\(\int x (d+e x) (d^2-e^2 x^2)^{3/2} \, dx\) [5]

Optimal result

Integrand size = 23, antiderivative size = 116 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \]

[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 = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {794, 201, 223, 209} \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e} \]

[In]

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

[Out]

(d^4*x*Sqrt[d^2 - e^2*x^2])/(16*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]

Rubi steps \begin{align*} \text {integral}& = -\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 = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (48 d^5+15 d^4 e x-96 d^3 e^2 x^2-70 d^2 e^3 x^3+48 d e^4 x^4+40 e^5 x^5\right )+30 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^2} \]

[In]

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

[Out]

-1/240*(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])])/e^2

Maple [A] (verified)

Time = 0.34 (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	\(e \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6 e^{2}}\right )-\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 e^{2}}\)	\(126\)

[In]

int(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),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)+1/16
*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 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {30 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 48 \, d e^{4} x^{4} - 70 \, d^{2} e^{3} x^{3} - 96 \, d^{3} e^{2} x^{2} + 15 \, d^{4} e x + 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{2}} \]

[In]

integrate(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),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 = 0.53 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.36 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{5}}{5 e^{2}} - \frac {d^{4} x}{16 e} + \frac {2 d^{3} x^{2}}{5} + \frac {7 d^{2} e x^{3}}{24} - \frac {d e^{2} x^{4}}{5} - \frac {e^{3} x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{2}}{2} + \frac {e x^{3}}{3}\right ) \left (d^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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) + sqrt(d**2 - e**2*x**2)*(-d**5/(5*e**2) - d**4*x/(16*e) + 2*d**3*x**2
/5 + 7*d**2*e*x**3/24 - d*e**2*x**4/5 - e**3*x**5/6), Ne(e**2, 0)), ((d*x**2/2 + e*x**3/3)*(d**2)**(3/2), True
))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{2}} \]

[In]

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

[Out]

1/16*d^6*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e) + 1/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e + 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 = 94, normalized size of antiderivative = 0.81 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e {\left | e \right |}} - \frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {48 \, d^{5}}{e^{2}} + {\left (\frac {15 \, d^{4}}{e} - 2 \, {\left (48 \, d^{3} + {\left (35 \, d^{2} e - 4 \, {\left (5 \, e^{3} x + 6 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[In]

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

[Out]

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