\(\int \frac {x (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\) [202]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {10 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[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] (verified)

Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {807, 677, 679, 201, 223, 209} \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {10 d x \sqrt {d^2-e^2 x^2}}{e} \]

[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 201

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 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 677

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 679

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 807

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (47 d^3+17 d^2 e x-5 d e^2 x^2+e^3 x^3\right )}{3 e^2 (d+e x)}+\frac {10 d^3 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(47*d^3 + 17*d^2*e*x - 5*d*e^2*x^2 + e^3*x^3))/(3*e^2*(d + e*x)) + (10*d^3*Sqrt[-e^2]*Log
[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^3

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.92

method result size
risch \(\frac {\left (e^{2} x^{2}-6 d e x +23 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}+\frac {10 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}+\frac {8 d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{3} \left (x +\frac {d}{e}\right )}\) \(119\)
default \(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 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}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}}{e^{4}}-\frac {d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}-\frac {3 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 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}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}\right )}{d}\right )}{e^{5}}\) \(644\)

[In]

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

[Out]

1/3*(e^2*x^2-6*d*e*x+23*d^2)/e^2*(-e^2*x^2+d^2)^(1/2)+10*d^3/e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)
^(1/2))+8*d^3/e^3/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.85 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {47 \, d^{3} e x + 47 \, d^{4} - 60 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (e^{3} x^{3} - 5 \, d e^{2} x^{2} + 17 \, d^{2} e x + 47 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{3} x + d e^{2}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.81 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{2 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{2 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{3} x + d e^{2}} + \frac {10 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{3 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{6 \, {\left (e^{3} x + d e^{2}\right )}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{2 \, e^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {10 \, d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e {\left | e \right |}} + \frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (x - \frac {6 \, d}{e}\right )} x + \frac {23 \, d^{2}}{e^{2}}\right )} - \frac {16 \, d^{3}}{e {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

10*d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/(e*abs(e)) + 1/3*sqrt(-e^2*x^2 + d^2)*((x - 6*d/e)*x + 23*d^2/e^2) - 16*d^3
/(e*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]

[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)