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

3.3.85.1 Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^2} \, dx=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}+\frac {e^4 (5-p) \left (d^2-e^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,-1+p,p,1-\frac {e^2 x^2}{d^2}\right )}{4 d^4 (1-p)} \]

output

-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)
 

3.3.85.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(389\) vs. \(2(145)=290\).

Time = 0.90 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.68 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^2} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (\frac {8 d^4 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x^3}+\frac {48 d^2 e^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\frac {18 d^3 e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {15\ 2^{1+p} e^4 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {6 d^5 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,-p,3-p,\frac {d^2}{e^2 x^2}\right )}{(-2+p) x^4}+\frac {3\ 2^p e^4 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {30 d e^4 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{12 d^7} \]

input

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

output

((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*Hypergeometr 
ic2F1[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)*Hypergeometric2F1[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)
 

3.3.85.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {570, 543, 27, 279, 278, 354, 87, 75}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^2} \, dx\)

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{p-2}}{x^5}dx\)

\(\Big \downarrow \) 543

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-2} \left (d^2+e^2 x^2\right )}{x^5}dx+\int -\frac {2 d e \left (d^2-e^2 x^2\right )^{p-2}}{x^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-2} \left (d^2+e^2 x^2\right )}{x^5}dx-2 d e \int \frac {\left (d^2-e^2 x^2\right )^{p-2}}{x^4}dx\)

\(\Big \downarrow \) 279

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-2} \left (d^2+e^2 x^2\right )}{x^5}dx-\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^{p-2}}{x^4}dx}{d^3}\)

\(\Big \downarrow \) 278

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{p-2} \left (d^2+e^2 x^2\right )}{x^5}dx+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (d^2-e^2 x^2\right )^{p-2} \left (d^2+e^2 x^2\right )}{x^6}dx^2+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} e^2 (5-p) \int \frac {\left (d^2-e^2 x^2\right )^{p-2}}{x^4}dx^2-\frac {\left (d^2-e^2 x^2\right )^{p-1}}{2 x^4}\right )+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {1}{2} \left (\frac {e^4 (5-p) \left (d^2-e^2 x^2\right )^{p-1} \operatorname {Hypergeometric2F1}\left (2,p-1,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^4 (1-p)}-\frac {\left (d^2-e^2 x^2\right )^{p-1}}{2 x^4}\right )+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^3 x^3}\)

input

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

output

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

3.3.85.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

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] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

rule 570

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

3.3.85.4 Maple [F]

\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{5} \left (e x +d \right )^{2}}d x\]

input

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

output

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

3.3.85.5 Fricas [F]

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

input

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

output

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

3.3.85.6 Sympy [F]

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

input

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

output

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

3.3.85.7 Maxima [F]

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

input

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

output

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

3.3.85.8 Giac [F]

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

input

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

output

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

3.3.85.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^5\,{\left (d+e\,x\right )}^2} \,d x \]

input

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

output

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