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

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}+\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {3 d^8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4} \]

[Out]

1/64*d^4*x*(-e^2*x^2+d^2)^(3/2)/e^3-1/7*d*x^2*(-e^2*x^2+d^2)^(5/2)/e^2-1/8*x^3*(-e^2*x^2+d^2)^(5/2)/e-1/560*d^
2*(35*e*x+32*d)*(-e^2*x^2+d^2)^(5/2)/e^4+3/128*d^8*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+3/128*d^6*x*(-e^2*x^2+
d^2)^(1/2)/e^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {847, 794, 201, 223, 209} \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {3 d^8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}+\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3} \]

[In]

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

[Out]

(3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) + (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/2
))/(7*e^2) - (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d + 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) + (3*d^
8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

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 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])

Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {\int x^2 \left (-3 d^2 e-8 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{8 e^2} \\ & = -\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac {\int x \left (16 d^3 e^2+21 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{56 e^4} \\ & = -\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {d^4 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{16 e^3} \\ & = \frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {\left (3 d^6\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{64 e^3} \\ & = \frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}+\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {\left (3 d^8\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{128 e^3} \\ & = \frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}+\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {\left (3 d^8\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^3} \\ & = \frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}+\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac {3 d^8 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.79 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (256 d^7+105 d^6 e x+128 d^5 e^2 x^2+70 d^4 e^3 x^3-1024 d^3 e^4 x^4-840 d^2 e^5 x^5+640 d e^6 x^6+560 e^7 x^7\right )+210 d^8 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{4480 e^4} \]

[In]

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

[Out]

-1/4480*(Sqrt[d^2 - e^2*x^2]*(256*d^7 + 105*d^6*e*x + 128*d^5*e^2*x^2 + 70*d^4*e^3*x^3 - 1024*d^3*e^4*x^4 - 84
0*d^2*e^5*x^5 + 640*d*e^6*x^6 + 560*e^7*x^7) + 210*d^8*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^4

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (560 e^{7} x^{7}+640 d \,e^{6} x^{6}-840 d^{2} e^{5} x^{5}-1024 d^{3} e^{4} x^{4}+70 d^{4} e^{3} x^{3}+128 d^{5} e^{2} x^{2}+105 d^{6} e x +256 d^{7}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{4480 e^{4}}+\frac {3 d^{8} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{128 e^{3} \sqrt {e^{2}}}\)	\(130\)
default	\(e \left (-\frac {x^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6 e^{2}}\right )}{8 e^{2}}\right )+d \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{7 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{35 e^{4}}\right )\)	\(184\)

[In]

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

[Out]

-1/4480*(560*e^7*x^7+640*d*e^6*x^6-840*d^2*e^5*x^5-1024*d^3*e^4*x^4+70*d^4*e^3*x^3+128*d^5*e^2*x^2+105*d^6*e*x
+256*d^7)/e^4*(-e^2*x^2+d^2)^(1/2)+3/128*d^8/e^3/(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 = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {210 \, d^{8} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (560 \, e^{7} x^{7} + 640 \, d e^{6} x^{6} - 840 \, d^{2} e^{5} x^{5} - 1024 \, d^{3} e^{4} x^{4} + 70 \, d^{4} e^{3} x^{3} + 128 \, d^{5} e^{2} x^{2} + 105 \, d^{6} e x + 256 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{4480 \, e^{4}} \]

[In]

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

[Out]

-1/4480*(210*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (560*e^7*x^7 + 640*d*e^6*x^6 - 840*d^2*e^5*x^5 -
1024*d^3*e^4*x^4 + 70*d^4*e^3*x^3 + 128*d^5*e^2*x^2 + 105*d^6*e*x + 256*d^7)*sqrt(-e^2*x^2 + d^2))/e^4

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.10 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 d^{8} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{128 e^{3}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{7}}{35 e^{4}} - \frac {3 d^{6} x}{128 e^{3}} - \frac {d^{5} x^{2}}{35 e^{2}} - \frac {d^{4} x^{3}}{64 e} + \frac {8 d^{3} x^{4}}{35} + \frac {3 d^{2} e x^{5}}{16} - \frac {d e^{2} x^{6}}{7} - \frac {e^{3} x^{7}}{8}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{4}}{4} + \frac {e x^{5}}{5}\right ) \left (d^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*d**8*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))/(128*e**3) + sqrt(d**2 - e**2*x**2)*(-2*d**7/(35*e**4) - 3*d**6*x/(128*e**3
) - d**5*x**2/(35*e**2) - d**4*x**3/(64*e) + 8*d**3*x**4/35 + 3*d**2*e*x**5/16 - d*e**2*x**6/7 - e**3*x**7/8),
Ne(e**2, 0)), ((d*x**4/4 + e*x**5/5)*(d**2)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.95 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {3 \, d^{8} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{128 \, \sqrt {e^{2}} e^{3}} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6} x}{128 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{3}}{8 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} x}{64 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x^{2}}{7 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x}{16 \, e^{3}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{35 \, e^{4}} \]

[In]

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

[Out]

3/128*d^8*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^3) + 3/128*sqrt(-e^2*x^2 + d^2)*d^6*x/e^3 - 1/8*(-e^2*x^2 +
d^2)^(5/2)*x^3/e + 1/64*(-e^2*x^2 + d^2)^(3/2)*d^4*x/e^3 - 1/7*(-e^2*x^2 + d^2)^(5/2)*d*x^2/e^2 - 1/16*(-e^2*
x^2 + d^2)^(5/2)*d^2*x/e^3 - 2/35*(-e^2*x^2 + d^2)^(5/2)*d^3/e^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {3 \, d^{8} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{128 \, e^{3} {\left | e \right |}} - \frac {1}{4480} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {256 \, d^{7}}{e^{4}} + {\left (\frac {105 \, d^{6}}{e^{3}} + 2 \, {\left (\frac {64 \, d^{5}}{e^{2}} + {\left (\frac {35 \, d^{4}}{e} - 4 \, {\left (128 \, d^{3} + 5 \, {\left (21 \, d^{2} e - 2 \, {\left (7 \, e^{3} x + 8 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

3/128*d^8*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 1/4480*sqrt(-e^2*x^2 + d^2)*(256*d^7/e^4 + (105*d^6/e^3 +
2*(64*d^5/e^2 + (35*d^4/e - 4*(128*d^3 + 5*(21*d^2*e - 2*(7*e^3*x + 8*d*e^2)*x)*x)*x)*x)*x)*x)

Mupad [F(-1)]

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

[In]

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

[Out]

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