\(\int \frac {(c^2-d^2 x^2)^p}{(c+d x)^3} \, dx\) [359]

Optimal result

Integrand size = 22, antiderivative size = 72 \[ \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^3} \, dx=-\frac {2^p c^2 \left (\frac {c-d x}{c}\right )^{2-p} \left (c^2-d^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (-2+p,-p,-1+p,\frac {c+d x}{2 c}\right )}{d (2-p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^3} \, dx=-\frac {2^{-3+p} (c-d x) \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c^3 d (1+p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {473, 79}

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.

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

-((2^(-3 + p)*(1 + (d*x)/c)^(-1 - p)*(c^2 - d^2*x^2)^(1 + p)*Hypergeometri 
c2F1[3 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c^4*d*(1 + p)))
 

Defintions of rubi rules used

rule 79

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

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^3} \, dx=\frac {-\left (-d^{2} x^{2}+c^{2}\right )^{p}-2 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) c^{2} d^{2} p -4 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) c \,d^{3} p x -2 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) d^{4} p \,x^{2}}{2 d \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

( - (c**2 - d**2*x**2)**p - 2*int(((c**2 - d**2*x**2)**p*x)/(c**4 + 2*c**3 
*d*x - 2*c*d**3*x**3 - d**4*x**4),x)*c**2*d**2*p - 4*int(((c**2 - d**2*x** 
2)**p*x)/(c**4 + 2*c**3*d*x - 2*c*d**3*x**3 - d**4*x**4),x)*c*d**3*p*x - 2 
*int(((c**2 - d**2*x**2)**p*x)/(c**4 + 2*c**3*d*x - 2*c*d**3*x**3 - d**4*x 
**4),x)*d**4*p*x**2)/(2*d*(c**2 + 2*c*d*x + d**2*x**2))