Integrand size = 20, antiderivative size = 109 \[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\frac {B \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac {C x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\left (A-\frac {a C}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \] Output:
1/2*B*(b*x^2+a)^(p+1)/b/(p+1)+C*x*(b*x^2+a)^(p+1)/b/(3+2*p)+(A-a*C/(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)
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05 \[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (3 B \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+6 A b (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+2 b C (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right )}{6 b (1+p)} \] Input:
Integrate[(a + b*x^2)^p*(A + B*x + C*x^2),x]
Output:
((a + b*x^2)^p*(3*B*(a + b*x^2)*(1 + (b*x^2)/a)^p + 6*A*b*(1 + p)*x*Hyperg eometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + 2*b*C*(1 + p)*x^3*Hypergeometric 2F1[3/2, -p, 5/2, -((b*x^2)/a)]))/(6*b*(1 + p)*(1 + (b*x^2)/a)^p)
Time = 0.26 (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.
\(\displaystyle \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\int -\left ((a C-A b (2 p+3)-b B (2 p+3) x) \left (b x^2+a\right )^p\right )dx}{b (2 p+3)}+\frac {C x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {C x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}-\frac {\int (a C-A b (2 p+3)-b B (2 p+3) x) \left (b x^2+a\right )^pdx}{b (2 p+3)}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {C x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}-\frac {(a C-A b (2 p+3)) \int \left (b x^2+a\right )^pdx-\frac {B (2 p+3) \left (a+b x^2\right )^{p+1}}{2 (p+1)}}{b (2 p+3)}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {C x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (a C-A b (2 p+3)) \int \left (\frac {b x^2}{a}+1\right )^pdx-\frac {B (2 p+3) \left (a+b x^2\right )^{p+1}}{2 (p+1)}}{b (2 p+3)}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {C x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (a C-A b (2 p+3)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )-\frac {B (2 p+3) \left (a+b x^2\right )^{p+1}}{2 (p+1)}}{b (2 p+3)}\) |
Input:
Int[(a + b*x^2)^p*(A + B*x + C*x^2),x]
Output:
(C*x*(a + b*x^2)^(1 + p))/(b*(3 + 2*p)) - (-1/2*(B*(3 + 2*p)*(a + b*x^2)^( 1 + p))/(1 + p) + ((a*C - A*b*(3 + 2*p))*x*(a + b*x^2)^p*Hypergeometric2F1 [1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p)/(b*(3 + 2*p))
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]
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]
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]
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]
\[\int \left (b \,x^{2}+a \right )^{p} \left (C \,x^{2}+B x +A \right )d x\]
Input:
int((b*x^2+a)^p*(C*x^2+B*x+A),x)
Output:
int((b*x^2+a)^p*(C*x^2+B*x+A),x)
\[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((b*x^2+a)^p*(C*x^2+B*x+A),x, algorithm="fricas")
Output:
integral((C*x^2 + B*x + A)*(b*x^2 + a)^p, x)
Time = 4.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=A a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + B \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {C a^{p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:
integrate((b*x**2+a)**p*(C*x**2+B*x+A),x)
Output:
A*a**p*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + B*Piecewise( (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)) + C*a**p*x**3*hyper((3/2, -p) , (5/2,), b*x**2*exp_polar(I*pi)/a)/3
\[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((b*x^2+a)^p*(C*x^2+B*x+A),x, algorithm="maxima")
Output:
integrate((C*x^2 + B*x + A)*(b*x^2 + a)^p, x)
\[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((b*x^2+a)^p*(C*x^2+B*x+A),x, algorithm="giac")
Output:
integrate((C*x^2 + B*x + A)*(b*x^2 + a)^p, x)
Timed out. \[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\int {\left (b\,x^2+a\right )}^p\,\left (C\,x^2+B\,x+A\right ) \,d x \] Input:
int((a + b*x^2)^p*(A + B*x + C*x^2),x)
Output:
int((a + b*x^2)^p*(A + B*x + C*x^2), x)
\[ \int \left (a+b x^2\right )^p \left (A+B x+C x^2\right ) \, dx=\frac {4 \left (b \,x^{2}+a \right )^{p} a b \,p^{2} x +4 \left (b \,x^{2}+a \right )^{p} a b \,p^{2}+10 \left (b \,x^{2}+a \right )^{p} a b p x +8 \left (b \,x^{2}+a \right )^{p} a b p +6 \left (b \,x^{2}+a \right )^{p} a b x +3 \left (b \,x^{2}+a \right )^{p} a b +4 \left (b \,x^{2}+a \right )^{p} a c \,p^{2} x +4 \left (b \,x^{2}+a \right )^{p} a c p x +4 \left (b \,x^{2}+a \right )^{p} b^{2} p^{2} x^{2}+8 \left (b \,x^{2}+a \right )^{p} b^{2} p \,x^{2}+3 \left (b \,x^{2}+a \right )^{p} b^{2} x^{2}+4 \left (b \,x^{2}+a \right )^{p} b c \,p^{2} x^{3}+6 \left (b \,x^{2}+a \right )^{p} b c p \,x^{3}+2 \left (b \,x^{2}+a \right )^{p} b c \,x^{3}+32 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} b \,p^{5}+144 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} b \,p^{4}+232 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} b \,p^{3}+156 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} b \,p^{2}+36 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} b p -16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} c \,p^{4}-48 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} c \,p^{3}-44 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} c \,p^{2}-12 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} c p}{2 b \left (4 p^{3}+12 p^{2}+11 p +3\right )} \] Input:
int((b*x^2+a)^p*(C*x^2+B*x+A),x)
Output:
(4*(a + b*x**2)**p*a*b*p**2*x + 4*(a + b*x**2)**p*a*b*p**2 + 10*(a + b*x** 2)**p*a*b*p*x + 8*(a + b*x**2)**p*a*b*p + 6*(a + b*x**2)**p*a*b*x + 3*(a + b*x**2)**p*a*b + 4*(a + b*x**2)**p*a*c*p**2*x + 4*(a + b*x**2)**p*a*c*p*x + 4*(a + b*x**2)**p*b**2*p**2*x**2 + 8*(a + b*x**2)**p*b**2*p*x**2 + 3*(a + b*x**2)**p*b**2*x**2 + 4*(a + b*x**2)**p*b*c*p**2*x**3 + 6*(a + b*x**2) **p*b*c*p*x**3 + 2*(a + b*x**2)**p*b*c*x**3 + 32*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*p**5 + 144*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*p**4 + 232*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*p**3 + 156*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*p**2 + 36*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*p - 16*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 *c*p**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*c*p**3 - 44*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*c*p**2 - 12 *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*c*p)/(2*b*(4*p**3 + 12*p**2 + 11*p + 3))