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

Optimal. Leaf size=130 \[ \frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[Out]

20/3*(-e^2*x^2+d^2)^(3/2)/e^2+8*(-e^2*x^2+d^2)^(5/2)/e^2/(e*x+d)^2+(-e^2*x^2+d^2)^(7/2)/e^2/(e*x+d)^4+10*d^3*a
rctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2+10*d*x*(-e^2*x^2+d^2)^(1/2)/e

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {793, 663, 665, 195, 217, 203} \[ \frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*d*x*Sqrt[d^2 - e^2*x^2])/e + (20*(d^2 - e^2*x^2)^(3/2))/(3*e^2) + (8*(d^2 - e^2*x^2)^(5/2))/(e^2*(d + e*x)
^2) + (d^2 - e^2*x^2)^(7/2)/(e^2*(d + e*x)^4) + (10*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 203

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

Rule 217

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 663

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {4 \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx}{e}\\ &=\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {20 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx}{e}\\ &=\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {(20 d) \int \sqrt {d^2-e^2 x^2} \, dx}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {\left (10 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {\left (10 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 83, normalized size = 0.64 \[ \frac {1}{3} \sqrt {d^2-e^2 x^2} \left (\frac {24 d^3}{e^2 (d+e x)}+\frac {23 d^2}{e^2}-\frac {6 d x}{e}+x^2\right )+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*((23*d^2)/e^2 - (6*d*x)/e + x^2 + (24*d^3)/(e^2*(d + e*x))))/3 + (10*d^3*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/e^2

________________________________________________________________________________________

fricas [A]  time = 0.99, size = 111, normalized size = 0.85 \[ \frac {47 \, d^{3} e x + 47 \, d^{4} - 60 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (e^{3} x^{3} - 5 \, d e^{2} x^{2} + 17 \, d^{2} e x + 47 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{3} x + d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/3*(47*d^3*e*x + 47*d^4 - 60*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (e^3*x^3 - 5*d*e^2*x
^2 + 17*d^2*e*x + 47*d^3)*sqrt(-e^2*x^2 + d^2))/(e^3*x + d*e^2)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (6*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^2+144*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^3*exp(1)^12*exp(2)^2+108*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^10*exp
(2)^3+18*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^4+6*d^3*(-1/2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^2-178*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^10*exp(2)^3-213*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/e
xp(2))^4*exp(1)^8*exp(2)^4-57*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(2
)^5+324*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^3-288*d^3*(-1/2*(-2
*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^4-432*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x
^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp(2)^5-108*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/e
xp(2))^5*exp(1)^4*exp(2)^6-336*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(
2)^4-516*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^5-258*d^3*(-1/2*(-2
*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^4*exp(2)^6-48*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^8-972*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*e
xp(1)^6*exp(2)^5-756*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^4*exp(2)^6+d^3*e
xp(1)^8*exp(2)^4-186*d^3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)^8-600*d^3*(-1/2*
(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^6+36*d^3*exp(1)^6*exp(2)^5-360*d^3*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8-46*d^3*exp(1)^4*exp(2)^6-372*d^3*(-1/2*(-2
*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8-186*d^3*exp(2)^8+4*d^3*(-1/2*(-2*d*exp(1)-2*sqrt
(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^14*exp(2)+156*d^3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(
2)^8/x/exp(2)+324*d^3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/exp(2)+219/2*d^3*(-2*d*e
xp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)-99*d^3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp
(1))*exp(1)^8*exp(2)^4/x/exp(2)-3*d^3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*exp(2)^3/x/exp(2))
/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp
(1))/x+exp(2))^3/(3*exp(1)^12+9*exp(1)^8*exp(2)^2+3*exp(1)^6*exp(2)^3+9*exp(1)^10*exp(2))+1/2*(12*d^3*exp(1)^1
0*exp(2)^2-86*d^3*exp(1)^8*exp(2)^3-136*d^3*exp(1)^6*exp(2)^4+64*d^3*exp(1)^4*exp(2)^5+272*d^3*exp(2)^7)*atan(
(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)
/(exp(1)^14+3*exp(1)^10*exp(2)^2+exp(1)^8*exp(2)^3+3*exp(1)^12*exp(2))+10*d^3*sign(d)*asin(x*exp(2)/d/exp(1))/
exp(1)^2+2*((2*exp(1)^4*1/12/exp(1)^4*x-12*exp(1)^3*d*1/12/exp(1)^4)*x+46*exp(1)^2*d^2*1/12/exp(1)^4)*sqrt(d^2
-x^2*exp(2))

________________________________________________________________________________________

maple [B]  time = 0.01, size = 290, normalized size = 2.23 \[ \frac {10 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e}+\frac {10 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d x}{e}+\frac {20 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{3 d e}+\frac {16 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{3 d^{2} e^{2}}+\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{3} d \,e^{5}}+\frac {16 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d^{2} e^{4}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{4} e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^6/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+4/d/e^5/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+16/3
/d^2/e^4/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+16/3/d^2/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+20/3/d
/e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+10*d/e*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+10*d^3/e/(e^2)^(1/2)*arc
tan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 235, normalized size = 1.81 \[ -\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{2 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{2 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{3} x + d e^{2}} + \frac {10 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{3 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{6 \, {\left (e^{3} x + d e^{2}\right )}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{2 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/2*(-e^2*x^2 + d^2)^(5/2)*d/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2) - 5/2*(-e^2*x^2 + d^2)^(3/2)*d^2
/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) + 15*sqrt(-e^2*x^2 + d^2)*d^3/(e^3*x + d*e^2) + 10*d^3*arcsin(e*x/d)/e^2 + 1/
3*(-e^2*x^2 + d^2)^(5/2)/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) + 5/6*(-e^2*x^2 + d^2)^(3/2)*d/(e^3*x + d*e^2) + 5/2*
sqrt(-e^2*x^2 + d^2)*d^2/e^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________