\(\int (a+b x+c x^2) (d+f x^2)^q \, dx\) [10]

Optimal result

Integrand size = 20, antiderivative size = 109 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^q \, dx=\frac {b \left (d+f x^2\right )^{1+q}}{2 f (1+q)}+\frac {c x \left (d+f x^2\right )^{1+q}}{f (3+2 q)}+\left (a-\frac {c d}{3 f+2 f q}\right ) x \left (d+f x^2\right )^q \left (1+\frac {f x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {f x^2}{d}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^q \, dx=\frac {\left (d+f x^2\right )^q \left (1+\frac {f x^2}{d}\right )^{-q} \left (3 b \left (d+f x^2\right ) \left (1+\frac {f x^2}{d}\right )^q+6 a f (1+q) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {f x^2}{d}\right )+2 c f (1+q) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {f x^2}{d}\right )\right )}{6 f (1+q)} \] Input:

Integrate[(a + b*x + c*x^2)*(d + f*x^2)^q,x]
 

Output:

((d + f*x^2)^q*(3*b*(d + f*x^2)*(1 + (f*x^2)/d)^q + 6*a*f*(1 + q)*x*Hyperg 
eometric2F1[1/2, -q, 3/2, -((f*x^2)/d)] + 2*c*f*(1 + q)*x^3*Hypergeometric 
2F1[3/2, -q, 5/2, -((f*x^2)/d)]))/(6*f*(1 + q)*(1 + (f*x^2)/d)^q)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2346, 25, 455, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

Input:

Int[(a + b*x + c*x^2)*(d + f*x^2)^q,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [F]

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

Input:

int((c*x^2+b*x+a)*(f*x^2+d)^q,x)
 

Output:

int((c*x^2+b*x+a)*(f*x^2+d)^q,x)
 

Fricas [F]

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

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

Output:

integral((c*x^2 + b*x + a)*(f*x^2 + d)^q, x)
 

Sympy [A] (verification not implemented)

Time = 4.73 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^q \, dx=a d^{q} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - q \\ \frac {3}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )} + b \left (\begin {cases} \frac {d^{q} x^{2}}{2} & \text {for}\: f = 0 \\\frac {\begin {cases} \frac {\left (d + f x^{2}\right )^{q + 1}}{q + 1} & \text {for}\: q \neq -1 \\\log {\left (d + f x^{2} \right )} & \text {otherwise} \end {cases}}{2 f} & \text {otherwise} \end {cases}\right ) + \frac {c d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )}}{3} \] Input:

integrate((c*x**2+b*x+a)*(f*x**2+d)**q,x)
 

Output:

a*d**q*x*hyper((1/2, -q), (3/2,), f*x**2*exp_polar(I*pi)/d) + b*Piecewise( 
(d**q*x**2/2, Eq(f, 0)), (Piecewise(((d + f*x**2)**(q + 1)/(q + 1), Ne(q, 
-1)), (log(d + f*x**2), True))/(2*f), True)) + c*d**q*x**3*hyper((3/2, -q) 
, (5/2,), f*x**2*exp_polar(I*pi)/d)/3
 

Maxima [F]

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

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

Output:

integrate((c*x^2 + b*x + a)*(f*x^2 + d)^q, x)
 

Giac [F]

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

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

Output:

integrate((c*x^2 + b*x + a)*(f*x^2 + d)^q, x)
 

Mupad [F(-1)]

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

int((d + f*x^2)^q*(a + b*x + c*x^2),x)
 

Output:

int((d + f*x^2)^q*(a + b*x + c*x^2), x)
 

Reduce [F]

\[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^q \, dx=\frac {4 \left (f \,x^{2}+d \right )^{q} a f \,q^{2} x +10 \left (f \,x^{2}+d \right )^{q} a f q x +6 \left (f \,x^{2}+d \right )^{q} a f x +4 \left (f \,x^{2}+d \right )^{q} b d \,q^{2}+8 \left (f \,x^{2}+d \right )^{q} b d q +3 \left (f \,x^{2}+d \right )^{q} b d +4 \left (f \,x^{2}+d \right )^{q} b f \,q^{2} x^{2}+8 \left (f \,x^{2}+d \right )^{q} b f q \,x^{2}+3 \left (f \,x^{2}+d \right )^{q} b f \,x^{2}+4 \left (f \,x^{2}+d \right )^{q} c d \,q^{2} x +4 \left (f \,x^{2}+d \right )^{q} c d q x +4 \left (f \,x^{2}+d \right )^{q} c f \,q^{2} x^{3}+6 \left (f \,x^{2}+d \right )^{q} c f q \,x^{3}+2 \left (f \,x^{2}+d \right )^{q} c f \,x^{3}+32 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) a d f \,q^{5}+144 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) a d f \,q^{4}+232 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) a d f \,q^{3}+156 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) a d f \,q^{2}+36 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) a d f q -16 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) c \,d^{2} q^{4}-48 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) c \,d^{2} q^{3}-44 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) c \,d^{2} q^{2}-12 \left (\int \frac {\left (f \,x^{2}+d \right )^{q}}{4 f \,q^{2} x^{2}+8 f q \,x^{2}+4 d \,q^{2}+3 f \,x^{2}+8 d q +3 d}d x \right ) c \,d^{2} q}{2 f \left (4 q^{3}+12 q^{2}+11 q +3\right )} \] Input:

int((c*x^2+b*x+a)*(f*x^2+d)^q,x)
 

Output:

(4*(d + f*x**2)**q*a*f*q**2*x + 10*(d + f*x**2)**q*a*f*q*x + 6*(d + f*x**2 
)**q*a*f*x + 4*(d + f*x**2)**q*b*d*q**2 + 8*(d + f*x**2)**q*b*d*q + 3*(d + 
 f*x**2)**q*b*d + 4*(d + f*x**2)**q*b*f*q**2*x**2 + 8*(d + f*x**2)**q*b*f* 
q*x**2 + 3*(d + f*x**2)**q*b*f*x**2 + 4*(d + f*x**2)**q*c*d*q**2*x + 4*(d 
+ f*x**2)**q*c*d*q*x + 4*(d + f*x**2)**q*c*f*q**2*x**3 + 6*(d + f*x**2)**q 
*c*f*q*x**3 + 2*(d + f*x**2)**q*c*f*x**3 + 32*int((d + f*x**2)**q/(4*d*q** 
2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x**2),x)*a*d*f*q**5 + 1 
44*int((d + f*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x** 
2 + 3*f*x**2),x)*a*d*f*q**4 + 232*int((d + f*x**2)**q/(4*d*q**2 + 8*d*q + 
3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x**2),x)*a*d*f*q**3 + 156*int((d + 
f*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x**2 
),x)*a*d*f*q**2 + 36*int((d + f*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*f*q** 
2*x**2 + 8*f*q*x**2 + 3*f*x**2),x)*a*d*f*q - 16*int((d + f*x**2)**q/(4*d*q 
**2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x**2),x)*c*d**2*q**4 
- 48*int((d + f*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x 
**2 + 3*f*x**2),x)*c*d**2*q**3 - 44*int((d + f*x**2)**q/(4*d*q**2 + 8*d*q 
+ 3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x**2),x)*c*d**2*q**2 - 12*int((d 
+ f*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*f*q**2*x**2 + 8*f*q*x**2 + 3*f*x* 
*2),x)*c*d**2*q)/(2*f*(4*q**3 + 12*q**2 + 11*q + 3))