\(\int (A+B x) (c+d x) (c^2-d^2 x^2)^p \, dx\) [116]

Optimal result

Integrand size = 25, antiderivative size = 123 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=-\frac {B (c+d x) \left (c^2-d^2 x^2\right )^{1+p}}{d^2 (3+2 p)}+\frac {2^p (B c+A d (3+2 p)) \left (\frac {c-d x}{c}\right )^{-2-p} \left (c^2-d^2 x^2\right )^{2+p} \operatorname {Hypergeometric2F1}\left (-p,2+p,3+p,\frac {c+d x}{2 c}\right )}{c^2 d^2 (2+p) (3+2 p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.16 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=\frac {\left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \left (-3 (B c+A d) \left (c^2-d^2 x^2\right ) \left (1-\frac {d^2 x^2}{c^2}\right )^p+6 A c d^2 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )+2 B d^3 (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {d^2 x^2}{c^2}\right )\right )}{6 d^2 (1+p)} \] Input:

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

Output:

((c^2 - d^2*x^2)^p*(-3*(B*c + A*d)*(c^2 - d^2*x^2)*(1 - (d^2*x^2)/c^2)^p + 
 6*A*c*d^2*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, (d^2*x^2)/c^2] + 2*B* 
d^3*(1 + p)*x^3*Hypergeometric2F1[3/2, -p, 5/2, (d^2*x^2)/c^2]))/(6*d^2*(1 
 + p)*(1 - (d^2*x^2)/c^2)^p)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {676, 238, 237}

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.

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.42 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=A c c^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + A d \left (\begin {cases} \frac {x^{2} \left (c^{2}\right )^{p}}{2} & \text {for}\: d^{2} = 0 \\- \frac {\begin {cases} \frac {\left (c^{2} - d^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (c^{2} - d^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 d^{2}} & \text {otherwise} \end {cases}\right ) + B c \left (\begin {cases} \frac {x^{2} \left (c^{2}\right )^{p}}{2} & \text {for}\: d^{2} = 0 \\- \frac {\begin {cases} \frac {\left (c^{2} - d^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (c^{2} - d^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 d^{2}} & \text {otherwise} \end {cases}\right ) + \frac {B c^{2 p} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{3} \] Input:

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

Output:

A*c*c**(2*p)*x*hyper((1/2, -p), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) 
+ A*d*Piecewise((x**2*(c**2)**p/2, Eq(d**2, 0)), (-Piecewise(((c**2 - d**2 
*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(c**2 - d**2*x**2), True))/(2*d** 
2), True)) + B*c*Piecewise((x**2*(c**2)**p/2, Eq(d**2, 0)), (-Piecewise((( 
c**2 - d**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(c**2 - d**2*x**2), Tr 
ue))/(2*d**2), True)) + B*c**(2*p)*d*x**3*hyper((3/2, -p), (5/2,), d**2*x* 
*2*exp_polar(2*I*pi)/c**2)/3
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 4*(c**2 - d**2*x**2)**p*a*c**2*d*p**2 - 8*(c**2 - d**2*x**2)**p*a*c**2 
*d*p - 3*(c**2 - d**2*x**2)**p*a*c**2*d + 4*(c**2 - d**2*x**2)**p*a*c*d**2 
*p**2*x + 10*(c**2 - d**2*x**2)**p*a*c*d**2*p*x + 6*(c**2 - d**2*x**2)**p* 
a*c*d**2*x + 4*(c**2 - d**2*x**2)**p*a*d**3*p**2*x**2 + 8*(c**2 - d**2*x** 
2)**p*a*d**3*p*x**2 + 3*(c**2 - d**2*x**2)**p*a*d**3*x**2 - 4*(c**2 - d**2 
*x**2)**p*b*c**3*p**2 - 8*(c**2 - d**2*x**2)**p*b*c**3*p - 3*(c**2 - d**2* 
x**2)**p*b*c**3 - 4*(c**2 - d**2*x**2)**p*b*c**2*d*p**2*x - 4*(c**2 - d**2 
*x**2)**p*b*c**2*d*p*x + 4*(c**2 - d**2*x**2)**p*b*c*d**2*p**2*x**2 + 8*(c 
**2 - d**2*x**2)**p*b*c*d**2*p*x**2 + 3*(c**2 - d**2*x**2)**p*b*c*d**2*x** 
2 + 4*(c**2 - d**2*x**2)**p*b*d**3*p**2*x**3 + 6*(c**2 - d**2*x**2)**p*b*d 
**3*p*x**3 + 2*(c**2 - d**2*x**2)**p*b*d**3*x**3 + 32*int((c**2 - d**2*x** 
2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 
- 3*d**2*x**2),x)*a*c**3*d**2*p**5 + 144*int((c**2 - d**2*x**2)**p/(4*c**2 
*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2 
),x)*a*c**3*d**2*p**4 + 232*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c** 
2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c**3*d 
**2*p**3 + 156*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 
- 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c**3*d**2*p**2 + 36 
*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2* 
x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c**3*d**2*p + 16*int((c**2 - d...