\(\int \frac {x^4}{(d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [128]

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

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x^3 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x (3 d-4 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}+\frac {8 \sqrt {d^2-e^2 x^2}}{3 e^5}+\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 113, 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, 655, 223, 209} \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {x^3 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 \sqrt {d^2-e^2 x^2}}{3 e^5}-\frac {x (3 d-4 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[x^4/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

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

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 655

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (8 d^3+5 d^2 e x-7 d e^2 x^2-3 e^3 x^3\right )}{(d-e x) (d+e x)^2}-6 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{3 e^5} \]

[In]

Integrate[x^4/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(8*d^3 + 5*d^2*e*x - 7*d*e^2*x^2 - 3*e^3*x^3))/((d - e*x)*(d + e*x)^2) - 6*d*ArcTan[(e*x
)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(3*e^5)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.65

method result size
risch \(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}+\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}+\frac {19 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{12 e^{6} \left (x +\frac {d}{e}\right )}-\frac {d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{4 e^{6} \left (x -\frac {d}{e}\right )}-\frac {d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{6 e^{7} \left (x +\frac {d}{e}\right )^{2}}\) \(186\)
default \(\frac {-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{e}+\frac {d^{2}}{e^{5} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d x}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{e^{2}}+\frac {d^{4} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{5}}\) \(257\)

[In]

int(x^4/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(-e^2*x^2+d^2)^(1/2)/e^5+d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+19/12*d/e^6/(x+d/e)*(-(x
+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/4*d/e^6/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)-1/6*d^2/e^7/(x+d/e)^2*
(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.55 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {8 \, d e^{3} x^{3} + 8 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x - 8 \, d^{4} - 6 \, {\left (d e^{3} x^{3} + d^{2} e^{2} x^{2} - d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{3} x^{3} + 7 \, d e^{2} x^{2} - 5 \, d^{2} e x - 8 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{8} x^{3} + d e^{7} x^{2} - d^{2} e^{6} x - d^{3} e^{5}\right )}} \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate(x**4/(e*x+d)/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(x**4/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.10 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d^{3}}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{5}\right )}} - \frac {x^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {4 \, d x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} + \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{5}} + \frac {3 \, d^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{5}} \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}} \,d x } \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^4/((-e^2*x^2 + d^2)^(3/2)*(e*x + d)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]

[In]

int(x^4/((d^2 - e^2*x^2)^(3/2)*(d + e*x)),x)

[Out]

int(x^4/((d^2 - e^2*x^2)^(3/2)*(d + e*x)), x)