\(\int \frac {(d^2-e^2 x^2)^p}{x (d+e x)^3} \, dx\) [291]

Optimal result

Integrand size = 25, antiderivative size = 175 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\frac {2 d \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}-\frac {2 e (4-3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^4 (3-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (1,-1+p,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1-p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {866, 1666, 457, 80, 67, 396, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^{p-1} \operatorname {Hypergeometric2F1}\left (1,p-1,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1-p)}-\frac {e x \left (d^2-e^2 x^2\right )^{p-2}}{3-2 p}+\frac {2 d \left (d^2-e^2 x^2\right )^{p-2}}{2-p}-\frac {2 e (4-3 p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^4 (3-2 p)} \]

[In]

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

[Out]

(2*d*(d^2 - e^2*x^2)^(-2 + p))/(2 - p) - (e*x*(d^2 - e^2*x^2)^(-2 + p))/(3 - 2*p) - (2*e*(4 - 3*p)*x*(d^2 - e^
2*x^2)^p*Hypergeometric2F1[1/2, 3 - p, 3/2, (e^2*x^2)/d^2])/(d^4*(3 - 2*p)*(1 - (e^2*x^2)/d^2)^p) + ((d^2 - e^
2*x^2)^(-1 + p)*Hypergeometric2F1[1, -1 + p, p, 1 - (e^2*x^2)/d^2])/(2*d*(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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c
, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 251

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

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 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 \frac {(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx \\ & = \int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (d^3+3 d e^2 x^2\right )}{x} \, dx+\int \left (d^2-e^2 x^2\right )^{-3+p} \left (-3 d^2 e-e^3 x^2\right ) \, dx \\ & = -\frac {e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}+\frac {1}{2} \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p} \left (d^3+3 d e^2 x\right )}{x} \, dx,x,x^2\right )-\frac {\left (2 d^2 e (4-3 p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{3-2 p} \\ & = \frac {2 d \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}+\frac {1}{2} d \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p}}{x} \, dx,x,x^2\right )-\frac {\left (2 e (4-3 p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^{-3+p} \, dx}{d^4 (3-2 p)} \\ & = \frac {2 d \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}-\frac {2 e (4-3 p) x \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},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^4 (3-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (1,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{2 d (1-p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.87 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\frac {2^{-3+p} \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )+2 p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+d p \left (1-\frac {d^2}{e^2 x^2}\right )^p \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )-e p \left (1-\frac {d^2}{e^2 x^2}\right )^p x \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+4 d \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )+4 d p \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )\right )}{d^4 p (1+p)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x \left (e x +d \right )^{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x \left (d + e x\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x\,{\left (d+e\,x\right )}^3} \,d x \]

[In]

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

[Out]

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