Integrand size = 16, antiderivative size = 148 \[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{p} \] Output:
2^(-1+2*p)*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p,-p,1-2*p,-1/2*(b-(-4*a*c+b^2)^ (1/2))/c/x,-1/2*(b+(-4*a*c+b^2)^(1/2))/c/x)/p/(((b-(-4*a*c+b^2)^(1/2)+2*c* x)/c/x)^p)/(((b+(-4*a*c+b^2)^(1/2)+2*c*x)/c/x)^p)
Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} (a+x (b+c x))^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )}{p} \] Input:
Integrate[(a + b*x + c*x^2)^p/x,x]
Output:
(2^(-1 + 2*p)*(a + x*(b + c*x))^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x)])/(p*((b - Sq rt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^ p)
Time = 0.50 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1178, 150}
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx\) |
\(\Big \downarrow \) 1178 |
\(\displaystyle -4^p \left (\frac {1}{x}\right )^{2 p} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \int \left (\frac {b-\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {b+\sqrt {b^2-4 a c}}{2 c x}+1\right )^p \left (\frac {1}{x}\right )^{-2 p-1}d\frac {1}{x}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2^{2 p-1} \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{p}\) |
Input:
Int[(a + b*x + c*x^2)^p/x,x]
Output:
(2^(-1 + 2*p)*(a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b - Sqrt[b^2 - 4*a*c])/(c*x), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x)])/(p*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x) )^p)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(1/(d + e*x))^(2*p))*((a + b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q + 2*c* x)/(2*c*(d + e*x))))^p)) Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b - q)/(2*c)))*x, x]^p*Simp[1 - (d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[m, 0]
\[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{x}d x\]
Input:
int((c*x^2+b*x+a)^p/x,x)
Output:
int((c*x^2+b*x+a)^p/x,x)
\[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{x} \,d x } \] Input:
integrate((c*x^2+b*x+a)^p/x,x, algorithm="fricas")
Output:
integral((c*x^2 + b*x + a)^p/x, x)
\[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{p}}{x}\, dx \] Input:
integrate((c*x**2+b*x+a)**p/x,x)
Output:
Integral((a + b*x + c*x**2)**p/x, x)
\[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{x} \,d x } \] Input:
integrate((c*x^2+b*x+a)^p/x,x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^p/x, x)
\[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{x} \,d x } \] Input:
integrate((c*x^2+b*x+a)^p/x,x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^p/x, x)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{x} \,d x \] Input:
int((a + b*x + c*x^2)^p/x,x)
Output:
int((a + b*x + c*x^2)^p/x, x)
\[ \int \frac {\left (a+b x+c x^2\right )^p}{x} \, dx=\frac {\left (c \,x^{2}+b x +a \right )^{p}+\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{c \,x^{3}+b \,x^{2}+a x}d x \right ) a p -\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x}{c \,x^{2}+b x +a}d x \right ) c p}{p} \] Input:
int((c*x^2+b*x+a)^p/x,x)
Output:
((a + b*x + c*x**2)**p + int((a + b*x + c*x**2)**p/(a*x + b*x**2 + c*x**3) ,x)*a*p - int(((a + b*x + c*x**2)**p*x)/(a + b*x + c*x**2),x)*c*p)/p