\(\int \frac {(c^2-d^2 x^2)^p}{(c+d x)^{3/2}} \, dx\) [369]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}} \, dx=-\frac {2^{-\frac {3}{2}+p} \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c^2 d (1+p) \sqrt {c+d x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}} \, dx=-\frac {2^{-\frac {3}{2}+p} (c-d x) \left (1+\frac {d x}{c}\right )^{\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c d (1+p) \sqrt {c+d x}} \] Input:

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

Output:

-((2^(-3/2 + p)*(c - d*x)*(1 + (d*x)/c)^(1/2 - p)*(c^2 - d^2*x^2)^p*Hyperg 
eometric2F1[3/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c*d*(1 + p)*Sqrt[c + 
 d*x]))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {474, 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.

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

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

-((2^(-3/2 + p)*(1 + (d*x)/c)^(-1/2 - p)*(c^2 - d^2*x^2)^(1 + p)*Hypergeom 
etric2F1[3/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c^2*d*(1 + p)*Sqrt[c + 
d*x]))
 

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]))
 

rule 474

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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