\(\int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [92]

Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {3 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5} \]

[Out]

3/8*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5+4/15*d^2*x^2*(-e^2*x^2+d^2)^(1/2)/e^3-1/4*d*x^3*(-e^2*x^2+d^2)^(1
/2)/e^2+1/5*x^4*(-e^2*x^2+d^2)^(1/2)/e+1/120*d^3*(-45*e*x+64*d)*(-e^2*x^2+d^2)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {864, 847, 794, 223, 209} \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {3 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5} \]

[In]

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

[Out]

(4*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (d*x^3*Sqrt[d^2 - e^2*x^2])/(4*e^2) + (x^4*Sqrt[d^2 - e^2*x^2])/(5*
e) + (d^3*(64*d - 45*e*x)*Sqrt[d^2 - e^2*x^2])/(120*e^5) + (3*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*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 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)}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x^3 \left (4 d^2 e-5 d e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{5 e^2} \\ & = -\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {\int \frac {x^2 \left (15 d^3 e^2-16 d^2 e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{20 e^4} \\ & = \frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x \left (32 d^4 e^3-45 d^3 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{60 e^6} \\ & = \frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^4} \\ & = \frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4} \\ & = \frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.70 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (64 d^4-45 d^3 e x+32 d^2 e^2 x^2-30 d e^3 x^3+24 e^4 x^4\right )-90 d^5 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{120 e^5} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(64*d^4 - 45*d^3*e*x + 32*d^2*e^2*x^2 - 30*d*e^3*x^3 + 24*e^4*x^4) - 90*d^5*ArcTan[(e*x)/
(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(120*e^5)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {\left (24 e^{4} x^{4}-30 d \,e^{3} x^{3}+32 d^{2} e^{2} x^{2}-45 d^{3} e x +64 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{120 e^{5}}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{4} \sqrt {e^{2}}}\)	\(97\)
default	\(\frac {-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 e^{4}}}{e}-\frac {d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{5}}-\frac {d^{3} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{4}}-\frac {d \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )}{e^{2}}+\frac {d^{4} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{5}}\)	\(296\)

[In]

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

[Out]

1/120*(24*e^4*x^4-30*d*e^3*x^3+32*d^2*e^2*x^2-45*d^3*e*x+64*d^4)/e^5*(-e^2*x^2+d^2)^(1/2)+3/8*d^5/e^4/(e^2)^(1
/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {90 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 32 \, d^{2} e^{2} x^{2} - 45 \, d^{3} e x + 64 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{5}} \]

[In]

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

[Out]

-1/120*(90*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*e^4*x^4 - 30*d*e^3*x^3 + 32*d^2*e^2*x^2 - 45*d^
3*e*x + 64*d^4)*sqrt(-e^2*x^2 + d^2))/e^5

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{5}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{5}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{4}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{15 \, e^{5}} \]

[In]

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

[Out]

3/8*d^5*arcsin(e*x/d)/e^5 - 5/8*sqrt(-e^2*x^2 + d^2)*d^3*x/e^4 - 1/5*(-e^2*x^2 + d^2)^(3/2)*x^2/e^3 + sqrt(-e^
2*x^2 + d^2)*d^4/e^5 + 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e^4 - 7/15*(-e^2*x^2 + d^2)^(3/2)*d^2/e^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{4} {\left | e \right |}} + \frac {1}{120} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, x {\left (\frac {4 \, x}{e} - \frac {5 \, d}{e^{2}}\right )} + \frac {16 \, d^{2}}{e^{3}}\right )} x - \frac {45 \, d^{3}}{e^{4}}\right )} x + \frac {64 \, d^{4}}{e^{5}}\right )} \]

[In]

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

[Out]

3/8*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^4*abs(e)) + 1/120*sqrt(-e^2*x^2 + d^2)*((2*(3*x*(4*x/e - 5*d/e^2) + 16*
d^2/e^3)*x - 45*d^3/e^4)*x + 64*d^4/e^5)

Mupad [F(-1)]

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

[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)