\(\int \frac {x (d^2-e^2 x^2)^{5/2}}{d+e x} \, dx\) [106]

Optimal result

Integrand size = 25, antiderivative size = 116 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}-\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \]

[Out]

-1/24*d^2*x*(-e^2*x^2+d^2)^(3/2)/e-1/30*(-5*e*x+6*d)*(-e^2*x^2+d^2)^(5/2)/e^2-1/16*d^6*arctan(e*x/(-e^2*x^2+d^
2)^(1/2))/e^2-1/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {799, 794, 201, 223, 209} \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}-\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e} \]

[In]

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

[Out]

-1/16*(d^4*x*Sqrt[d^2 - e^2*x^2])/e - (d^2*x*(d^2 - e^2*x^2)^(3/2))/(24*e) - ((6*d - 5*e*x)*(d^2 - e^2*x^2)^(5
/2))/(30*e^2) - (d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*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 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 799

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int x \left (d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{d e} \\ & = -\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^2 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e} \\ & = -\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^4 \int \sqrt {d^2-e^2 x^2} \, dx}{8 e} \\ & = -\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}-\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^6 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e} \\ & = -\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}-\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^6 \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e} \\ & = -\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}-\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}-\frac {d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-48 d^5+15 d^4 e x+96 d^3 e^2 x^2-70 d^2 e^3 x^3-48 d e^4 x^4+40 e^5 x^5\right )+30 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^2} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-48*d^5 + 15*d^4*e*x + 96*d^3*e^2*x^2 - 70*d^2*e^3*x^3 - 48*d*e^4*x^4 + 40*e^5*x^5) + 30
*d^6*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(240*e^2)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {\left (-40 e^{5} x^{5}+48 d \,e^{4} x^{4}+70 d^{2} e^{3} x^{3}-96 d^{3} e^{2} x^{2}-15 d^{4} e x +48 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{2}}-\frac {d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e \sqrt {e^{2}}}\)	\(108\)
default	\(\frac {\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}}{e}-\frac {d \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{e^{2}}\)	\(295\)

[In]

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

[Out]

-1/240*(-40*e^5*x^5+48*d*e^4*x^4+70*d^2*e^3*x^3-96*d^3*e^2*x^2-15*d^4*e*x+48*d^5)/e^2*(-e^2*x^2+d^2)^(1/2)-1/1
6*d^6/e/(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 = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {30 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} - 48 \, d e^{4} x^{4} - 70 \, d^{2} e^{3} x^{3} + 96 \, d^{3} e^{2} x^{2} + 15 \, d^{4} e x - 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{2}} \]

[In]

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

[Out]

1/240*(30*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 - 48*d*e^4*x^4 - 70*d^2*e^3*x^3 + 96*d^3
*e^2*x^2 + 15*d^4*e*x - 48*d^5)*sqrt(-e^2*x^2 + d^2))/e^2

Sympy [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.96 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=d^{3} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2}}{3 e^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{2} \sqrt {d^{2}}}{2} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2} x}{8 e^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{4}} - \frac {d^{2} x^{2}}{15 e^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e^{4}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{4} x}{16 e^{4}} - \frac {d^{2} x^{3}}{24 e^{2}} + \frac {x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{5} \sqrt {d^{2}}}{5} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**3*Piecewise((sqrt(d**2 - e**2*x**2)*(-d**2/(3*e**2) + x**2/3), Ne(e**2, 0)), (x**2*sqrt(d**2)/2, True)) - d
**2*e*Piecewise((d**4*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0
)), (x*log(x)/sqrt(-e**2*x**2), True))/(8*e**2) + sqrt(d**2 - e**2*x**2)*(-d**2*x/(8*e**2) + x**3/4), Ne(e**2,
0)), (x**3*sqrt(d**2)/3, True)) - d*e**2*Piecewise((sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**4) - d**2*x**2/(15
*e**2) + x**4/5), Ne(e**2, 0)), (x**4*sqrt(d**2)/4, True)) + e**3*Piecewise((d**6*Piecewise((log(-2*e**2*x + 2
*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(16*e**4) +
sqrt(d**2 - e**2*x**2)*(-d**4*x/(16*e**4) - d**2*x**3/(24*e**2) + x**5/6), Ne(e**2, 0)), (x**5*sqrt(d**2)/5,
True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.52 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {3 i \, d^{6} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{2}} + \frac {5 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{2}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x}{8 \, e} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{4 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{2}} \]

[In]

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

[Out]

3/8*I*d^6*arcsin(e*x/d + 2)/e^2 + 5/16*d^6*arcsin(e*x/d)/e^2 - 3/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4*x/e + 5
/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e - 3/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^5/e^2 - 1/24*(-e^2*x^2 + d^2)^(3/2)*d
^2*x/e + 1/6*(-e^2*x^2 + d^2)^(5/2)*x/e - 1/5*(-e^2*x^2 + d^2)^(5/2)*d/e^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {d^{6} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e {\left | e \right |}} - \frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {48 \, d^{5}}{e^{2}} - {\left (\frac {15 \, d^{4}}{e} + 2 \, {\left (48 \, d^{3} - {\left (35 \, d^{2} e - 4 \, {\left (5 \, e^{3} x - 6 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

-1/16*d^6*arcsin(e*x/d)*sgn(d)*sgn(e)/(e*abs(e)) - 1/240*sqrt(-e^2*x^2 + d^2)*(48*d^5/e^2 - (15*d^4/e + 2*(48*
d^3 - (35*d^2*e - 4*(5*e^3*x - 6*d*e^2)*x)*x)*x)*x)

Mupad [F(-1)]

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

[In]

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

[Out]

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