\(\int \frac {x^4 (d^2-e^2 x^2)^p}{(d+e x)^2} \, dx\) [276]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 184 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=-\frac {d^5 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}-\frac {2 d^3 \left (d^2-e^2 x^2\right )^p}{e^5 p}+\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{e^5 (1+p)}+\frac {2 (4+p) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2 (3+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {866, 1666, 470, 372, 371, 12, 272, 45} \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {2 (p+4) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2 (2 p+3)}-\frac {x^5 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+3}+\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{e^5 (p+1)}-\frac {d^5 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)}-\frac {2 d^3 \left (d^2-e^2 x^2\right )^p}{e^5 p} \]

[In]

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

[Out]

-((d^5*(d^2 - e^2*x^2)^(-1 + p))/(e^5*(1 - p))) - (x^5*(d^2 - e^2*x^2)^(-1 + p))/(3 + 2*p) - (2*d^3*(d^2 - e^2
*x^2)^p)/(e^5*p) + (d*(d^2 - e^2*x^2)^(1 + p))/(e^5*(1 + p)) + (2*(4 + p)*x^5*(d^2 - e^2*x^2)^p*Hypergeometric
2F1[5/2, 2 - p, 7/2, (e^2*x^2)/d^2])/(5*d^2*(3 + 2*p)*(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 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 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 470

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

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.36 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {x^5 (d-e x)^p (d+e x)^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {AppellF1}\left (5,-p,2-p,6,\frac {e x}{d},-\frac {e x}{d}\right )}{5 d^2} \]

[In]

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

[Out]

(x^5*(d - e*x)^p*(d + e*x)^p*AppellF1[5, -p, 2 - p, 6, (e*x)/d, -((e*x)/d)])/(5*d^2*(1 - (e^2*x^2)/d^2)^p)

Maple [F]

\[\int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((-e^2*x^2 + d^2)^p*x^4/(e^2*x^2 + 2*d*e*x + d^2), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]

[In]

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

[Out]

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