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

Optimal result

Integrand size = 25, antiderivative size = 159 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \]

[Out]

1/24*d^3*x*(-e^2*x^2+d^2)^(3/2)/e^2-1/5*d^2*(-e^2*x^2+d^2)^(5/2)/e^3-1/6*d*x*(-e^2*x^2+d^2)^(5/2)/e^2+1/7*(-e^
2*x^2+d^2)^(7/2)/e^3+1/16*d^7*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/16*d^5*x*(-e^2*x^2+d^2)^(1/2)/e^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 655, 201, 223, 209} \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^7 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2} \]

[In]

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

[Out]

(d^5*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (d^3*x*(d^2 - e^2*x^2)^(3/2))/(24*e^2) - (d^2*(d^2 - e^2*x^2)^(5/2))/(5
*e^3) - (d*x*(d^2 - e^2*x^2)^(5/2))/(6*e^2) + (d^2 - e^2*x^2)^(7/2)/(7*e^3) + (d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2
*x^2]])/(16*e^3)

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 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 811

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int (d+e x) \left (d^2-e^2 x^2\right )^{5/2} \, dx}{e^2}+\frac {d^2 \int (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {d \int \left (d^2-e^2 x^2\right )^{5/2} \, dx}{e^2}+\frac {d^3 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2} \\ & = \frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\left (5 d^3\right ) \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e^2}+\frac {\left (3 d^5\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{4 e^2} \\ & = \frac {3 d^5 x \sqrt {d^2-e^2 x^2}}{8 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\left (5 d^5\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2}+\frac {\left (3 d^7\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2} \\ & = \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\left (5 d^7\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2}+\frac {\left (3 d^7\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^2} \\ & = \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {3 d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {\left (5 d^7\right ) \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^2} \\ & = \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{5/2}}{6 e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.79 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (96 d^6+105 d^5 e x+48 d^4 e^2 x^2-490 d^3 e^3 x^3-384 d^2 e^4 x^4+280 d e^5 x^5+240 e^6 x^6\right )+210 d^7 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{1680 e^3} \]

[In]

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

[Out]

-1/1680*(Sqrt[d^2 - e^2*x^2]*(96*d^6 + 105*d^5*e*x + 48*d^4*e^2*x^2 - 490*d^3*e^3*x^3 - 384*d^2*e^4*x^4 + 280*
d*e^5*x^5 + 240*e^6*x^6) + 210*d^7*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^3

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (240 e^{6} x^{6}+280 d \,e^{5} x^{5}-384 d^{2} e^{4} x^{4}-490 d^{3} x^{3} e^{3}+48 d^{4} e^{2} x^{2}+105 d^{5} e x +96 d^{6}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{1680 e^{3}}+\frac {d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e^{2} \sqrt {e^{2}}}\)	\(119\)
default	\(e \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{7 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{35 e^{4}}\right )+d \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6 e^{2}}\right )\)	\(153\)

[In]

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

[Out]

-1/1680*(240*e^6*x^6+280*d*e^5*x^5-384*d^2*e^4*x^4-490*d^3*e^3*x^3+48*d^4*e^2*x^2+105*d^5*e*x+96*d^6)/e^3*(-e^
2*x^2+d^2)^(1/2)+1/16*d^7/e^2/(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 = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.73 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {210 \, d^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (240 \, e^{6} x^{6} + 280 \, d e^{5} x^{5} - 384 \, d^{2} e^{4} x^{4} - 490 \, d^{3} e^{3} x^{3} + 48 \, d^{4} e^{2} x^{2} + 105 \, d^{5} e x + 96 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, e^{3}} \]

[In]

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

[Out]

-1/1680*(210*d^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (240*e^6*x^6 + 280*d*e^5*x^5 - 384*d^2*e^4*x^4 -
490*d^3*e^3*x^3 + 48*d^4*e^2*x^2 + 105*d^5*e*x + 96*d^6)*sqrt(-e^2*x^2 + d^2))/e^3

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {d^{7} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{6}}{35 e^{3}} - \frac {d^{5} x}{16 e^{2}} - \frac {d^{4} x^{2}}{35 e} + \frac {7 d^{3} x^{3}}{24} + \frac {8 d^{2} e x^{4}}{35} - \frac {d e^{2} x^{5}}{6} - \frac {e^{3} x^{6}}{7}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{3}}{3} + \frac {e x^{4}}{4}\right ) \left (d^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d**7*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**2) + sqrt(d**2 - e**2*x**2)*(-2*d**6/(35*e**3) - d**5*x/(16*e**2) - d*
*4*x**2/(35*e) + 7*d**3*x**3/24 + 8*d**2*e*x**4/35 - d*e**2*x**5/6 - e**3*x**6/7), Ne(e**2, 0)), ((d*x**3/3 +
e*x**4/4)*(d**2)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{7} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{5} x}{16 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} x}{24 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}}{7 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x}{6 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{35 \, e^{3}} \]

[In]

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

[Out]

1/16*d^7*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/16*sqrt(-e^2*x^2 + d^2)*d^5*x/e^2 + 1/24*(-e^2*x^2 +
d^2)^(3/2)*d^3*x/e^2 - 1/7*(-e^2*x^2 + d^2)^(5/2)*x^2/e - 1/6*(-e^2*x^2 + d^2)^(5/2)*d*x/e^2 - 2/35*(-e^2*x^2
+ d^2)^(5/2)*d^2/e^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.67 \[ \int x^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{7} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e^{2} {\left | e \right |}} - \frac {1}{1680} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {96 \, d^{6}}{e^{3}} + {\left (\frac {105 \, d^{5}}{e^{2}} + 2 \, {\left (\frac {24 \, d^{4}}{e} - {\left (245 \, d^{3} + 4 \, {\left (48 \, d^{2} e - 5 \, {\left (6 \, e^{3} x + 7 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

1/16*d^7*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) - 1/1680*sqrt(-e^2*x^2 + d^2)*(96*d^6/e^3 + (105*d^5/e^2 + 2
*(24*d^4/e - (245*d^3 + 4*(48*d^2*e - 5*(6*e^3*x + 7*d*e^2)*x)*x)*x)*x)*x)

Mupad [F(-1)]

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

[In]

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

[Out]

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