\(\int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) [33]

Optimal result

Integrand size = 27, antiderivative size = 173 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {11 d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5} \]

[Out]

11/16*d^6*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5-8/15*d^3*x^2*(-e^2*x^2+d^2)^(1/2)/e^3-11/24*d^2*x^3*(-e^2*x^2+d
^2)^(1/2)/e^2-2/5*d*x^4*(-e^2*x^2+d^2)^(1/2)/e-1/6*x^5*(-e^2*x^2+d^2)^(1/2)-1/240*d^4*(165*e*x+256*d)*(-e^2*x^
2+d^2)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 173, 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 = {1823, 847, 794, 223, 209} \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {11 d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3} \]

[In]

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

[Out]

(-8*d^3*x^2*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (11*d^2*x^3*Sqrt[d^2 - e^2*x^2])/(24*e^2) - (2*d*x^4*Sqrt[d^2 - e^
2*x^2])/(5*e) - (x^5*Sqrt[d^2 - e^2*x^2])/6 - (d^4*(256*d + 165*e*x)*Sqrt[d^2 - e^2*x^2])/(240*e^5) + (11*d^6*
ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*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 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^4 \left (-11 d^2 e^2-12 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{6 e^2} \\ & = -\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x^3 \left (48 d^3 e^3+55 d^2 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{30 e^4} \\ & = -\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^2 \left (-165 d^4 e^4-192 d^3 e^5 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{120 e^6} \\ & = -\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x \left (384 d^5 e^5+495 d^4 e^6 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{360 e^8} \\ & = -\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^4} \\ & = -\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\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^4} \\ & = -\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (256 d^5+165 d^4 e x+128 d^3 e^2 x^2+110 d^2 e^3 x^3+96 d e^4 x^4+40 e^5 x^5\right )+330 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^5} \]

[In]

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

[Out]

-1/240*(Sqrt[d^2 - e^2*x^2]*(256*d^5 + 165*d^4*e*x + 128*d^3*e^2*x^2 + 110*d^2*e^3*x^3 + 96*d*e^4*x^4 + 40*e^5
*x^5) + 330*d^6*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^5

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {\left (40 e^{5} x^{5}+96 d \,e^{4} x^{4}+110 d^{2} e^{3} x^{3}+128 d^{3} e^{2} x^{2}+165 d^{4} e x +256 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{5}}+\frac {11 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e^{4} \sqrt {e^{2}}}\)	\(108\)
default	\(e^{2} \left (-\frac {x^{5} \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{2}}+\frac {5 d^{2} \left (-\frac {x^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )}{6 e^{2}}\right )+d^{2} \left (-\frac {x^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )+2 d e \left (-\frac {x^{4} \sqrt {-e^{2} x^{2}+d^{2}}}{5 e^{2}}+\frac {4 d^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )}{5 e^{2}}\right )\)	\(295\)

[In]

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

[Out]

-1/240*(40*e^5*x^5+96*d*e^4*x^4+110*d^2*e^3*x^3+128*d^3*e^2*x^2+165*d^4*e*x+256*d^5)/e^5*(-e^2*x^2+d^2)^(1/2)+
11/16*d^6/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 = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {330 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 96 \, d e^{4} x^{4} + 110 \, d^{2} e^{3} x^{3} + 128 \, d^{3} e^{2} x^{2} + 165 \, d^{4} e x + 256 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{5}} \]

[In]

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

[Out]

-1/240*(330*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 + 96*d*e^4*x^4 + 110*d^2*e^3*x^3 + 128
*d^3*e^2*x^2 + 165*d^4*e*x + 256*d^5)*sqrt(-e^2*x^2 + d^2))/e^5

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {11 d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e^{4}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {16 d^{5}}{15 e^{5}} - \frac {11 d^{4} x}{16 e^{4}} - \frac {8 d^{3} x^{2}}{15 e^{3}} - \frac {11 d^{2} x^{3}}{24 e^{2}} - \frac {2 d x^{4}}{5 e} - \frac {x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d^{2} x^{5}}{5} + \frac {d e x^{6}}{3} + \frac {e^{2} x^{7}}{7}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((11*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)*(-16*d**5/(15*e**5) - 11*d**4*x/(16*e**
4) - 8*d**3*x**2/(15*e**3) - 11*d**2*x**3/(24*e**2) - 2*d*x**4/(5*e) - x**5/6), Ne(e**2, 0)), ((d**2*x**5/5 +
d*e*x**6/3 + e**2*x**7/7)/sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{5} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d x^{4}}{5 \, e} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x^{3}}{24 \, e^{2}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x^{2}}{15 \, e^{3}} + \frac {11 \, d^{6} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e^{4}} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{4}} - \frac {16 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{15 \, e^{5}} \]

[In]

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

[Out]

-1/6*sqrt(-e^2*x^2 + d^2)*x^5 - 2/5*sqrt(-e^2*x^2 + d^2)*d*x^4/e - 11/24*sqrt(-e^2*x^2 + d^2)*d^2*x^3/e^2 - 8/
15*sqrt(-e^2*x^2 + d^2)*d^3*x^2/e^3 + 11/16*d^6*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^4) - 11/16*sqrt(-e^2*
x^2 + d^2)*d^4*x/e^4 - 16/15*sqrt(-e^2*x^2 + d^2)*d^5/e^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55 \[ \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {11 \, d^{6} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e^{4} {\left | e \right |}} - \frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, x + \frac {12 \, d}{e}\right )} x + \frac {55 \, d^{2}}{e^{2}}\right )} x + \frac {64 \, d^{3}}{e^{3}}\right )} x + \frac {165 \, d^{4}}{e^{4}}\right )} x + \frac {256 \, d^{5}}{e^{5}}\right )} \]

[In]

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

[Out]

11/16*d^6*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^4*abs(e)) - 1/240*sqrt(-e^2*x^2 + d^2)*((2*((4*(5*x + 12*d/e)*x + 55*
d^2/e^2)*x + 64*d^3/e^3)*x + 165*d^4/e^4)*x + 256*d^5/e^5)

Mupad [F(-1)]

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

[In]

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

[Out]

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