Integrand size = 15, antiderivative size = 45 \[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=-\frac {\left (a x+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,-1+2 p,-1+p,-\frac {b x}{a}\right )}{a (2-p) x^3} \] Output:
-(b*x^2+a*x)^(p+1)*hypergeom([1, -1+2*p],[-1+p],-b*x/a)/a/(2-p)/x^3
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\frac {(x (a+b x))^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2+p,-p,-1+p,-\frac {b x}{a}\right )}{(-2+p) x^2} \] Input:
Integrate[(a*x + b*x^2)^p/x^3,x]
Output:
((x*(a + b*x))^p*Hypergeometric2F1[-2 + p, -p, -1 + p, -((b*x)/a)])/((-2 + p)*x^2*(1 + (b*x)/a)^p)
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1137, 76, 74}
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.
\(\displaystyle \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx\) |
\(\Big \downarrow \) 1137 |
\(\displaystyle x^{-p} (a+b x)^{-p} \left (a x+b x^2\right )^p \int x^{p-3} (a+b x)^pdx\) |
\(\Big \downarrow \) 76 |
\(\displaystyle x^{-p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x+b x^2\right )^p \int x^{p-3} \left (\frac {b x}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 74 |
\(\displaystyle -\frac {\left (\frac {b x}{a}+1\right )^{-p} \left (a x+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (p-2,-p,p-1,-\frac {b x}{a}\right )}{(2-p) x^2}\) |
Input:
Int[(a*x + b*x^2)^p/x^3,x]
Output:
-(((a*x + b*x^2)^p*Hypergeometric2F1[-2 + p, -p, -1 + p, -((b*x)/a)])/((2 - p)*x^2*(1 + (b*x)/a)^p))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart [n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d* (x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) Int[x^(m + p)*(b + c*x)^ p, x], x] /; FreeQ[{b, c, e, m}, x]
\[\int \frac {\left (b \,x^{2}+a x \right )^{p}}{x^{3}}d x\]
Input:
int((b*x^2+a*x)^p/x^3,x)
Output:
int((b*x^2+a*x)^p/x^3,x)
\[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a x\right )}^{p}}{x^{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^p/x^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a*x)^p/x^3, x)
\[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{p}}{x^{3}}\, dx \] Input:
integrate((b*x**2+a*x)**p/x**3,x)
Output:
Integral((x*(a + b*x))**p/x**3, x)
\[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a x\right )}^{p}}{x^{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^p/x^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a*x)^p/x^3, x)
\[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a x\right )}^{p}}{x^{3}} \,d x } \] Input:
integrate((b*x^2+a*x)^p/x^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a*x)^p/x^3, x)
Timed out. \[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^p}{x^3} \,d x \] Input:
int((a*x + b*x^2)^p/x^3,x)
Output:
int((a*x + b*x^2)^p/x^3, x)
\[ \int \frac {\left (a x+b x^2\right )^p}{x^3} \, dx=\frac {\left (b \,x^{2}+a x \right )^{p}+\left (\int \frac {\left (b \,x^{2}+a x \right )^{p}}{b p \,x^{4}+a p \,x^{3}-b \,x^{4}-a \,x^{3}}d x \right ) a \,p^{2} x^{2}-\left (\int \frac {\left (b \,x^{2}+a x \right )^{p}}{b p \,x^{4}+a p \,x^{3}-b \,x^{4}-a \,x^{3}}d x \right ) a p \,x^{2}}{2 x^{2} \left (p -1\right )} \] Input:
int((b*x^2+a*x)^p/x^3,x)
Output:
((a*x + b*x**2)**p + int((a*x + b*x**2)**p/(a*p*x**3 - a*x**3 + b*p*x**4 - b*x**4),x)*a*p**2*x**2 - int((a*x + b*x**2)**p/(a*p*x**3 - a*x**3 + b*p*x **4 - b*x**4),x)*a*p*x**2)/(2*x**2*(p - 1))