\(\int \frac {x^5}{(d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [127]

Optimal result

Integrand size = 27, antiderivative size = 128 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {5 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6} \]

[Out]

1/3*x^4*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(3/2)-5/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^6-1/3*x^2*(-5*e*x+4*d)/e^
4/(-e^2*x^2+d^2)^(1/2)-1/6*(-15*e*x+16*d)*(-e^2*x^2+d^2)^(1/2)/e^6

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 128, 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, 794, 223, 209} \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {5 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6}+\frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[x^5/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

(x^4*(d - e*x))/(3*e^2*(d^2 - e^2*x^2)^(3/2)) - (x^2*(4*d - 5*e*x))/(3*e^4*Sqrt[d^2 - e^2*x^2]) - ((16*d - 15*
e*x)*Sqrt[d^2 - e^2*x^2])/(6*e^6) - (5*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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 )^{5/2}} \, dx \\ & = \frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {x^3 \left (4 d^3-5 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d^2 e^2} \\ & = \frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {x \left (8 d^5-15 d^4 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 d^4 e^4} \\ & = \frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {\left (5 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^5} \\ & = \frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \\ & = \frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {5 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (16 d^4+d^3 e x-23 d^2 e^2 x^2-3 d e^3 x^3+3 e^4 x^4\right )}{6 e^6 (-d+e x) (d+e x)^2}+\frac {5 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^6} \]

[In]

Integrate[x^5/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(16*d^4 + d^3*e*x - 23*d^2*e^2*x^2 - 3*d*e^3*x^3 + 3*e^4*x^4))/(6*e^6*(-d + e*x)*(d + e*x
)^2) + (5*d^2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^6

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.58

method	result	size
risch	\(-\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{6}}-\frac {5 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{5} \sqrt {e^{2}}}+\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{6 e^{8} \left (x +\frac {d}{e}\right )^{2}}-\frac {25 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{12 e^{7} \left (x +\frac {d}{e}\right )}-\frac {d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{4 e^{7} \left (x -\frac {d}{e}\right )}\)	\(202\)
default	\(\frac {-\frac {x^{3}}{2 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 d^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{2 e^{2}}}{e}+\frac {d^{2} x}{e^{5} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {d^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{e^{3}}-\frac {d \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2}}-\frac {d^{3}}{e^{6} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d^{5} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{6}}\)	\(350\)

[In]

int(x^5/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(-e*x+2*d)/e^6*(-e^2*x^2+d^2)^(1/2)-5/2*d^2/e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+1/
6*d^3/e^8/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-25/12*d^2/e^7/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^
(1/2)-1/4*d^2/e^7/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.48 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} - 16 \, d^{4} e x - 16 \, d^{5} - 30 \, {\left (d^{2} e^{3} x^{3} + d^{3} e^{2} x^{2} - d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (3 \, e^{4} x^{4} - 3 \, d e^{3} x^{3} - 23 \, d^{2} e^{2} x^{2} + d^{3} e x + 16 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{9} x^{3} + d e^{8} x^{2} - d^{2} e^{7} x - d^{3} e^{6}\right )}} \]

[In]

integrate(x^5/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")

[Out]

-1/6*(16*d^2*e^3*x^3 + 16*d^3*e^2*x^2 - 16*d^4*e*x - 16*d^5 - 30*(d^2*e^3*x^3 + d^3*e^2*x^2 - d^4*e*x - d^5)*a
rctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (3*e^4*x^4 - 3*d*e^3*x^3 - 23*d^2*e^2*x^2 + d^3*e*x + 16*d^4)*sqrt(
-e^2*x^2 + d^2))/(e^9*x^3 + d*e^8*x^2 - d^2*e^7*x - d^3*e^6)

Sympy [F]

\[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate(x**5/(e*x+d)/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(x**5/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.18 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^{4}}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{7} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{6}\right )}} - \frac {x^{3}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} + \frac {d x^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{4}} + \frac {17 \, d^{2} x}{6 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {5 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{6}} - \frac {3 \, d^{3}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{6}} \]

[In]

integrate(x^5/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")

[Out]

1/3*d^4/(sqrt(-e^2*x^2 + d^2)*e^7*x + sqrt(-e^2*x^2 + d^2)*d*e^6) - 1/2*x^3/(sqrt(-e^2*x^2 + d^2)*e^3) + d*x^2
/(sqrt(-e^2*x^2 + d^2)*e^4) + 17/6*d^2*x/(sqrt(-e^2*x^2 + d^2)*e^5) - 5/2*d^2*arcsin(e*x/d)/e^6 - 3*d^3/(sqrt(
-e^2*x^2 + d^2)*e^6)

Giac [F]

\[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{5}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}} \,d x } \]

[In]

integrate(x^5/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^5/((-e^2*x^2 + d^2)^(3/2)*(e*x + d)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]

[In]

int(x^5/((d^2 - e^2*x^2)^(3/2)*(d + e*x)),x)

[Out]

int(x^5/((d^2 - e^2*x^2)^(3/2)*(d + e*x)), x)