3.305 \(\int \frac{(d^2-e^2 x^2)^p}{x^5 (d+e x)^4} \, dx\)

Optimal. Leaf size=216 \[ \frac{e^4 \left (p^2-21 p+70\right ) \left (d^2-e^2 x^2\right )^{p-3} \, _2F_1\left (1,p-3;p-2;1-\frac{e^2 x^2}{d^2}\right )}{4 d^2 (3-p)}+\frac{8 e^3 (6-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},4-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^7 x}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{p-3}}{4 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3 x^3}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{4 x^4} \]

[Out]

-(d^2*(d^2 - e^2*x^2)^(-3 + p))/(4*x^4) + (4*d*e*(d^2 - e^2*x^2)^(-3 + p))/(3*x^3) - (e^2*(17 - p)*(d^2 - e^2*
x^2)^(-3 + p))/(4*x^2) + (8*e^3*(6 - p)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[-1/2, 4 - p, 1/2, (e^2*x^2)/d^2])/
(3*d^7*x*(1 - (e^2*x^2)/d^2)^p) + (e^4*(70 - 21*p + p^2)*(d^2 - e^2*x^2)^(-3 + p)*Hypergeometric2F1[1, -3 + p,
-2 + p, 1 - (e^2*x^2)/d^2])/(4*d^2*(3 - p))

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Rubi [A]  time = 0.414663, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {852, 1807, 764, 365, 364, 266, 65} \[ \frac{e^4 \left (p^2-21 p+70\right ) \left (d^2-e^2 x^2\right )^{p-3} \, _2F_1\left (1,p-3;p-2;1-\frac{e^2 x^2}{d^2}\right )}{4 d^2 (3-p)}+\frac{8 e^3 (6-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},4-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^7 x}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{p-3}}{4 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3 x^3}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^2*(d^2 - e^2*x^2)^(-3 + p))/(4*x^4) + (4*d*e*(d^2 - e^2*x^2)^(-3 + p))/(3*x^3) - (e^2*(17 - p)*(d^2 - e^2*
x^2)^(-3 + p))/(4*x^2) + (8*e^3*(6 - p)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[-1/2, 4 - p, 1/2, (e^2*x^2)/d^2])/
(3*d^7*x*(1 - (e^2*x^2)/d^2)^p) + (e^4*(70 - 21*p + p^2)*(d^2 - e^2*x^2)^(-3 + p)*Hypergeometric2F1[1, -3 + p,
-2 + p, 1 - (e^2*x^2)/d^2])/(4*d^2*(3 - p))

Rule 852

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 1807

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])

Rule 764

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 365

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 266

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 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^5} \, dx\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (16 d^5 e-2 d^4 e^2 (17-p) x+16 d^3 e^3 x^2-4 d^2 e^4 x^3\right )}{x^4} \, dx}{4 d^2}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (6 d^6 e^2 (17-p)-32 d^5 e^3 (6-p) x+12 d^4 e^4 x^2\right )}{x^3} \, dx}{12 d^4}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^2}-\frac{\int \frac{\left (64 d^7 e^3 (6-p)-12 d^6 e^4 \left (70-21 p+p^2\right ) x\right ) \left (d^2-e^2 x^2\right )^{-4+p}}{x^2} \, dx}{24 d^6}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^2}-\frac{1}{3} \left (8 d e^3 (6-p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-4+p}}{x^2} \, dx+\frac{1}{2} \left (e^4 \left (70-21 p+p^2\right )\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-4+p}}{x} \, dx\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^2}+\frac{1}{4} \left (e^4 \left (70-21 p+p^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-4+p}}{x} \, dx,x,x^2\right )-\frac{\left (8 e^3 (6-p) \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 )^{-4+p}}{x^2} \, dx}{3 d^7}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^4}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}-\frac{e^2 (17-p) \left (d^2-e^2 x^2\right )^{-3+p}}{4 x^2}+\frac{8 e^3 (6-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},4-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^7 x}+\frac{e^4 \left (70-21 p+p^2\right ) \left (d^2-e^2 x^2\right )^{-3+p} \, _2F_1\left (1,-3+p;-2+p;1-\frac{e^2 x^2}{d^2}\right )}{4 d^2 (3-p)}\\ \end{align*}

Mathematica [B]  time = 0.833583, size = 505, normalized size = 2.34 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{240 d^3 e^2 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{24 d^5 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (2-p,-p;3-p;\frac{d^2}{e^2 x^2}\right )}{(p-2) x^4}+\frac{840 d e^4 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}+\frac{64 d^4 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x^3}+\frac{960 d^2 e^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{105 e^4 2^{p+3} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{45 e^4 2^{p+2} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{15 e^4 2^{p+1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^4 2^p (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{48 d^9} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((64*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) +
(960*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) + (240*d^3*e^2*Hyperge
ometric2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (105*2^(3 + p)*e^4*(d - e*
x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (24*d^5*Hypergeometric
2F1[2 - p, -p, 3 - p, d^2/(e^2*x^2)])/((-2 + p)*(1 - d^2/(e^2*x^2))^p*x^4) + (45*2^(2 + p)*e^4*(d - e*x)*Hyper
geometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (15*2^(1 + p)*e^4*(d - e*x)*Hy
pergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (3*2^p*e^4*(d - e*x)*Hyperg
eometric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (840*d*e^4*Hypergeometric2F1[-p
, -p, 1 - p, d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(48*d^9)

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Maple [F]  time = 0.681, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{5} \left ( ex+d \right ) ^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^p/x^5/(e*x+d)^4,x)

[Out]

int((-e^2*x^2+d^2)^p/x^5/(e*x+d)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^p/x^5/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*x^5), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{4} x^{9} + 4 \, d e^{3} x^{8} + 6 \, d^{2} e^{2} x^{7} + 4 \, d^{3} e x^{6} + d^{4} x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^p/x^5/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((-e^2*x^2 + d^2)^p/(e^4*x^9 + 4*d*e^3*x^8 + 6*d^2*e^2*x^7 + 4*d^3*e*x^6 + d^4*x^5), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{5} \left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**p/x**5/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**p/(x**5*(d + e*x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^p/x^5/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*x^5), x)