\(\int \frac {x^4}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\) [139]

Optimal result

Integrand size = 27, antiderivative size = 85 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4}{5 e^5 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {864, 819, 272, 45} \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4}{5 e^5 \sqrt {d^2-e^2 x^2}}+\frac {4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rule 819

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

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^4 (d-e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = -\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 \int \frac {x^3}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 e} \\ & = -\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 \text {Subst}\left (\int \frac {x}{\left (d^2-e^2 x\right )^{5/2}} \, dx,x,x^2\right )}{5 e} \\ & = -\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 \text {Subst}\left (\int \left (\frac {d^2}{e^2 \left (d^2-e^2 x\right )^{5/2}}-\frac {1}{e^2 \left (d^2-e^2 x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{5 e} \\ & = -\frac {x^4 (d-e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 d^2}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4}{5 e^5 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-8 d^4-8 d^3 e x+12 d^2 e^2 x^2+12 d e^3 x^3-3 e^4 x^4\right )}{15 d e^5 (d-e x)^2 (d+e x)^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-8*d^4 - 8*d^3*e*x + 12*d^2*e^2*x^2 + 12*d*e^3*x^3 - 3*e^4*x^4))/(15*d*e^5*(d - e*x)^2*(
d + e*x)^3)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (3 e^{4} x^{4}-12 d \,e^{3} x^{3}-12 d^{2} e^{2} x^{2}+8 d^{3} e x +8 d^{4}\right )}{15 d \,e^{5} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(70\)
trager	\(-\frac {\left (3 e^{4} x^{4}-12 d \,e^{3} x^{3}-12 d^{2} e^{2} x^{2}+8 d^{3} e x +8 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 e^{5} d \left (e x +d \right )^{3} \left (-e x +d \right )^{2}}\)	\(79\)
default	\(\frac {\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}}{e}+\frac {d^{2}}{3 e^{5} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{3} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4}}-\frac {d \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )}{e^{2}}+\frac {d^{4} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{5}}\)	\(363\)

[In]

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

[Out]

-1/15*(-e*x+d)*(3*e^4*x^4-12*d*e^3*x^3-12*d^2*e^2*x^2+8*d^3*e*x+8*d^4)/d/e^5/(-e^2*x^2+d^2)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (74) = 148\).

Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.98 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {8 \, e^{5} x^{5} + 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} - 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x + 8 \, d^{5} + {\left (3 \, e^{4} x^{4} - 12 \, d e^{3} x^{3} - 12 \, d^{2} e^{2} x^{2} + 8 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{10} x^{5} + d^{2} e^{9} x^{4} - 2 \, d^{3} e^{8} x^{3} - 2 \, d^{4} e^{7} x^{2} + d^{5} e^{6} x + d^{6} e^{5}\right )}} \]

[In]

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

[Out]

-1/15*(8*e^5*x^5 + 8*d*e^4*x^4 - 16*d^2*e^3*x^3 - 16*d^3*e^2*x^2 + 8*d^4*e*x + 8*d^5 + (3*e^4*x^4 - 12*d*e^3*x
^3 - 12*d^2*e^2*x^2 + 8*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*e^10*x^5 + d^2*e^9*x^4 - 2*d^3*e^8*x^3 - 2*d
^4*e^7*x^2 + d^5*e^6*x + d^6*e^5)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.58 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {d^{3}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{5}\right )}} + \frac {x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {2 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {d^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} + \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.74 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (8\,d^4+8\,d^3\,e\,x-12\,d^2\,e^2\,x^2-12\,d\,e^3\,x^3+3\,e^4\,x^4\right )}{15\,d\,e^5\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \]

[In]

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

[Out]

-((d^2 - e^2*x^2)^(1/2)*(8*d^4 + 3*e^4*x^4 - 12*d*e^3*x^3 - 12*d^2*e^2*x^2 + 8*d^3*e*x))/(15*d*e^5*(d + e*x)^3
*(d - e*x)^2)