\(\int (c+d x)^3 (a+b x^2)^p \, dx\) [24]

Optimal result

Integrand size = 17, antiderivative size = 169 \[ \int (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {d (c+d x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac {d \left ((3+2 p) \left (a d^2-b c^2 (5+2 p)\right )-2 b c d (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+c \left (c^2-\frac {3 a d^2}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \] Output:

1/2*d*(d*x+c)^2*(b*x^2+a)^(p+1)/b/(2+p)-1/2*d*((3+2*p)*(a*d^2-b*c^2*(5+2*p 
))-2*b*c*d*(p+1)*(3+p)*x)*(b*x^2+a)^(p+1)/b^2/(2+p)/(2*p^2+5*p+3)+c*(c^2-3 
*a*d^2/(2*b*p+3*b))*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/((1+ 
b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.32 \[ \int (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (2 b^2 c^3 \left (2+3 p+p^2\right ) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d \left (b^2 x^2 \left (1+\frac {b x^2}{a}\right )^p \left (3 c^2 (2+p)+d^2 (1+p) x^2\right )-a^2 d^2 \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+a b \left (d^2 p x^2 \left (1+\frac {b x^2}{a}\right )^p+3 c^2 (2+p) \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+2 b^2 c d \left (2+3 p+p^2\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right )\right )}{2 b^2 (1+p) (2+p)} \] Input:

Integrate[(c + d*x)^3*(a + b*x^2)^p,x]
 

Output:

((a + b*x^2)^p*(2*b^2*c^3*(2 + 3*p + p^2)*x*Hypergeometric2F1[1/2, -p, 3/2 
, -((b*x^2)/a)] + d*(b^2*x^2*(1 + (b*x^2)/a)^p*(3*c^2*(2 + p) + d^2*(1 + p 
)*x^2) - a^2*d^2*(-1 + (1 + (b*x^2)/a)^p) + a*b*(d^2*p*x^2*(1 + (b*x^2)/a) 
^p + 3*c^2*(2 + p)*(-1 + (1 + (b*x^2)/a)^p)) + 2*b^2*c*d*(2 + 3*p + p^2)*x 
^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(2*b^2*(1 + p)*(2 + p) 
*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {497, 27, 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.

\(\displaystyle \int (c+d x)^3 \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 497

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 676

\(\displaystyle \frac {d (c+d x)^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)}-\frac {\frac {c (p+2) \left (3 a d^2-b c^2 (2 p+3)\right ) \int \left (b x^2+a\right )^pdx}{2 p+3}+\frac {d \left (a+b x^2\right )^{p+1} \left (a d^2-b c^2 (2 p+5)\right )}{2 b (p+1)}-\frac {c d^2 (p+3) x \left (a+b x^2\right )^{p+1}}{2 p+3}}{b (p+2)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {d (c+d x)^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)}-\frac {\frac {c (p+2) \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a d^2-b c^2 (2 p+3)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{2 p+3}+\frac {d \left (a+b x^2\right )^{p+1} \left (a d^2-b c^2 (2 p+5)\right )}{2 b (p+1)}-\frac {c d^2 (p+3) x \left (a+b x^2\right )^{p+1}}{2 p+3}}{b (p+2)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {d (c+d x)^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)}-\frac {\frac {c (p+2) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a d^2-b c^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{2 p+3}+\frac {d \left (a+b x^2\right )^{p+1} \left (a d^2-b c^2 (2 p+5)\right )}{2 b (p+1)}-\frac {c d^2 (p+3) x \left (a+b x^2\right )^{p+1}}{2 p+3}}{b (p+2)}\)

Input:

Int[(c + d*x)^3*(a + b*x^2)^p,x]
 

Output:

(d*(c + d*x)^2*(a + b*x^2)^(1 + p))/(2*b*(2 + p)) - ((d*(a*d^2 - b*c^2*(5 
+ 2*p))*(a + b*x^2)^(1 + p))/(2*b*(1 + p)) - (c*d^2*(3 + p)*x*(a + b*x^2)^ 
(1 + p))/(3 + 2*p) + (c*(2 + p)*(3*a*d^2 - b*c^2*(3 + 2*p))*x*(a + b*x^2)^ 
p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/((3 + 2*p)*(1 + (b*x^2)/a 
)^p))/(b*(2 + p))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 497

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b 
*(n + 2*p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 
 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n 
, p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p 
+ 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

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]

\[\int \left (d x +c \right )^{3} \left (b \,x^{2}+a \right )^{p}d x\]

Input:

int((d*x+c)^3*(b*x^2+a)^p,x)
 

Output:

int((d*x+c)^3*(b*x^2+a)^p,x)
 

Fricas [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p,x, algorithm="fricas")
 

Output:

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(b*x^2 + a)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (146) = 292\).

Time = 6.09 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.59 \[ \int (c+d x)^3 \left (a+b x^2\right )^p \, dx=a^{p} c^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + a^{p} c d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + 3 c^{2} d \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) + d^{3} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)**3*(b*x**2+a)**p,x)
 

Output:

a**p*c**3*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*c*d* 
*2*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a) + 3*c**2*d*Piec 
ewise((a**p*x**2/2, Eq(b, 0)), (Piecewise(((a + b*x**2)**(p + 1)/(p + 1), 
Ne(p, -1)), (log(a + b*x**2), True))/(2*b), True)) + d**3*Piecewise((a**p* 
x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a*log 
(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b**3*x**2) + b 
*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(x + sqrt(- 
a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - sqrt(-a/b))/(2*b** 
2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)), (-a**2*(a + 
b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2*(a + b*x**2)**p/ 
(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p/(2*b**2*p* 
*2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2* 
p + 4*b**2), True))
 

Maxima [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p, x)
 

Giac [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p, x)
 

Mupad [F(-1)]

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

int((a + b*x^2)^p*(c + d*x)^3,x)
 

Output:

int((a + b*x^2)^p*(c + d*x)^3, x)
 

Reduce [F]

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

int((d*x+c)^3*(b*x^2+a)^p,x)
 

Output:

( - 4*(a + b*x**2)**p*a**2*d**3*p**2 - 8*(a + b*x**2)**p*a**2*d**3*p - 3*( 
a + b*x**2)**p*a**2*d**3 + 12*(a + b*x**2)**p*a*b*c**2*d*p**3 + 48*(a + b* 
x**2)**p*a*b*c**2*d*p**2 + 57*(a + b*x**2)**p*a*b*c**2*d*p + 18*(a + b*x** 
2)**p*a*b*c**2*d + 12*(a + b*x**2)**p*a*b*c*d**2*p**3*x + 36*(a + b*x**2)* 
*p*a*b*c*d**2*p**2*x + 24*(a + b*x**2)**p*a*b*c*d**2*p*x + 4*(a + b*x**2)* 
*p*a*b*d**3*p**3*x**2 + 8*(a + b*x**2)**p*a*b*d**3*p**2*x**2 + 3*(a + b*x* 
*2)**p*a*b*d**3*p*x**2 + 4*(a + b*x**2)**p*b**2*c**3*p**3*x + 18*(a + b*x* 
*2)**p*b**2*c**3*p**2*x + 26*(a + b*x**2)**p*b**2*c**3*p*x + 12*(a + b*x** 
2)**p*b**2*c**3*x + 12*(a + b*x**2)**p*b**2*c**2*d*p**3*x**2 + 48*(a + b*x 
**2)**p*b**2*c**2*d*p**2*x**2 + 57*(a + b*x**2)**p*b**2*c**2*d*p*x**2 + 18 
*(a + b*x**2)**p*b**2*c**2*d*x**2 + 12*(a + b*x**2)**p*b**2*c*d**2*p**3*x* 
*3 + 42*(a + b*x**2)**p*b**2*c*d**2*p**2*x**3 + 42*(a + b*x**2)**p*b**2*c* 
d**2*p*x**3 + 12*(a + b*x**2)**p*b**2*c*d**2*x**3 + 4*(a + b*x**2)**p*b**2 
*d**3*p**3*x**4 + 12*(a + b*x**2)**p*b**2*d**3*p**2*x**4 + 11*(a + b*x**2) 
**p*b**2*d**3*p*x**4 + 3*(a + b*x**2)**p*b**2*d**3*x**4 - 48*int((a + b*x* 
*2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x) 
*a**2*b*c*d**2*p**5 - 240*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4* 
b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*b*c*d**2*p**4 - 420*int((a + 
b*x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2 
),x)*a**2*b*c*d**2*p**3 - 300*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 3...