3.3.85 \(\int \frac {(d^2-e^2 x^2)^p}{x^5 (d+e x)^2} \, dx\) [285]

Optimal. Leaf size=145 \[ -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {3}{2},2-p;-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}+\frac {e^4 (5-p) \left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (2,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{4 d^4 (1-p)} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1821, 778, 372, 371, 272, 67} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right )^{p-1}}{4 x^4}+\frac {e^4 (5-p) \left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (2,p-1;p;1-\frac {e^2 x^2}{d^2}\right )}{4 d^4 (1-p)}+\frac {2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac {3}{2},2-p;-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 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 778

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

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 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps

\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^2} \, dx &=\int \frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p}}{x^5} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}-\frac {\int \frac {\left (8 d^3 e-2 d^2 e^2 (5-p) x\right ) \left (d^2-e^2 x^2\right )^{-2+p}}{x^4} \, dx}{4 d^2}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}-(2 d e) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x^4} \, dx+\frac {1}{2} \left (e^2 (5-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x^3} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}+\frac {1}{4} \left (e^2 (5-p)\right ) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p}}{x^2} \, 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 \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^{-2+p}}{x^4} \, dx}{d^3}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {3}{2},2-p;-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}+\frac {e^4 (5-p) \left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (2,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{4 d^4 (1-p)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(389\) vs. \(2(145)=290\).
time = 0.60, size = 389, normalized size = 2.68 \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^p \left (\frac {8 d^4 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {3}{2},-p;-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x^3}+\frac {48 d^2 e^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}+\frac {18 d^3 e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {15\ 2^{1+p} e^4 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (1-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{1+p}+\frac {6 d^5 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (2-p,-p;3-p;\frac {d^2}{e^2 x^2}\right )}{(-2+p) x^4}+\frac {3\ 2^p e^4 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{1+p}+\frac {30 d e^4 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )}{p}\right )}{12 d^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((8*d^4*e*Hypergeometric2F1[-3/2, -p, -1/2, (e^2*x^2)/d^2])/(x^3*(1 - (e^2*x^2)/d^2)^p) + (
48*d^2*e^3*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) + (18*d^3*e^2*Hypergeome
tric2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (15*2^(1 + p)*e^4*(d - e*x)*H
ypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (6*d^5*Hypergeometric2F1[2
- p, -p, 3 - p, d^2/(e^2*x^2)])/((-2 + p)*(1 - d^2/(e^2*x^2))^p*x^4) + (3*2^p*e^4*(d - e*x)*Hypergeometric2F1
[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (30*d*e^4*Hypergeometric2F1[-p, -p, 1 - p,
d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(12*d^7)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{5} \left (e x +d \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^p/x^5/(e*x+d)^2,x)

[Out]

int((-e^2*x^2+d^2)^p/x^5/(e*x+d)^2,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((-e^2*x^2+d^2)^p/x^5/(e*x+d)^2,x, algorithm="maxima")

[Out]

integrate((-x^2*e^2 + d^2)^p/((x*e + d)^2*x^5), 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((-e^2*x^2+d^2)^p/x^5/(e*x+d)^2,x, algorithm="fricas")

[Out]

integral((-x^2*e^2 + d^2)^p/(x^7*e^2 + 2*d*x^6*e + d^2*x^5), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{5} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**p/x**5/(e*x+d)**2,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**p/(x**5*(d + e*x)**2), 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((-e^2*x^2+d^2)^p/x^5/(e*x+d)^2,x, algorithm="giac")

[Out]

integrate((-x^2*e^2 + d^2)^p/((x*e + d)^2*x^5), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^5\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^p/(x^5*(d + e*x)^2),x)

[Out]

int((d^2 - e^2*x^2)^p/(x^5*(d + e*x)^2), x)

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