\(\int (1-\sqrt {b} x)^m (1+\sqrt {b} x)^m (c+d x)^p \, dx\) [1835]

Optimal result

Integrand size = 31, antiderivative size = 125 \[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\frac {(c+d x)^{1+p} \left (1-b x^2\right )^m \left (1-\frac {c+d x}{c-\frac {d}{\sqrt {b}}}\right )^{-m} \left (1-\frac {c+d x}{c+\frac {d}{\sqrt {b}}}\right )^{-m} \operatorname {AppellF1}\left (1+p,-m,-m,2+p,\frac {c+d x}{c-\frac {d}{\sqrt {b}}},\frac {c+d x}{c+\frac {d}{\sqrt {b}}}\right )}{d (1+p)} \] Output:

(d*x+c)^(p+1)*(-b*x^2+1)^m*AppellF1(p+1,-m,-m,2+p,(d*x+c)/(c-d/b^(1/2)),(d 
*x+c)/(c+d/b^(1/2)))/d/(p+1)/((1-(d*x+c)/(c-d/b^(1/2)))^m)/((1-(d*x+c)/(c+ 
d/b^(1/2)))^m)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.12 \[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\frac {\left (\frac {d \left (\sqrt {\frac {1}{b}}-x\right )}{c+\sqrt {\frac {1}{b}} d}\right )^{-m} \left (\frac {d \left (\sqrt {\frac {1}{b}}+x\right )}{-c+\sqrt {\frac {1}{b}} d}\right )^{-m} (c+d x)^{1+p} \left (1-b x^2\right )^m \operatorname {AppellF1}\left (1+p,-m,-m,2+p,\frac {c+d x}{c-\sqrt {\frac {1}{b}} d},\frac {c+d x}{c+\sqrt {\frac {1}{b}} d}\right )}{d (1+p)} \] Input:

Integrate[(1 - Sqrt[b]*x)^m*(1 + Sqrt[b]*x)^m*(c + d*x)^p,x]
 

Output:

((c + d*x)^(1 + p)*(1 - b*x^2)^m*AppellF1[1 + p, -m, -m, 2 + p, (c + d*x)/ 
(c - Sqrt[b^(-1)]*d), (c + d*x)/(c + Sqrt[b^(-1)]*d)])/(d*(1 + p)*((d*(Sqr 
t[b^(-1)] - x))/(c + Sqrt[b^(-1)]*d))^m*((d*(Sqrt[b^(-1)] + x))/(-c + Sqrt 
[b^(-1)]*d))^m)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {156, 155}

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.

\(\displaystyle \int \left (1-\sqrt {b} x\right )^m \left (\sqrt {b} x+1\right )^m (c+d x)^p \, dx\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle -\frac {2^m \left (1-\sqrt {b} x\right )^{m+1} (c+d x)^p \left (\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+d}\right )^{-p} \operatorname {AppellF1}\left (m+1,-m,-p,m+2,\frac {1}{2} \left (1-\sqrt {b} x\right ),\frac {d \left (1-\sqrt {b} x\right )}{\sqrt {b} c+d}\right )}{\sqrt {b} (m+1)}\)

Input:

Int[(1 - Sqrt[b]*x)^m*(1 + Sqrt[b]*x)^m*(c + d*x)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

Maple [F]

\[\int \left (1-\sqrt {b}\, x \right )^{m} \left (1+\sqrt {b}\, x \right )^{m} \left (x d +c \right )^{p}d x\]

Input:

int((1-b^(1/2)*x)^m*(1+b^(1/2)*x)^m*(d*x+c)^p,x)
 

Output:

int((1-b^(1/2)*x)^m*(1+b^(1/2)*x)^m*(d*x+c)^p,x)
 

Fricas [F]

\[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\int { {\left (d x + c\right )}^{p} {\left (\sqrt {b} x + 1\right )}^{m} {\left (-\sqrt {b} x + 1\right )}^{m} \,d x } \] Input:

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

Output:

integral((d*x + c)^p*(sqrt(b)*x + 1)^m*(-sqrt(b)*x + 1)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\text {Timed out} \] Input:

integrate((1-b**(1/2)*x)**m*(1+b**(1/2)*x)**m*(d*x+c)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\int { {\left (d x + c\right )}^{p} {\left (\sqrt {b} x + 1\right )}^{m} {\left (-\sqrt {b} x + 1\right )}^{m} \,d x } \] Input:

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

Output:

integrate((d*x + c)^p*(sqrt(b)*x + 1)^m*(-sqrt(b)*x + 1)^m, x)
 

Giac [F]

\[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\int { {\left (d x + c\right )}^{p} {\left (\sqrt {b} x + 1\right )}^{m} {\left (-\sqrt {b} x + 1\right )}^{m} \,d x } \] Input:

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

Output:

integrate((d*x + c)^p*(sqrt(b)*x + 1)^m*(-sqrt(b)*x + 1)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\int {\left (1-\sqrt {b}\,x\right )}^m\,{\left (\sqrt {b}\,x+1\right )}^m\,{\left (c+d\,x\right )}^p \,d x \] Input:

int((1 - b^(1/2)*x)^m*(b^(1/2)*x + 1)^m*(c + d*x)^p,x)
 

Output:

int((1 - b^(1/2)*x)^m*(b^(1/2)*x + 1)^m*(c + d*x)^p, x)
 

Reduce [F]

\[ \int \left (1-\sqrt {b} x\right )^m \left (1+\sqrt {b} x\right )^m (c+d x)^p \, dx=\text {too large to display} \] Input:

int((1-b^(1/2)*x)^m*(1+b^(1/2)*x)^m*(d*x+c)^p,x)
 

Output:

((c + d*x)**p*(sqrt(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*b*c*x - (c + d*x)** 
p*(sqrt(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*d + 2*int(((c + d*x)**p*(sqrt(b 
)*x + 1)**m*( - sqrt(b)*x + 1)**m*x**2)/(2*b*c*m*x**2 + b*c*p*x**2 + b*c*x 
**2 + 2*b*d*m*x**3 + b*d*p*x**3 + b*d*x**3 - 2*c*m - c*p - c - 2*d*m*x - d 
*p*x - d*x),x)*b**2*c**2*m*p + int(((c + d*x)**p*(sqrt(b)*x + 1)**m*( - sq 
rt(b)*x + 1)**m*x**2)/(2*b*c*m*x**2 + b*c*p*x**2 + b*c*x**2 + 2*b*d*m*x**3 
 + b*d*p*x**3 + b*d*x**3 - 2*c*m - c*p - c - 2*d*m*x - d*p*x - d*x),x)*b** 
2*c**2*p**2 + int(((c + d*x)**p*(sqrt(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*x 
**2)/(2*b*c*m*x**2 + b*c*p*x**2 + b*c*x**2 + 2*b*d*m*x**3 + b*d*p*x**3 + b 
*d*x**3 - 2*c*m - c*p - c - 2*d*m*x - d*p*x - d*x),x)*b**2*c**2*p + 4*int( 
((c + d*x)**p*(sqrt(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*x**2)/(2*b*c*m*x**2 
 + b*c*p*x**2 + b*c*x**2 + 2*b*d*m*x**3 + b*d*p*x**3 + b*d*x**3 - 2*c*m - 
c*p - c - 2*d*m*x - d*p*x - d*x),x)*b*d**2*m**2 + 4*int(((c + d*x)**p*(sqr 
t(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*x**2)/(2*b*c*m*x**2 + b*c*p*x**2 + b* 
c*x**2 + 2*b*d*m*x**3 + b*d*p*x**3 + b*d*x**3 - 2*c*m - c*p - c - 2*d*m*x 
- d*p*x - d*x),x)*b*d**2*m*p + 2*int(((c + d*x)**p*(sqrt(b)*x + 1)**m*( - 
sqrt(b)*x + 1)**m*x**2)/(2*b*c*m*x**2 + b*c*p*x**2 + b*c*x**2 + 2*b*d*m*x* 
*3 + b*d*p*x**3 + b*d*x**3 - 2*c*m - c*p - c - 2*d*m*x - d*p*x - d*x),x)*b 
*d**2*m + int(((c + d*x)**p*(sqrt(b)*x + 1)**m*( - sqrt(b)*x + 1)**m*x**2) 
/(2*b*c*m*x**2 + b*c*p*x**2 + b*c*x**2 + 2*b*d*m*x**3 + b*d*p*x**3 + b*...