3.3.86 \(\int \frac {x^4 (d^2-e^2 x^2)^p}{(d+e x)^3} \, dx\) [286]

Optimal. Leaf size=220 \[ -\frac {2 d^6 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{2 e^5 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^5 (1+p)}+\frac {2 (8+p) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (1+2 p)} \]

[Out]

-2*d^6*(-e^2*x^2+d^2)^(-2+p)/e^5/(2-p)-3*d*x^5*(-e^2*x^2+d^2)^(-2+p)/(1+2*p)+9/2*d^4*(-e^2*x^2+d^2)^(-1+p)/e^5
/(1-p)+3*d^2*(-e^2*x^2+d^2)^p/e^5/p-1/2*(-e^2*x^2+d^2)^(1+p)/e^5/(1+p)+2/5*(8+p)*x^5*(-e^2*x^2+d^2)^p*hypergeo
m([5/2, 3-p],[7/2],e^2*x^2/d^2)/d^3/(1+2*p)/((1-e^2*x^2/d^2)^p)

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Rubi [A]
time = 0.15, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1666, 470, 372, 371, 457, 78} \begin {gather*} -\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p-2}}{2 p+1}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^5 (p+1)}-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p-1}}{2 e^5 (1-p)}+\frac {2 (p+8) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {5}{2},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(d^2 - e^2*x^2)^p)/(d + e*x)^3,x]

[Out]

(-2*d^6*(d^2 - e^2*x^2)^(-2 + p))/(e^5*(2 - p)) - (3*d*x^5*(d^2 - e^2*x^2)^(-2 + p))/(1 + 2*p) + (9*d^4*(d^2 -
e^2*x^2)^(-1 + p))/(2*e^5*(1 - p)) + (3*d^2*(d^2 - e^2*x^2)^p)/(e^5*p) - (d^2 - e^2*x^2)^(1 + p)/(2*e^5*(1 +
p)) + (2*(8 + p)*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, 3 - p, 7/2, (e^2*x^2)/d^2])/(5*d^3*(1 + 2*p)*(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 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*} \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx &=\int x^4 (d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p} \, dx\\ &=\int x^4 \left (d^2-e^2 x^2\right )^{-3+p} \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^5 \left (d^2-e^2 x^2\right )^{-3+p} \left (-3 d^2 e-e^3 x^2\right ) \, dx\\ &=-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-3+p} \left (-3 d^2 e-e^3 x\right ) \, dx,x,x^2\right )+\frac {\left (2 d^3 (8+p)\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{1+2 p}\\ &=-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int \left (-\frac {4 d^6 \left (d^2-e^2 x\right )^{-3+p}}{e^3}+\frac {9 d^4 \left (d^2-e^2 x\right )^{-2+p}}{e^3}-\frac {6 d^2 \left (d^2-e^2 x\right )^{-1+p}}{e^3}+\frac {\left (d^2-e^2 x\right )^p}{e^3}\right ) \, dx,x,x^2\right )+\frac {\left (2 (8+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 )^{-3+p} \, dx}{d^3 (1+2 p)}\\ &=-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{2 e^5 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^5 (1+p)}+\frac {2 (8+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},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (1+2 p)}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 245, normalized size = 1.11 \begin {gather*} -\frac {2^{-3+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 (24 d e (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \, _2F_1\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 (4 d \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p+4 e x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p+24 d \, _2F_1\left (1-p,1+p;2+p;\frac {d-e x}{2 d}\right )-8 d \, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )+d \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )\right )\right )}{e^5 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d^2 - e^2*x^2)^p)/(d + e*x)^3,x]

[Out]

-((2^(-3 + p)*(d^2 - e^2*x^2)^p*(24*d*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*(4*d*(1/2 + (e*x)/(2*d))^p + 4*e*x*(1/2 + (e*x)/(2*d))^p + 24*d*Hyp
ergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - 8*d*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2
*d)] + d*Hypergeometric2F1[3 - 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))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(-e**2*x**2+d**2)**p/(e*x+d)**3,x)

[Out]

Integral(x**4*(-(-d + e*x)*(d + e*x))**p/(d + e*x)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^p/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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