\(\int (c+d x)^{3/2} (c^2-d^2 x^2)^p \, dx\) [366]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^p \, dx=-\frac {2^{\frac {3}{2}+p} \sqrt {c+d x} \left (1+\frac {d x}{c}\right )^{-\frac {3}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d (1+p)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.96 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.22 \[ \int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^p \, dx=\frac {2^{-1+p} \sqrt {c+d x} \left (1-\frac {d x}{c}\right )^{-p} \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-2 p} \left (d^2 (1+p) x^2 (c-d x)^p (c+d x)^p \left (\frac {1}{2}+\frac {d x}{2 c}\right )^p \operatorname {AppellF1}\left (2,-p,-\frac {1}{2}-p,3,\frac {d x}{c},-\frac {d x}{c}\right )-2 \sqrt {2} c (c-d x) \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{d (1+p)} \] Input:

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

Output:

(2^(-1 + p)*Sqrt[c + d*x]*(1 + (d*x)/c)^(-1/2 - 2*p)*(d^2*(1 + p)*x^2*(c - 
 d*x)^p*(c + d*x)^p*(1/2 + (d*x)/(2*c))^p*AppellF1[2, -p, -1/2 - p, 3, (d* 
x)/c, -((d*x)/c)] - 2*Sqrt[2]*c*(c - d*x)*(c^2 - d^2*x^2)^p*(1 - (d^2*x^2) 
/c^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)]))/(d*(1 
 + p)*(1 - (d*x)/c)^p)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 85, 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.

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

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^p \, dx=\frac {32 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2} p^{2}+48 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2} p +6 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2}+8 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c d p x +12 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} c d x +8 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{2} p \,x^{2}+6 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{2} x^{2}+1024 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-16 d^{2} p^{2} x^{2}-32 d^{2} p \,x^{2}+16 c^{2} p^{2}-15 d^{2} x^{2}+32 c^{2} p +15 c^{2}}d x \right ) c^{2} d^{2} p^{5}+4096 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-16 d^{2} p^{2} x^{2}-32 d^{2} p \,x^{2}+16 c^{2} p^{2}-15 d^{2} x^{2}+32 c^{2} p +15 c^{2}}d x \right ) c^{2} d^{2} p^{4}+5824 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-16 d^{2} p^{2} x^{2}-32 d^{2} p \,x^{2}+16 c^{2} p^{2}-15 d^{2} x^{2}+32 c^{2} p +15 c^{2}}d x \right ) c^{2} d^{2} p^{3}+3456 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-16 d^{2} p^{2} x^{2}-32 d^{2} p \,x^{2}+16 c^{2} p^{2}-15 d^{2} x^{2}+32 c^{2} p +15 c^{2}}d x \right ) c^{2} d^{2} p^{2}+720 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-16 d^{2} p^{2} x^{2}-32 d^{2} p \,x^{2}+16 c^{2} p^{2}-15 d^{2} x^{2}+32 c^{2} p +15 c^{2}}d x \right ) c^{2} d^{2} p}{d \left (16 p^{2}+32 p +15\right )} \] Input:

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

Output:

(2*(16*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**2*p**2 + 24*sqrt(c + d*x)*(c 
**2 - d**2*x**2)**p*c**2*p + 3*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**2 + 
4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c*d*p*x + 6*sqrt(c + d*x)*(c**2 - d* 
*2*x**2)**p*c*d*x + 4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d**2*p*x**2 + 3* 
sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d**2*x**2 + 512*int((sqrt(c + d*x)*(c* 
*2 - d**2*x**2)**p*x)/(16*c**2*p**2 + 32*c**2*p + 15*c**2 - 16*d**2*p**2*x 
**2 - 32*d**2*p*x**2 - 15*d**2*x**2),x)*c**2*d**2*p**5 + 2048*int((sqrt(c 
+ d*x)*(c**2 - d**2*x**2)**p*x)/(16*c**2*p**2 + 32*c**2*p + 15*c**2 - 16*d 
**2*p**2*x**2 - 32*d**2*p*x**2 - 15*d**2*x**2),x)*c**2*d**2*p**4 + 2912*in 
t((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(16*c**2*p**2 + 32*c**2*p + 15*c 
**2 - 16*d**2*p**2*x**2 - 32*d**2*p*x**2 - 15*d**2*x**2),x)*c**2*d**2*p**3 
 + 1728*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(16*c**2*p**2 + 32*c** 
2*p + 15*c**2 - 16*d**2*p**2*x**2 - 32*d**2*p*x**2 - 15*d**2*x**2),x)*c**2 
*d**2*p**2 + 360*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(16*c**2*p**2 
 + 32*c**2*p + 15*c**2 - 16*d**2*p**2*x**2 - 32*d**2*p*x**2 - 15*d**2*x**2 
),x)*c**2*d**2*p))/(d*(16*p**2 + 32*p + 15))