\(\int \frac {(A+B x) (c^2-d^2 x^2)^p}{\sqrt {c+d x}} \, dx\) [124]

Optimal result

Integrand size = 29, antiderivative size = 146 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{\sqrt {c+d x}} \, dx=-\frac {2 B \left (c^2-d^2 x^2\right )^{1+p}}{d^2 (3+4 p) \sqrt {c+d x}}+\frac {2^{-\frac {1}{2}+p} (B c-A d (3+4 p)) \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c d^2 (1+p) (3+4 p) \sqrt {c+d x}} \] Output:

-2*B*(-d^2*x^2+c^2)^(p+1)/d^2/(3+4*p)/(d*x+c)^(1/2)+2^(-1/2+p)*(B*c-A*d*(3 
+4*p))*(1+d*x/c)^(-1/2-p)*(-d^2*x^2+c^2)^(p+1)*hypergeom([p+1, 1/2-p],[2+p 
],1/2*(-d*x+c)/c)/c/d^2/(p+1)/(3+4*p)/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{\sqrt {c+d x}} \, dx=-\frac {(c-d x) \sqrt {c+d x} \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^p \left (\sqrt {2} (B c-A d) (1+p) \left (1+\frac {d x}{c}\right )^{\frac {1}{2}+p}-2^p (B c-A d (3+4 p)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{\sqrt {2} c d^2 (1+p) (1+2 p)} \] Input:

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

Output:

-(((c - d*x)*Sqrt[c + d*x]*(1 + (d*x)/c)^(-1/2 - p)*(c^2 - d^2*x^2)^p*(Sqr 
t[2]*(B*c - A*d)*(1 + p)*(1 + (d*x)/c)^(1/2 + p) - 2^p*(B*c - A*d*(3 + 4*p 
))*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)]))/(Sqrt[2]*c 
*d^2*(1 + p)*(1 + 2*p)))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {672, 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.

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

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

(-2*B*(c^2 - d^2*x^2)^(1 + p))/(d^2*(3 + 4*p)*Sqrt[c + d*x]) - (2^(-1/2 + 
p)*(A - (B*c)/(3*d + 4*d*p))*(1 + (d*x)/c)^(-1/2 - p)*(c^2 - d^2*x^2)^(1 + 
 p)*Hypergeometric2F1[1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c*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])
 

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^ 
2 + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
 

Maple [F]

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

Input:

int((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x)
 

Output:

int((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x)
 

Fricas [F]

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

integrate((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((B*x+A)*(-d**2*x**2+c**2)**p/(d*x+c)**(1/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**p*(A + B*x)/sqrt(c + d*x), x)
 

Maxima [F]

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

integrate((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate((B*x + A)*(-d^2*x^2 + c^2)^p/sqrt(d*x + c), x)
 

Giac [F]

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

integrate((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate((B*x + A)*(-d^2*x^2 + c^2)^p/sqrt(d*x + c), x)
 

Mupad [F(-1)]

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

int(((c^2 - d^2*x^2)^p*(A + B*x))/(c + d*x)^(1/2),x)
 

Output:

int(((c^2 - d^2*x^2)^p*(A + B*x))/(c + d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{\sqrt {c+d x}} \, dx=\frac {8 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} a d p +6 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} a d -4 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} b c +2 \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} b d x +64 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) a \,d^{3} p^{3}+96 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) a \,d^{3} p^{2}+36 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) a \,d^{3} p -16 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) b c \,d^{2} p^{2}-12 \left (\int \frac {\sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right )^{p} x}{-4 d^{2} p \,x^{2}-3 d^{2} x^{2}+4 c^{2} p +3 c^{2}}d x \right ) b c \,d^{2} p}{d^{2} \left (4 p +3\right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^p/(d*x+c)^(1/2),x)
 

Output:

(2*(4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d*p + 3*sqrt(c + d*x)*(c**2 - 
d**2*x**2)**p*a*d - 2*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*b*c + sqrt(c + d 
*x)*(c**2 - d**2*x**2)**p*b*d*x + 32*int((sqrt(c + d*x)*(c**2 - d**2*x**2) 
**p*x)/(4*c**2*p + 3*c**2 - 4*d**2*p*x**2 - 3*d**2*x**2),x)*a*d**3*p**3 + 
48*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d**2 
*p*x**2 - 3*d**2*x**2),x)*a*d**3*p**2 + 18*int((sqrt(c + d*x)*(c**2 - d**2 
*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d**2*p*x**2 - 3*d**2*x**2),x)*a*d**3*p 
 - 8*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d* 
*2*p*x**2 - 3*d**2*x**2),x)*b*c*d**2*p**2 - 6*int((sqrt(c + d*x)*(c**2 - d 
**2*x**2)**p*x)/(4*c**2*p + 3*c**2 - 4*d**2*p*x**2 - 3*d**2*x**2),x)*b*c*d 
**2*p))/(d**2*(4*p + 3))