\(\int \frac {x^3 (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\) [23]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d+3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{5 d^2 e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {833, 792, 197} \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d+3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{5 d^2 e^3 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(x^2*(d + e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*d + 3*e*x)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + x/(5*d^2*e^3*Sq
rt[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 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])

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^4+2 d^3 e x+3 d^2 e^2 x^2-3 d e^3 x^3+3 e^4 x^4\right )}{15 d^2 e^4 (d-e x)^3 (d+e x)^2} \]

[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^4 + 2*d^3*e*x + 3*d^2*e^2*x^2 - 3*d*e^3*x^3 + 3*e^4*x^4))/(15*d^2*e^4*(d - e*x)^3*(
d + e*x)^2)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (-3 e^{4} x^{4}+3 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}-2 d^{3} e x +2 d^{4}\right )}{15 d^{2} e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(77\)
trager	\(-\frac {\left (-3 e^{4} x^{4}+3 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}-2 d^{3} e x +2 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{2} e^{4} \left (-e x +d \right )^{3} \left (e x +d \right )^{2}}\)	\(79\)
default	\(e \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+d \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )\)	\(178\)

[In]

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

[Out]

-1/15*(-e*x+d)*(e*x+d)^2*(-3*e^4*x^4+3*d*e^3*x^3-3*d^2*e^2*x^2-2*d^3*e*x+2*d^4)/d^2/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. 172 vs. \(2 (78) = 156\).

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.91 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, e^{5} x^{5} - 2 \, d e^{4} x^{4} - 4 \, d^{2} e^{3} x^{3} + 4 \, d^{3} e^{2} x^{2} + 2 \, d^{4} e x - 2 \, d^{5} + {\left (3 \, e^{4} x^{4} - 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 2 \, d^{3} e x - 2 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{9} x^{5} - d^{3} e^{8} x^{4} - 2 \, d^{4} e^{7} x^{3} + 2 \, d^{5} e^{6} x^{2} + d^{6} e^{5} x - d^{7} e^{4}\right )}} \]

[In]

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

[Out]

-1/15*(2*e^5*x^5 - 2*d*e^4*x^4 - 4*d^2*e^3*x^3 + 4*d^3*e^2*x^2 + 2*d^4*e*x - 2*d^5 + (3*e^4*x^4 - 3*d*e^3*x^3
+ 3*d^2*e^2*x^2 + 2*d^3*e*x - 2*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2*e^9*x^5 - d^3*e^8*x^4 - 2*d^4*e^7*x^3 + 2*d^5*
e^6*x^2 + d^6*e^5*x - d^7*e^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (78) = 156\).

Time = 6.83 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.74 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=d \left (\begin {cases} - \frac {2 d^{2}}{15 d^{4} e^{4} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{8} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} + \frac {5 e^{2} x^{2}}{15 d^{4} e^{4} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{8} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {i x^{5}}{5 d^{7} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 10 d^{5} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 5 d^{3} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {x^{5}}{5 d^{7} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 10 d^{5} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 5 d^{3} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d*Piecewise((-2*d**2/(15*d**4*e**4*sqrt(d**2 - e**2*x**2) - 30*d**2*e**6*x**2*sqrt(d**2 - e**2*x**2) + 15*e**8
*x**4*sqrt(d**2 - e**2*x**2)) + 5*e**2*x**2/(15*d**4*e**4*sqrt(d**2 - e**2*x**2) - 30*d**2*e**6*x**2*sqrt(d**2
- e**2*x**2) + 15*e**8*x**4*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**4/(4*(d**2)**(7/2)), True)) + e*Piecewise
((-I*x**5/(5*d**7*sqrt(-1 + e**2*x**2/d**2) - 10*d**5*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 5*d**3*e**4*x**4*s
qrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (x**5/(5*d**7*sqrt(1 - e**2*x**2/d**2) - 10*d**5*e**2*x**
2*sqrt(1 - e**2*x**2/d**2) + 5*d**3*e**4*x**4*sqrt(1 - e**2*x**2/d**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.49 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {3 \, d^{2} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {2 \, d^{3}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{3}} \]

[In]

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

[Out]

1/2*x^3/((-e^2*x^2 + d^2)^(5/2)*e) + 1/3*d*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2) - 3/10*d^2*x/((-e^2*x^2 + d^2)^(5/
2)*e^3) - 2/15*d^3/((-e^2*x^2 + d^2)^(5/2)*e^4) + 1/10*x/((-e^2*x^2 + d^2)^(3/2)*e^3) + 1/5*x/(sqrt(-e^2*x^2 +
d^2)*d^2*e^3)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.60 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (-2\,d^4+2\,d^3\,e\,x+3\,d^2\,e^2\,x^2-3\,d\,e^3\,x^3+3\,e^4\,x^4\right )}{15\,d^2\,e^4\,{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^3} \]

[In]

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

[Out]

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