Integrand size = 15, antiderivative size = 70 \[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\frac {B \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+A x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \] Output:
1/2*B*(b*x^2+a)^(p+1)/b/(p+1)+A*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b *x^2/a)/((1+b*x^2/a)^p)
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.40 \[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (b B x^2 \left (1+\frac {b x^2}{a}\right )^p+a B \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+2 A b (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{2 b (1+p)} \] Input:
Integrate[(A + B*x)*(a + b*x^2)^p,x]
Output:
((a + b*x^2)^p*(b*B*x^2*(1 + (b*x^2)/a)^p + a*B*(-1 + (1 + (b*x^2)/a)^p) + 2*A*b*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)]))/(2*b*(1 + p)*(1 + (b*x^2)/a)^p)
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int (A+B x) \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \left (b x^2+a\right )^pdx+\frac {B \left (a+b x^2\right )^{p+1}}{2 b (p+1)}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle A \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \left (\frac {b x^2}{a}+1\right )^pdx+\frac {B \left (a+b x^2\right )^{p+1}}{2 b (p+1)}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle A x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+\frac {B \left (a+b x^2\right )^{p+1}}{2 b (p+1)}\) |
Input:
Int[(A + B*x)*(a + b*x^2)^p,x]
Output:
(B*(a + b*x^2)^(1 + p))/(2*b*(1 + p)) + (A*x*(a + b*x^2)^p*Hypergeometric2 F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^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]
\[\int \left (B x +A \right ) \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int((B*x+A)*(b*x^2+a)^p,x)
Output:
int((B*x+A)*(b*x^2+a)^p,x)
\[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((B*x + A)*(b*x^2 + a)^p, x)
Time = 2.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int (A+B x) \left (a+b x^2\right )^p \, dx=A a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + B \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((B*x+A)*(b*x**2+a)**p,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))
\[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((B*x + A)*(b*x^2 + a)^p, x)
\[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((B*x + A)*(b*x^2 + a)^p, x)
Time = 1.00 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{p+1}}{2\,b\,\left (p+1\right )}+\frac {A\,x\,{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^p} \] Input:
int((a + b*x^2)^p*(A + B*x),x)
Output:
(B*(a + b*x^2)^(p + 1))/(2*b*(p + 1)) + (A*x*(a + b*x^2)^p*hypergeom([1/2, -p], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^p
\[ \int (A+B x) \left (a+b x^2\right )^p \, dx=\frac {2 \left (b \,x^{2}+a \right )^{p} a p x +2 \left (b \,x^{2}+a \right )^{p} a p +2 \left (b \,x^{2}+a \right )^{p} a x +\left (b \,x^{2}+a \right )^{p} a +2 \left (b \,x^{2}+a \right )^{p} b p \,x^{2}+\left (b \,x^{2}+a \right )^{p} b \,x^{2}+8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a^{2} p^{3}+12 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a^{2} p^{2}+4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a^{2} p}{4 p^{2}+6 p +2} \] Input:
int((B*x+A)*(b*x^2+a)^p,x)
Output:
(2*(a + b*x**2)**p*a*p*x + 2*(a + b*x**2)**p*a*p + 2*(a + b*x**2)**p*a*x + (a + b*x**2)**p*a + 2*(a + b*x**2)**p*b*p*x**2 + (a + b*x**2)**p*b*x**2 + 8*int((a + b*x**2)**p/(2*a*p + a + 2*b*p*x**2 + b*x**2),x)*a**2*p**3 + 12 *int((a + b*x**2)**p/(2*a*p + a + 2*b*p*x**2 + b*x**2),x)*a**2*p**2 + 4*in t((a + b*x**2)**p/(2*a*p + a + 2*b*p*x**2 + b*x**2),x)*a**2*p)/(2*(2*p**2 + 3*p + 1))