\(\int \frac {(A+B x) (c^2-d^2 x^2)^p}{(c+d x)^2} \, dx\) [119]

Optimal result

Integrand size = 27, antiderivative size = 114 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^2} \, dx=-\frac {B \left (c^2-d^2 x^2\right )^{1+p}}{2 d^2 p (c+d x)^2}+\frac {2^p c (B c-A d p) \left (\frac {c-d x}{c}\right )^{1-p} \left (c^2-d^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (-1+p,-p,p,\frac {c+d x}{2 c}\right )}{d^2 (1-p) p} \] Output:

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^2} \, dx=-\frac {2^{-2+p} (c-d x) \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \left (2 B c \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {c-d x}{2 c}\right )+(-B c+A d) \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{c^2 d^2 (1+p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.21, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(1 + p))/(2*c*d^2*(1 - p)*(c + d*x)^2) - (2^( 
-1 + p)*(B*c - A*d*p)*(1 + (d*x)/c)^(-1 - p)*(c^2 - d^2*x^2)^(1 + p)*Hyper 
geometric2F1[1 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c^3*d^2*(1 - 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^p}{(c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,1,1,1,0,0]%%%}+%%%{1,[0,0,0,1,1,1]%%%} / %%%{1,[0,0, 
0,0,1,0]%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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