\(\int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [119]

Optimal result

Integrand size = 27, antiderivative size = 118 \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {3 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \]

[Out]

-3/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5+x^3*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(1/2)-4/3*x^2*(-e^2*x^2+d^2)^(1/
2)/e^3-1/6*d*(-9*e*x+16*d)*(-e^2*x^2+d^2)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {864, 833, 847, 794, 223, 209} \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}+\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3} \]

[In]

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

[Out]

(x^3*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) - (4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^3) - (d*(16*d - 9*e*x)*Sqrt[d^2 -
e^2*x^2])/(6*e^5) - (3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 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 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 )^{3/2}} \, dx \\ & = \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x^2 \left (3 d^3-4 d^2 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2} \\ & = \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}+\frac {\int \frac {x \left (8 d^4 e-9 d^3 e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 d^2 e^4} \\ & = \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^4} \\ & = \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 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 )}{2 e^4} \\ & = \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-7 d^2 e x+d e^2 x^2-2 e^3 x^3\right )}{6 e^5 (d+e x)}+\frac {3 d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 7*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3))/(6*e^5*(d + e*x)) + (3*d^3*ArcTan[(e*x)/(S
qrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^5

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {\left (2 e^{2} x^{2}-3 d e x +10 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{5}}-\frac {3 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{4} \sqrt {e^{2}}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}\)	\(120\)
default	\(\frac {-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}}{e}-\frac {d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {d \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{e^{2}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}\)	\(215\)

[In]

int(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(2*e^2*x^2-3*d*e*x+10*d^2)/e^5*(-e^2*x^2+d^2)^(1/2)-3/2*d^3/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^
2+d^2)^(1/2))-d^3/e^6/(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 = 112, normalized size of antiderivative = 0.95 \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {16 \, d^{3} e x + 16 \, d^{4} - 18 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{3} x^{3} - d e^{2} x^{2} + 7 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{6} x + d e^{5}\right )}} \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(16*d^3*e*x + 16*d^4 - 18*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^3*x^3 - d*e^2*
x^2 + 7*d^2*e*x + 16*d^3)*sqrt(-e^2*x^2 + d^2))/(e^6*x + d*e^5)

Sympy [F]

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

[In]

integrate(x**4/(e*x+d)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{6} x + d e^{5}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e^{3}} - \frac {3 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{5}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{4}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{5}} \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x {\left (\frac {2 \, x}{e^{3}} - \frac {3 \, d}{e^{4}}\right )} + \frac {10 \, d^{2}}{e^{5}}\right )} - \frac {3 \, d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{4} {\left | e \right |}} + \frac {2 \, d^{3}}{e^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

integrate(x^4/(e*x+d)/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^4/((d^2 - e^2*x^2)^(1/2)*(d + e*x)),x)

[Out]

int(x^4/((d^2 - e^2*x^2)^(1/2)*(d + e*x)), x)