\(\int \frac {x^2}{(d+e x) (d^2-e^2 x^2)^{7/2}} \, dx\) [149]

Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+2 e x)}{35 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 x}{105 d^3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 x}{105 d^5 e^2 \sqrt {d^2-e^2 x^2}} \]

[Out]

-1/7*x^2/d/e/(e*x+d)/(-e^2*x^2+d^2)^(5/2)+2/35*(2*e*x+d)/d/e^3/(-e^2*x^2+d^2)^(5/2)-4/105*x/d^3/e^2/(-e^2*x^2+
d^2)^(3/2)-8/105*x/d^5/e^2/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {869, 792, 198, 197} \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+2 e x)}{35 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 x}{105 d^5 e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x}{105 d^3 e^2 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

-1/7*x^2/(d*e*(d + e*x)*(d^2 - e^2*x^2)^(5/2)) + (2*(d + 2*e*x))/(35*d*e^3*(d^2 - e^2*x^2)^(5/2)) - (4*x)/(105
*d^3*e^2*(d^2 - e^2*x^2)^(3/2)) - (8*x)/(105*d^5*e^2*Sqrt[d^2 - e^2*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

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 869

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

Rubi steps \begin{align*} \text {integral}& = -\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {x (2 d+4 e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx}{7 d e} \\ & = -\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+2 e x)}{35 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{35 d e^2} \\ & = -\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+2 e x)}{35 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 x}{105 d^3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{105 d^3 e^2} \\ & = -\frac {x^2}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+2 e x)}{35 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 x}{105 d^3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 x}{105 d^5 e^2 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (6 d^6+6 d^5 e x-15 d^4 e^2 x^2+20 d^3 e^3 x^3+20 d^2 e^4 x^4-8 d e^5 x^5-8 e^6 x^6\right )}{105 d^5 e^3 (d-e x)^3 (d+e x)^4} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(6*d^6 + 6*d^5*e*x - 15*d^4*e^2*x^2 + 20*d^3*e^3*x^3 + 20*d^2*e^4*x^4 - 8*d*e^5*x^5 - 8*e
^6*x^6))/(105*d^5*e^3*(d - e*x)^3*(d + e*x)^4)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (-8 e^{6} x^{6}-8 d \,e^{5} x^{5}+20 d^{2} e^{4} x^{4}+20 d^{3} x^{3} e^{3}-15 d^{4} e^{2} x^{2}+6 d^{5} e x +6 d^{6}\right )}{105 d^{5} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(92\)
trager	\(\frac {\left (-8 e^{6} x^{6}-8 d \,e^{5} x^{5}+20 d^{2} e^{4} x^{4}+20 d^{3} x^{3} e^{3}-15 d^{4} e^{2} x^{2}+6 d^{5} e x +6 d^{6}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{105 d^{5} \left (e x +d \right )^{4} \left (-e x +d \right )^{3} e^{3}}\)	\(101\)
default	\(\frac {1}{5 e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{e^{2}}+\frac {d^{2} \left (-\frac {1}{7 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {6 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{10 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right )}{15 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {4 \left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right )}{15 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{d^{2}}\right )}{7 d}\right )}{e^{3}}\)	\(319\)

[In]

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

[Out]

1/105*(-e*x+d)*(-8*e^6*x^6-8*d*e^5*x^5+20*d^2*e^4*x^4+20*d^3*e^3*x^3-15*d^4*e^2*x^2+6*d^5*e*x+6*d^6)/d^5/e^3/(
-e^2*x^2+d^2)^(7/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (107) = 214\).

Time = 0.35 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.93 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {6 \, e^{7} x^{7} + 6 \, d e^{6} x^{6} - 18 \, d^{2} e^{5} x^{5} - 18 \, d^{3} e^{4} x^{4} + 18 \, d^{4} e^{3} x^{3} + 18 \, d^{5} e^{2} x^{2} - 6 \, d^{6} e x - 6 \, d^{7} + {\left (8 \, e^{6} x^{6} + 8 \, d e^{5} x^{5} - 20 \, d^{2} e^{4} x^{4} - 20 \, d^{3} e^{3} x^{3} + 15 \, d^{4} e^{2} x^{2} - 6 \, d^{5} e x - 6 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{105 \, {\left (d^{5} e^{10} x^{7} + d^{6} e^{9} x^{6} - 3 \, d^{7} e^{8} x^{5} - 3 \, d^{8} e^{7} x^{4} + 3 \, d^{9} e^{6} x^{3} + 3 \, d^{10} e^{5} x^{2} - d^{11} e^{4} x - d^{12} e^{3}\right )}} \]

[In]

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

[Out]

1/105*(6*e^7*x^7 + 6*d*e^6*x^6 - 18*d^2*e^5*x^5 - 18*d^3*e^4*x^4 + 18*d^4*e^3*x^3 + 18*d^5*e^2*x^2 - 6*d^6*e*x
- 6*d^7 + (8*e^6*x^6 + 8*d*e^5*x^5 - 20*d^2*e^4*x^4 - 20*d^3*e^3*x^3 + 15*d^4*e^2*x^2 - 6*d^5*e*x - 6*d^6)*sq
rt(-e^2*x^2 + d^2))/(d^5*e^10*x^7 + d^6*e^9*x^6 - 3*d^7*e^8*x^5 - 3*d^8*e^7*x^4 + 3*d^9*e^6*x^3 + 3*d^10*e^5*x
^2 - d^11*e^4*x - d^12*e^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.08 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {d}{7 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{3}\right )}} - \frac {x}{35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{2}} + \frac {1}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {4 \, x}{105 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2}} - \frac {8 \, x}{105 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} e^{2}} \]

[In]

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

[Out]

-1/7*d/((-e^2*x^2 + d^2)^(5/2)*e^4*x + (-e^2*x^2 + d^2)^(5/2)*d*e^3) - 1/35*x/((-e^2*x^2 + d^2)^(5/2)*d*e^2) +
1/5/((-e^2*x^2 + d^2)^(5/2)*e^3) - 4/105*x/((-e^2*x^2 + d^2)^(3/2)*d^3*e^2) - 8/105*x/(sqrt(-e^2*x^2 + d^2)*d
^5*e^2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.76 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.31 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {1}{56\,d^2\,e^3}-\frac {4\,x}{105\,d^3\,e^2}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}+\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {2}{35\,e^3}+\frac {3\,x}{70\,d\,e^2}\right )}{{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^3}-\frac {\sqrt {d^2-e^2\,x^2}}{56\,d^2\,e^3\,{\left (d+e\,x\right )}^4}-\frac {8\,x\,\sqrt {d^2-e^2\,x^2}}{105\,d^5\,e^2\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \]

[In]

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

[Out]

((d^2 - e^2*x^2)^(1/2)*(1/(56*d^2*e^3) - (4*x)/(105*d^3*e^2)))/((d + e*x)^2*(d - e*x)^2) + ((d^2 - e^2*x^2)^(1
/2)*(2/(35*e^3) + (3*x)/(70*d*e^2)))/((d + e*x)^3*(d - e*x)^3) - (d^2 - e^2*x^2)^(1/2)/(56*d^2*e^3*(d + e*x)^4
) - (8*x*(d^2 - e^2*x^2)^(1/2))/(105*d^5*e^2*(d + e*x)*(d - e*x))