\(\int \frac {x^3}{(d+e x) (d^2-e^2 x^2)^{7/2}} \, dx\) [148]

Optimal result

Integrand size = 27, antiderivative size = 118 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 d-3 e x}{35 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{35 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{35 d^4 e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {864, 833, 792, 198, 197} \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 d-3 e x}{35 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{35 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{35 d^4 e^3 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(x^2*(d - e*x))/(7*e^2*(d^2 - e^2*x^2)^(7/2)) - (2*d - 3*e*x)/(35*e^4*(d^2 - e^2*x^2)^(5/2)) - x/(35*d^2*e^3*(
d^2 - e^2*x^2)^(3/2)) - (2*x)/(35*d^4*e^3*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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 864

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 (d-e x)}{\left (d^2-e^2 x^2\right )^{9/2}} \, dx \\ & = \frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {\int \frac {x \left (2 d^3-3 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx}{7 d^2 e^2} \\ & = \frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 d-3 e x}{35 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {3 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{35 e^3} \\ & = \frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 d-3 e x}{35 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{35 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{35 d^2 e^3} \\ & = \frac {x^2 (d-e x)}{7 e^2 \left (d^2-e^2 x^2\right )^{7/2}}-\frac {2 d-3 e x}{35 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{35 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{35 d^4 e^3 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^6-2 d^5 e x+5 d^4 e^2 x^2+5 d^3 e^3 x^3+5 d^2 e^4 x^4-2 d e^5 x^5-2 e^6 x^6\right )}{35 d^4 e^4 (d-e x)^3 (d+e x)^4} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-2*d^6 - 2*d^5*e*x + 5*d^4*e^2*x^2 + 5*d^3*e^3*x^3 + 5*d^2*e^4*x^4 - 2*d*e^5*x^5 - 2*e^6
*x^6))/(35*d^4*e^4*(d - e*x)^3*(d + e*x)^4)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (2 e^{6} x^{6}+2 d \,e^{5} x^{5}-5 d^{2} e^{4} x^{4}-5 d^{3} x^{3} e^{3}-5 d^{4} e^{2} x^{2}+2 d^{5} e x +2 d^{6}\right )}{35 d^{4} e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(92\)
trager	\(-\frac {\left (2 e^{6} x^{6}+2 d \,e^{5} x^{5}-5 d^{2} e^{4} x^{4}-5 d^{3} x^{3} e^{3}-5 d^{4} e^{2} x^{2}+2 d^{5} e x +2 d^{6}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{35 d^{4} e^{4} \left (e x +d \right )^{4} \left (-e x +d \right )^{3}}\)	\(101\)
default	\(\frac {\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}}{e}+\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{e^{3}}-\frac {d}{5 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{3} \left (-\frac {1}{7 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {6 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{10 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right )}{15 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {4 \left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right )}{15 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{d^{2}}\right )}{7 d}\right )}{e^{4}}\)	\(422\)

[In]

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

[Out]

-1/35*(-e*x+d)*(2*e^6*x^6+2*d*e^5*x^5-5*d^2*e^4*x^4-5*d^3*e^3*x^3-5*d^4*e^2*x^2+2*d^5*e*x+2*d^6)/d^4/e^4/(-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. 239 vs. \(2 (103) = 206\).

Time = 0.32 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.03 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, e^{7} x^{7} + 2 \, d e^{6} x^{6} - 6 \, d^{2} e^{5} x^{5} - 6 \, d^{3} e^{4} x^{4} + 6 \, d^{4} e^{3} x^{3} + 6 \, d^{5} e^{2} x^{2} - 2 \, d^{6} e x - 2 \, d^{7} - {\left (2 \, e^{6} x^{6} + 2 \, d e^{5} x^{5} - 5 \, d^{2} e^{4} x^{4} - 5 \, d^{3} e^{3} x^{3} - 5 \, d^{4} e^{2} x^{2} + 2 \, d^{5} e x + 2 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{35 \, {\left (d^{4} e^{11} x^{7} + d^{5} e^{10} x^{6} - 3 \, d^{6} e^{9} x^{5} - 3 \, d^{7} e^{8} x^{4} + 3 \, d^{8} e^{7} x^{3} + 3 \, d^{9} e^{6} x^{2} - d^{10} e^{5} x - d^{11} e^{4}\right )}} \]

[In]

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

[Out]

-1/35*(2*e^7*x^7 + 2*d*e^6*x^6 - 6*d^2*e^5*x^5 - 6*d^3*e^4*x^4 + 6*d^4*e^3*x^3 + 6*d^5*e^2*x^2 - 2*d^6*e*x - 2
*d^7 - (2*e^6*x^6 + 2*d*e^5*x^5 - 5*d^2*e^4*x^4 - 5*d^3*e^3*x^3 - 5*d^4*e^2*x^2 + 2*d^5*e*x + 2*d^6)*sqrt(-e^2
*x^2 + d^2))/(d^4*e^11*x^7 + d^5*e^10*x^6 - 3*d^6*e^9*x^5 - 3*d^7*e^8*x^4 + 3*d^8*e^7*x^3 + 3*d^9*e^6*x^2 - d^
10*e^5*x - d^11*e^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.13 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^{2}}{7 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{4}\right )}} + \frac {8 \, x}{35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {d}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} - \frac {x}{35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3}} - \frac {2 \, x}{35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.92 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}}{56\,d\,e^4\,{\left (d+e\,x\right )}^4}-\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {1}{56\,d\,e^4}+\frac {x}{35\,d^2\,e^3}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}-\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {2\,d}{35\,e^4}-\frac {11\,x}{70\,e^3}\right )}{{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^3}-\frac {2\,x\,\sqrt {d^2-e^2\,x^2}}{35\,d^4\,e^3\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \]

[In]

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

[Out]

(d^2 - e^2*x^2)^(1/2)/(56*d*e^4*(d + e*x)^4) - ((d^2 - e^2*x^2)^(1/2)*(1/(56*d*e^4) + x/(35*d^2*e^3)))/((d + e
*x)^2*(d - e*x)^2) - ((d^2 - e^2*x^2)^(1/2)*((2*d)/(35*e^4) - (11*x)/(70*e^3)))/((d + e*x)^3*(d - e*x)^3) - (2
*x*(d^2 - e^2*x^2)^(1/2))/(35*d^4*e^3*(d + e*x)*(d - e*x))