\(\int \frac {x^4 (d^2-e^2 x^2)^p}{(d+e x)^4} \, dx\) [296]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 265 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 \left (16+15 p+p^2\right ) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )} \]

[Out]

-4*d^7*(-e^2*x^2+d^2)^(-3+p)/e^5/(3-p)+d^2*(13+12*p)*x^5*(-e^2*x^2+d^2)^(-3+p)/(-4*p^2+1)-e^2*x^7*(-e^2*x^2+d^
2)^(-3+p)/(1+2*p)+10*d^5*(-e^2*x^2+d^2)^(-2+p)/e^5/(2-p)-8*d^3*(-e^2*x^2+d^2)^(-1+p)/e^5/(1-p)-2*d*(-e^2*x^2+d
^2)^p/e^5/p-4/5*(p^2+15*p+16)*x^5*(-e^2*x^2+d^2)^p*hypergeom([5/2, 4-p],[7/2],e^2*x^2/d^2)/d^4/(-4*p^2+1)/((1-
e^2*x^2/d^2)^p)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 265, 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, 1281, 470, 372, 371, 457, 78} \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac {4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)} \]

[In]

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

[Out]

(-4*d^7*(d^2 - e^2*x^2)^(-3 + p))/(e^5*(3 - p)) + (d^2*(13 + 12*p)*x^5*(d^2 - e^2*x^2)^(-3 + p))/(1 - 4*p^2) -
 (e^2*x^7*(d^2 - e^2*x^2)^(-3 + p))/(1 + 2*p) + (10*d^5*(d^2 - e^2*x^2)^(-2 + p))/(e^5*(2 - p)) - (8*d^3*(d^2
- e^2*x^2)^(-1 + p))/(e^5*(1 - p)) - (2*d*(d^2 - e^2*x^2)^p)/(e^5*p) - (4*(16 + 15*p + p^2)*x^5*(d^2 - e^2*x^2
)^p*Hypergeometric2F1[5/2, 4 - p, 7/2, (e^2*x^2)/d^2])/(5*d^4*(1 - 4*p^2)*(1 - (e^2*x^2)/d^2)^p)

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 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 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^(q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
 - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
 IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

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)^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx \\ & = \int x^5 \left (d^2-e^2 x^2\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x^2\right ) \, dx+\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (d^4+6 d^2 e^2 x^2+e^4 x^4\right ) \, dx \\ & = -\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x\right ) \, dx,x,x^2\right )-\frac {\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (1+2 p)-d^2 e^4 (13+12 p) x^2\right ) \, dx}{e^2 (1+2 p)} \\ & = \frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int \left (-\frac {8 d^7 \left (d^2-e^2 x\right )^{-4+p}}{e^3}+\frac {20 d^5 \left (d^2-e^2 x\right )^{-3+p}}{e^3}-\frac {16 d^3 \left (d^2-e^2 x\right )^{-2+p}}{e^3}+\frac {4 d \left (d^2-e^2 x\right )^{-1+p}}{e^3}\right ) \, dx,x,x^2\right )-\frac {\left (4 d^4 \left (16+15 p+p^2\right )\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{1-4 p^2} \\ & = -\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (4 \left (16+15 p+p^2\right ) \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 )^{-4+p} \, dx}{d^4 \left (1-4 p^2\right )} \\ & = -\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 \left (16+15 p+p^2\right ) 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},4-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {2^{-4+p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (16 e (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (32 \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )-24 \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+8 \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )-\operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )\right )}{e^5 (1+p)} \]

[In]

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

[Out]

(2^(-4 + p)*(d^2 - e^2*x^2)^p*(16*e*(1 + p)*x*(1/2 + (e*x)/(2*d))^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/
d^2] + (d - e*x)*(1 - (e^2*x^2)/d^2)^p*(32*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - 24*Hyperg
eometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] + 8*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]
- Hypergeometric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])))/(e^5*(1 + p)*(1 + (e*x)/d)^p*(1 - (e^2*x^2)/d^2)^
p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((-e^2*x^2 + d^2)^p*x^4/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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