\(\int \frac {x^3 (d^2-e^2 x^2)^p}{(d+e x)^2} \, dx\) [277]

Optimal result

Integrand size = 25, antiderivative size = 150 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{e^4 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}-\frac {2 e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^3} \]

[Out]

d^4*(-e^2*x^2+d^2)^(-1+p)/e^4/(1-p)+3/2*d^2*(-e^2*x^2+d^2)^p/e^4/p-1/2*(-e^2*x^2+d^2)^(p+1)/e^4/(p+1)-2/5*e*x^
5*(-e^2*x^2+d^2)^p*hypergeom([5/2, 2-p],[7/2],e^2*x^2/d^2)/d^3/((1-e^2*x^2/d^2)^p)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1666, 457, 78, 12, 372, 371} \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^4 (p+1)}+\frac {d^4 \left (d^2-e^2 x^2\right )^{p-1}}{e^4 (1-p)}-\frac {2 e x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^3} \]

[In]

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

[Out]

(d^4*(d^2 - e^2*x^2)^(-1 + p))/(e^4*(1 - p)) + (3*d^2*(d^2 - e^2*x^2)^p)/(2*e^4*p) - (d^2 - e^2*x^2)^(1 + p)/(
2*e^4*(1 + p)) - (2*e*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, 2 - p, 7/2, (e^2*x^2)/d^2])/(5*d^3*(1 - (e^
2*x^2)/d^2)^p)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 866

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

Rule 1666

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \int x^3 (d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx \\ & = \int -2 d e x^4 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\int x^3 \left (d^2-e^2 x^2\right )^{-2+p} \left (d^2+e^2 x^2\right ) \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int x \left (d^2-e^2 x\right )^{-2+p} \left (d^2+e^2 x\right ) \, dx,x,x^2\right )-(2 d e) \int x^4 \left (d^2-e^2 x^2\right )^{-2+p} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {2 d^4 \left (d^2-e^2 x\right )^{-2+p}}{e^2}-\frac {3 d^2 \left (d^2-e^2 x\right )^{-1+p}}{e^2}+\frac {\left (d^2-e^2 x\right )^p}{e^2}\right ) \, dx,x,x^2\right )-\frac {\left (2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^3} \\ & = \frac {d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{e^4 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}-\frac {2 e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},2-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(332\) vs. \(2(150)=300\).

Time = 0.39 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.21 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {2^{-2+p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (2 d^2 \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p-2 d^2 \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^p+2 e^2 x^2 \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^p-8 d e (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-6 d (d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )+d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )-d e x \left (1-\frac {e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{e^4 (1+p)} \]

[In]

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

[Out]

(2^(-2 + p)*(d^2 - e^2*x^2)^p*(2*d^2*(1/2 + (e*x)/(2*d))^p - 2*d^2*(1/2 + (e*x)/(2*d))^p*(1 - (e^2*x^2)/d^2)^p
+ 2*e^2*x^2*(1/2 + (e*x)/(2*d))^p*(1 - (e^2*x^2)/d^2)^p - 8*d*e*(1 + p)*x*(1/2 + (e*x)/(2*d))^p*Hypergeometri
c2F1[1/2, -p, 3/2, (e^2*x^2)/d^2] - 6*d*(d - e*x)*(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[1 - p, 1 + p, 2 + p,
(d - e*x)/(2*d)] + d^2*(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - d*e*x*
(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/(e^4*(1 + p)*(1 + (e*x)/d)^p*(
1 - (e^2*x^2)/d^2)^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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