\(\int \frac {x (d^2-e^2 x^2)^p}{(d+e x)^3} \, dx\) [289]

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (2-p) (d+e x)^3}-\frac {3\ 2^{-3+p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^3 e^2 (2-p) (1+p)} \]

[Out]

1/2*(-e^2*x^2+d^2)^(p+1)/e^2/(2-p)/(e*x+d)^3-3*2^(-3+p)*(1+e*x/d)^(-1-p)*(-e^2*x^2+d^2)^(p+1)*hypergeom([p+1,
2-p],[2+p],1/2*(-e*x+d)/d)/d^3/e^2/(-p^2+p+2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {807, 692, 71} \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (2-p) (d+e x)^3}-\frac {3\ 2^{p-3} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d^3 e^2 (2-p) (p+1)} \]

[In]

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

[Out]

(d^2 - e^2*x^2)^(1 + p)/(2*e^2*(2 - p)*(d + e*x)^3) - (3*2^(-3 + p)*(1 + (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1
+ p)*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(d^3*e^2*(2 - p)*(1 + p))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(m - 1)*((a + c*x^2)^(p + 1)/((1
+ e*(x/d))^(p + 1)*(a/d + (c*x)/e)^(p + 1))), Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 807

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (2-p) (d+e x)^3}+\frac {3 \int \frac {\left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx}{2 e (2-p)} \\ & = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (2-p) (d+e x)^3}+\frac {\left (3 (d-e x)^{-1-p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p}\right ) \int (d-e x)^p \left (1+\frac {e x}{d}\right )^{-2+p} \, dx}{2 d^3 e (2-p)} \\ & = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (2-p) (d+e x)^3}-\frac {3\ 2^{-3+p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^3 e^2 (2-p) (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {2^{-3+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (-2 \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+\operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{d^2 e^2 (1+p)} \]

[In]

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

[Out]

(2^(-3 + p)*(d - e*x)*(d^2 - e^2*x^2)^p*(-2*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] + Hypergeo
metric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/(d^2*e^2*(1 + p)*(1 + (e*x)/d)^p)

Maple [F]

\[\int \frac {x \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((-e^2*x^2 + d^2)^p*x/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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