3.3.93 \(\int \frac {(d^2-e^2 x^2)^p}{x^3 (d+e x)^3} \, dx\) [293]

3.3.93.1 Optimal result

Integrand size = 25, antiderivative size = 173 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {2 e^3 (8-3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6}+\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (1,-2+p,-1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)} \]

output

-1/2*d*(-e^2*x^2+d^2)^(-2+p)/x^2+3*e*(-e^2*x^2+d^2)^(-2+p)/x-2*e^3*(8-3*p) 
*x*(-e^2*x^2+d^2)^p*hypergeom([1/2, 3-p],[3/2],e^2*x^2/d^2)/d^6/((1-e^2*x^ 
2/d^2)^p)+1/2*e^2*(6-p)*(-e^2*x^2+d^2)^(-2+p)*hypergeom([1, -2+p],[-1+p],1 
-e^2*x^2/d^2)/d/(2-p)
 

3.3.93.2 Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.97 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (\frac {24 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\frac {4 d^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {3\ 2^{3+p} e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^{1+p} e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {2^p e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {24 d e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{8 d^6} \]

input

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

output

((d^2 - e^2*x^2)^p*((24*d^2*e*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d 
^2])/(x*(1 - (e^2*x^2)/d^2)^p) + (4*d^3*Hypergeometric2F1[1 - p, -p, 2 - p 
, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (3*2^(3 + p)*e^2* 
(d - e*x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p 
)*(1 + (e*x)/d)^p) + (3*2^(1 + p)*e^2*(d - e*x)*Hypergeometric2F1[2 - p, 1 
 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (2^p*e^2*(d - e 
*x)*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + 
 (e*x)/d)^p) + (24*d*e^2*Hypergeometric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/ 
(p*(1 - d^2/(e^2*x^2))^p)))/(8*d^6)
 

3.3.93.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {570, 543, 354, 27, 87, 75, 359, 238, 237}

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.

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 359

\(\displaystyle -2 e^3 (8-3 p) \int \left (d^2-e^2 x^2\right )^{p-3}dx+\frac {1}{2} d \left (\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {\left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}\)

\(\Big \downarrow \) 238

\(\displaystyle -\frac {2 e^3 (8-3 p) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \int \left (1-\frac {e^2 x^2}{d^2}\right )^{p-3}dx}{d^6}+\frac {1}{2} d \left (\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {\left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {1}{2} d \left (\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d^2 (2-p)}-\frac {\left (d^2-e^2 x^2\right )^{p-2}}{x^2}\right )+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac {2 e^3 (8-3 p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6}\)

input

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

output

(3*e*(d^2 - e^2*x^2)^(-2 + p))/x - (2*e^3*(8 - 3*p)*x*(d^2 - e^2*x^2)^p*Hy 
pergeometric2F1[1/2, 3 - p, 3/2, (e^2*x^2)/d^2])/(d^6*(1 - (e^2*x^2)/d^2)^ 
p) + (d*(-((d^2 - e^2*x^2)^(-2 + p)/x^2) + (e^2*(6 - p)*(d^2 - e^2*x^2)^(- 
2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 - (e^2*x^2)/d^2])/(d^2*(2 - 
p))))/2
 

3.3.93.3.1 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 237

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

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

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])
 

3.3.93.4 Maple [F]

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

input

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

output

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

3.3.93.5 Fricas [F]

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

input

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

output

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

3.3.93.6 Sympy [F]

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

input

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

output

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

3.3.93.7 Maxima [F]

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

input

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

output

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

3.3.93.8 Giac [F]

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

input

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

output

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

3.3.93.9 Mupad [F(-1)]

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

input

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

output

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