\(\int \frac {x^3 (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\) [200]

Optimal result

Integrand size = 27, antiderivative size = 192 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {27 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \]

[Out]

27/2*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+d^2*(-e*x+d)^4/e^4/(-e^2*x^2+d^2)^(1/2)+101/5*d^4*(-e^2*x^2+d^2)
^(1/2)/e^4-19/2*d^3*x*(-e^2*x^2+d^2)^(1/2)/e^3+18/5*d^2*x^2*(-e^2*x^2+d^2)^(1/2)/e^2-d*x^3*(-e^2*x^2+d^2)^(1/2
)/e+1/5*x^4*(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {866, 1649, 1829, 655, 223, 209} \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {27 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3} \]

[In]

Int[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(d^2*(d - e*x)^4)/(e^4*Sqrt[d^2 - e^2*x^2]) + (101*d^4*Sqrt[d^2 - e^2*x^2])/(5*e^4) - (19*d^3*x*Sqrt[d^2 - e^2
*x^2])/(2*e^3) + (18*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(5*e^2) - (d*x^3*Sqrt[d^2 - e^2*x^2])/e + (x^4*Sqrt[d^2 - e^
2*x^2])/5 + (27*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 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 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[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]

Rule 1829

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*(q + 2*p + 1))), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {(d-e x)^3 \left (-\frac {4 d^3}{e^3}+\frac {d^2 x}{e^2}-\frac {d x^2}{e}\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {\frac {20 d^6}{e}-65 d^5 x+80 d^4 e x^2-54 d^3 e^2 x^3+20 d^2 e^3 x^4}{\sqrt {d^2-e^2 x^2}} \, dx}{5 d e^2} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-80 d^6 e+260 d^5 e^2 x-380 d^4 e^3 x^2+216 d^3 e^4 x^3}{\sqrt {d^2-e^2 x^2}} \, dx}{20 d e^4} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {240 d^6 e^3-1212 d^5 e^4 x+1140 d^4 e^5 x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{60 d e^6} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-1620 d^6 e^5+2424 d^5 e^6 x}{\sqrt {d^2-e^2 x^2}} \, dx}{120 d e^8} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (27 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^3} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (27 d^5\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \\ & = \frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {27 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.67 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\frac {e \sqrt {d^2-e^2 x^2} \left (212 d^5+77 d^4 e x-29 d^3 e^2 x^2+16 d^2 e^3 x^3-8 d e^4 x^4+2 e^5 x^5\right )}{d+e x}+135 d^5 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{10 e^5} \]

[In]

Integrate[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

((e*Sqrt[d^2 - e^2*x^2]*(212*d^5 + 77*d^4*e*x - 29*d^3*e^2*x^2 + 16*d^2*e^3*x^3 - 8*d*e^4*x^4 + 2*e^5*x^5))/(d
+ e*x) + 135*d^5*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(10*e^5)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (2 e^{4} x^{4}-10 d \,e^{3} x^{3}+26 d^{2} e^{2} x^{2}-55 d^{3} e x +132 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{10 e^{4}}+\frac {27 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{3} \sqrt {e^{2}}}+\frac {8 d^{5} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\)	\(142\)
default	\(\text {Expression too large to display}\)	\(1086\)

[In]

int(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

1/10*(2*e^4*x^4-10*d*e^3*x^3+26*d^2*e^2*x^2-55*d^3*e*x+132*d^4)/e^4*(-e^2*x^2+d^2)^(1/2)+27/2*d^5/e^3/(e^2)^(1
/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+8*d^5/e^5/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {212 \, d^{5} e x + 212 \, d^{6} - 270 \, {\left (d^{5} e x + d^{6}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} + 16 \, d^{2} e^{3} x^{3} - 29 \, d^{3} e^{2} x^{2} + 77 \, d^{4} e x + 212 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (e^{5} x + d e^{4}\right )}} \]

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/10*(212*d^5*e*x + 212*d^6 - 270*(d^5*e*x + d^6)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^5*x^5 - 8*d
*e^4*x^4 + 16*d^2*e^3*x^3 - 29*d^3*e^2*x^2 + 77*d^4*e*x + 212*d^5)*sqrt(-e^2*x^2 + d^2))/(e^5*x + d*e^4)

Sympy [F]

\[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate(x**3*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral(x**3*(-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.12 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{2 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{e^{5} x + d e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x + d e^{4}\right )}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{4 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {3 i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{2 \, e^{4}} + \frac {15 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{e^{4}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{2 \, e^{3}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{e^{4}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{2 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{3}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{4 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{4}} \]

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/2*(-e^2*x^2 + d^2)^(5/2)*d^3/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) - 5/2*(-e^2*x^2 + d^2)^(3/2)*d
^4/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 15*sqrt(-e^2*x^2 + d^2)*d^5/(e^5*x + d*e^4) + (-e^2*x^2 + d^2)^(5/2)*d^2/
(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d^3/(e^5*x + d*e^4) - 3/4*(-e^2*x^2 + d^2)^(5/2)*
d/(e^5*x + d*e^4) + 3/2*I*d^5*arcsin(e*x/d + 2)/e^4 + 15*d^5*arcsin(e*x/d)/e^4 - 3/2*sqrt(e^2*x^2 + 4*d*e*x +
3*d^2)*d^3*x/e^3 - 3*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4/e^4 + 15/2*sqrt(-e^2*x^2 + d^2)*d^4/e^4 + 1/4*(-e^2*x
^2 + d^2)^(3/2)*d*x/e^3 - 5/4*(-e^2*x^2 + d^2)^(3/2)*d^2/e^4 + 1/5*(-e^2*x^2 + d^2)^(5/2)/e^4

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.66 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {27 \, d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{3} {\left | e \right |}} + \frac {1}{10} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left ({\left (x - \frac {5 \, d}{e}\right )} x + \frac {13 \, d^{2}}{e^{2}}\right )} x - \frac {55 \, d^{3}}{e^{3}}\right )} x + \frac {132 \, d^{4}}{e^{4}}\right )} - \frac {16 \, d^{5}}{e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

27/2*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) + 1/10*sqrt(-e^2*x^2 + d^2)*((2*((x - 5*d/e)*x + 13*d^2/e^2)
*x - 55*d^3/e^3)*x + 132*d^4/e^4) - 16*d^5/(e^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]

[In]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x)

[Out]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4, x)