\(\int \frac {x^3 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [85]

Optimal result

Integrand size = 27, antiderivative size = 118 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/5*d^2*(e*x+d)^3/e^4/(-e^2*x^2+d^2)^(5/2)-13/15*d*(e*x+d)^2/e^4/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2
)^(1/2))/e^4+32/15*(e*x+d)/e^4/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1649, 792, 223, 209} \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^2*(d + e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d + e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d +
e*x))/(15*e^4*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^4

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 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 d^3}{e^3}+\frac {5 d^2 x}{e^2}+\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\left (\frac {17 d^3}{e^3}+\frac {15 d^2 x}{e^2}\right ) (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2} \\ & = \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3} \\ & = \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \\ & = \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2-51 d e x+32 e^2 x^2\right )}{(d-e x)^3}+30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^4} \]

[In]

Integrate[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(22*d^2 - 51*d*e*x + 32*e^2*x^2))/(d - e*x)^3 + 30*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 -
e^2*x^2])])/(15*e^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(376\) vs. \(2(104)=208\).

Time = 0.40 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.19

method	result	size
default	\(e^{3} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+d^{3} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+3 d \,e^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+3 d^{2} e \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\)	\(377\)

[In]

int(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(1/5*x^5/e^2/(-e^2*x^2+d^2)^(5/2)-1/e^2*(1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2
)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))))+d^3*(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d
^2/e^4/(-e^2*x^2+d^2)^(5/2))+3*d*e^2*(x^4/e^2/(-e^2*x^2+d^2)^(5/2)-4*d^2/e^2*(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)
-2/15*d^2/e^4/(-e^2*x^2+d^2)^(5/2)))+3*d^2*e*(1/2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/2*d^2/e^2*(1/4*x/e^2/(-e^2*x^
2+d^2)^(5/2)-1/4*d^2/e^2*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e
^2*x^2+d^2)^(1/2)))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {22 \, e^{3} x^{3} - 66 \, d e^{2} x^{2} + 66 \, d^{2} e x - 22 \, d^{3} + 30 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (32 \, e^{2} x^{2} - 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} - 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x - d^{3} e^{4}\right )}} \]

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(22*e^3*x^3 - 66*d*e^2*x^2 + 66*d^2*e*x - 22*d^3 + 30*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (32*e^2*x^2 - 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 - 3*d*e^6*x
^2 + 3*d^2*e^5*x - d^3*e^4)

Sympy [F]

\[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{3} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(x**3*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**3*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (104) = 208\).

Time = 0.30 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.61 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{15} \, e^{3} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, e x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {3 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {11 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {9 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {22 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {17 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{3}} \]

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^3*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 1/3*e*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) + 3*d*x^
4/(-e^2*x^2 + d^2)^(5/2) + 3/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e) - 11/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2)
- 9/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)*e^3) + 22/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^4) + 17/30*d^2*x/((-e^2*x^2 +
d^2)^(3/2)*e^3) + 2/15*x/(sqrt(-e^2*x^2 + d^2)*e^3) - arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.58 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{3} {\left | e \right |}} - \frac {2 \, {\left (\frac {95 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {145 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {75 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} - 22\right )}}{15 \, e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 2/15*(95*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 145*(d*e +
sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 75*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) - 15*(d*e + sqrt
(-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) - 22)/(e^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[In]

int((x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2), x)