\(\int \frac {x^4 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [84]

Optimal result

Integrand size = 27, antiderivative size = 142 \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1649, 655, 223, 209} \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}+\frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]

[In]

Int[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^3*(d + e*x)^3)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) - (6*d^2*(d + e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) + (24*d*(d
+ e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/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 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 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^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 d^4}{e^4}+\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}+\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d+e x) \left (\frac {27 d^4}{e^4}+\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2} \\ & = \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {45 d^4}{e^4}+\frac {15 d^3 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3} \\ & = \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4} \\ & = \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \\ & = \frac {d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {24 d (d+e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69 \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (24 d^3-57 d^2 e x+39 d e^2 x^2-5 e^3 x^3\right )}{(d-e x)^3}+30 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{5 e^5} \]

[In]

Integrate[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(24*d^3 - 57*d^2*e*x + 39*d*e^2*x^2 - 5*e^3*x^3))/(d - e*x)^3 + 30*d*ArcTan[(e*x)/(Sqrt[
d^2] - Sqrt[d^2 - e^2*x^2])])/(5*e^5)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {d^{3} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 e^{8} \left (x -\frac {d}{e}\right )^{3}}-\frac {6 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 e^{7} \left (x -\frac {d}{e}\right )^{2}}-\frac {24 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{5 e^{6} \left (x -\frac {d}{e}\right )}\)	\(195\)
default	\(e^{3} \left (-\frac {x^{6}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 d^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )}{e^{2}}\right )+d^{3} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+3 d \,e^{2} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+3 d^{2} e \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )\)	\(438\)

[In]

int(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-e^2*x^2+d^2)^(1/2)/e^5-3*d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/5*d^3/e^8/(x-d/e)^3*
(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)-6/5*d^2/e^7/(x-d/e)^2*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)-24/5*d/e^6/(x-
d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {24 \, d e^{3} x^{3} - 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x - 24 \, d^{4} + 30 \, {\left (d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (5 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 57 \, d^{2} e x - 24 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{8} x^{3} - 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x - d^{3} e^{5}\right )}} \]

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/5*(24*d*e^3*x^3 - 72*d^2*e^2*x^2 + 72*d^3*e*x - 24*d^4 + 30*(d*e^3*x^3 - 3*d^2*e^2*x^2 + 3*d^3*e*x - d^4)*ar
ctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^3*x^3 - 39*d*e^2*x^2 + 57*d^2*e*x - 24*d^3)*sqrt(-e^2*x^2 + d^2
))/(e^8*x^3 - 3*d*e^7*x^2 + 3*d^2*e^6*x - d^3*e^5)

Sympy [F]

\[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{4} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(x**4*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**4*(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. 336 vs. \(2 (126) = 252\).

Time = 0.29 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.37 \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{5} \, d e^{2} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {e x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - d x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {9 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{3} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {12 \, d^{4} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {3 \, d^{5} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {24 \, d^{6}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {9 \, d^{3} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {6 \, d x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {3 \, d \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{4}} \]

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/5*d*e^2*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)) - e*x^6/(-e^2*x^2 + d^2)^(5/2) - d*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)) + 9*d^2*x^4/((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^3*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 12*d^4
*x^2/((-e^2*x^2 + d^2)^(5/2)*e^3) - 3/10*d^5*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 24/5*d^6/((-e^2*x^2 + d^2)^(5/2)
*e^5) + 9/10*d^3*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 6/5*d*x/(sqrt(-e^2*x^2 + d^2)*e^4) - 3*d*arcsin(e^2*x/(d*sqr
t(e^2)))/(sqrt(e^2)*e^4)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.49 \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {3 \, d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{4} {\left | e \right |}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{5}} + \frac {2 \, {\left (19 \, d - \frac {80 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d}{e^{2} x} + \frac {120 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{e^{4} x^{2}} - \frac {70 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d}{e^{8} x^{4}}\right )}}{5 \, e^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-3*d*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^4*abs(e)) + sqrt(-e^2*x^2 + d^2)/e^5 + 2/5*(19*d - 80*(d*e + sqrt(-e^2*x^2
+ d^2)*abs(e))*d/(e^2*x) + 120*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d/(e^4*x^2) - 70*(d*e + sqrt(-e^2*x^2 +
d^2)*abs(e))^3*d/(e^6*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*d/(e^8*x^4))/(e^4*((d*e + sqrt(-e^2*x^2
+ d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^4\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[In]

int((x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2), x)