4} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{5}}+\frac {d^{2} \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )}{e^{3}}-\frac {d \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}-\frac {d^{3}}{3 e^{6} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{5} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{6}}\)	\(450\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (109) = 218\).

Time = 0.34 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.98 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {8 \, e^{5} x^{5} + 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} - 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x + 8 \, d^{5} - 30 \, {\left (e^{5} x^{5} + d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} - 2 \, d^{3} e^{2} x^{2} + d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (23 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} - 7 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{11} x^{5} + d e^{10} x^{4} - 2 \, d^{2} e^{9} x^{3} - 2 \, d^{3} e^{8} x^{2} + d^{4} e^{7} x + d^{5} e^{6}\right )}} \]

[In]

integrate(x^5/(e*x+d)/(-e^2*x^2+d^2)^(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 - 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 [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (109) = 218\).

Time = 0.32 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.92 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {d^{4}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{6}\right )}} + \frac {x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {8 \, d x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {4 \, d^{2} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {x^{2}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}} + \frac {2 \, d^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6}} - \frac {8 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{6}} - \frac {4 \, d}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{3 \, d e^{6}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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