\(\int \frac {(A+B x) (c^2-d^2 x^2)^p}{(c+d x)^3} \, dx\) [120]

Optimal result

Integrand size = 27, antiderivative size = 128 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^3} \, dx=\frac {B \left (c^2-d^2 x^2\right )^{1+p}}{d^2 (1-2 p) (c+d x)^3}-\frac {2^p c^2 (3 B c+A d (1-2 p)) \left (\frac {c-d x}{c}\right )^{2-p} \left (c^2-d^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (-2+p,-p,-1+p,\frac {c+d x}{2 c}\right )}{d^2 (1-2 p) (2-p)} \] Output:

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

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^3} \, dx=-\frac {2^{-3+p} (c-d x) \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \left (2 B c \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {c-d x}{2 c}\right )+(-B c+A d) \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{c^3 d^2 (1+p)} \] Input:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((B*x + A)*(-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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*(c**2 - d**2*x**2)**p*a*d*p + (c**2 - d**2*x**2)**p*a*d + (c**2 - d* 
*2*x**2)**p*b*c + 2*(c**2 - d**2*x**2)**p*b*d*x - 8*int(((c**2 - d**2*x**2 
)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4*c*d**3*p*x**3 + 2 
*c*d**3*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a*c**2*d**3*p**3 + 8*int(((c* 
*2 - d**2*x**2)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4*c*d 
**3*p*x**3 + 2*c*d**3*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a*c**2*d**3*p** 
2 - 2*int(((c**2 - d**2*x**2)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c* 
*3*d*x - 4*c*d**3*p*x**3 + 2*c*d**3*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a 
*c**2*d**3*p - 16*int(((c**2 - d**2*x**2)**p*x)/(2*c**4*p - c**4 + 4*c**3* 
d*p*x - 2*c**3*d*x - 4*c*d**3*p*x**3 + 2*c*d**3*x**3 - 2*d**4*p*x**4 + d** 
4*x**4),x)*a*c*d**4*p**3*x + 16*int(((c**2 - d**2*x**2)**p*x)/(2*c**4*p - 
c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4*c*d**3*p*x**3 + 2*c*d**3*x**3 - 2*d** 
4*p*x**4 + d**4*x**4),x)*a*c*d**4*p**2*x - 4*int(((c**2 - d**2*x**2)**p*x) 
/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4*c*d**3*p*x**3 + 2*c*d**3 
*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a*c*d**4*p*x - 8*int(((c**2 - d**2*x 
**2)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4*c*d**3*p*x**3 
+ 2*c*d**3*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a*d**5*p**3*x**2 + 8*int(( 
(c**2 - d**2*x**2)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x - 2*c**3*d*x - 4* 
c*d**3*p*x**3 + 2*c*d**3*x**3 - 2*d**4*p*x**4 + d**4*x**4),x)*a*d**5*p**2* 
x**2 - 2*int(((c**2 - d**2*x**2)**p*x)/(2*c**4*p - c**4 + 4*c**3*d*p*x ...