\(\int \frac {(A+B x) (c^2-d^2 x^2)^p}{(c+d x)^{3/2}} \, dx\) [125]

Optimal result

Integrand size = 29, antiderivative size = 147 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}} \, dx=-\frac {2 B \left (c^2-d^2 x^2\right )^{1+p}}{d^2 (1+4 p) (c+d x)^{3/2}}+\frac {2^{-\frac {3}{2}+p} (3 B c-A d (1+4 p)) \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c^2 d^2 (1+p) (1+4 p) \sqrt {c+d x}} \] Output:

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

Mathematica [A] (verified)

Time = 1.89 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}} \, dx=-\frac {(c-d x) \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \left (2 \sqrt {2} (B c-A d) (1+p) \left (1+\frac {d x}{c}\right )^p-2^p (3 B c-A d (1+4 p)) \sqrt {1+\frac {d x}{c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{2 \sqrt {2} c d^2 (1+p) (-1+2 p) \sqrt {c+d x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(1 + p))/(c*d^2*(1 - 2*p)*(c + d*x)^(3/2)) - 
(2^(-3/2 + p)*(3*B*c - A*d*(1 + 4*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 
^2*d^2*(1 - 2*p)*(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 671

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*( - 4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d*p - 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**3*p + c**3 + 4*c**2*d*p*x + c**2*d*x - 4*c*d**2*p*x**2 - c*d 
**2*x**2 - 4*d**3*p*x**3 - d**3*x**3),x)*a*c*d**3*p**3 - 16*int((sqrt(c + 
d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**3*p + c**3 + 4*c**2*d*p*x + c**2*d*x - 
 4*c*d**2*p*x**2 - c*d**2*x**2 - 4*d**3*p*x**3 - d**3*x**3),x)*a*c*d**3*p* 
*2 - 2*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**3*p + c**3 + 4*c* 
*2*d*p*x + c**2*d*x - 4*c*d**2*p*x**2 - c*d**2*x**2 - 4*d**3*p*x**3 - d**3 
*x**3),x)*a*c*d**3*p - 32*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c 
**3*p + c**3 + 4*c**2*d*p*x + c**2*d*x - 4*c*d**2*p*x**2 - c*d**2*x**2 - 4 
*d**3*p*x**3 - d**3*x**3),x)*a*d**4*p**3*x - 16*int((sqrt(c + d*x)*(c**2 - 
 d**2*x**2)**p*x)/(4*c**3*p + c**3 + 4*c**2*d*p*x + c**2*d*x - 4*c*d**2*p* 
x**2 - c*d**2*x**2 - 4*d**3*p*x**3 - d**3*x**3),x)*a*d**4*p**2*x - 2*int(( 
sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**3*p + c**3 + 4*c**2*d*p*x + c 
**2*d*x - 4*c*d**2*p*x**2 - c*d**2*x**2 - 4*d**3*p*x**3 - d**3*x**3),x)*a* 
d**4*p*x + 24*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(4*c**3*p + c**3 
 + 4*c**2*d*p*x + c**2*d*x - 4*c*d**2*p*x**2 - c*d**2*x**2 - 4*d**3*p*x**3 
 - d**3*x**3),x)*b*c**2*d**2*p**2 + 6*int((sqrt(c + d*x)*(c**2 - d**2*x**2 
)**p*x)/(4*c**3*p + c**3 + 4*c**2*d*p*x + c**2*d*x - 4*c*d**2*p*x**2 - ...