Integrand size = 19, antiderivative size = 48 \[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\frac {x^{-1-3 p} \left (a x^2+b x^3\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,2,2-p,-\frac {b x}{a}\right )}{a (1-p)} \] Output:
x^(-1-3*p)*(b*x^3+a*x^2)^(p+1)*hypergeom([1, 2],[2-p],-b*x/a)/a/(1-p)
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\frac {x^{1-3 p} \left (x^2 (a+b x)\right )^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x}{a}\right )}{1-p} \] Input:
Integrate[(a*x^2 + b*x^3)^p/x^(3*p),x]
Output:
(x^(1 - 3*p)*(x^2*(a + b*x))^p*Hypergeometric2F1[1 - p, -p, 2 - p, -((b*x) /a)])/((1 - p)*(1 + (b*x)/a)^p)
Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1938, 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.
\(\displaystyle \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle x^{-2 p} (a+b x)^{-p} \left (a x^2+b x^3\right )^p \int x^{-p} (a+b x)^pdx\) |
\(\Big \downarrow \) 76 |
\(\displaystyle x^{-2 p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \int x^{-p} \left (\frac {b x}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {x^{1-3 p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x}{a}\right )}{1-p}\) |
Input:
Int[(a*x^2 + b*x^3)^p/x^(3*p),x]
Output:
(x^(1 - 3*p)*(a*x^2 + b*x^3)^p*Hypergeometric2F1[1 - p, -p, 2 - p, -((b*x) /a)])/((1 - p)*(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[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])) Int[x^(m + j* p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !Inte gerQ[p] && NeQ[n, j] && PosQ[n - j]
\[\int \left (b \,x^{3}+a \,x^{2}\right )^{p} x^{-3 p}d x\]
Input:
int((b*x^3+a*x^2)^p/(x^(3*p)),x)
Output:
int((b*x^3+a*x^2)^p/(x^(3*p)),x)
\[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{p}}{x^{3 \, p}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p/(x^(3*p)),x, algorithm="fricas")
Output:
integral((b*x^3 + a*x^2)^p/x^(3*p), x)
\[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\int x^{- 3 p} \left (x^{2} \left (a + b x\right )\right )^{p}\, dx \] Input:
integrate((b*x**3+a*x**2)**p/(x**(3*p)),x)
Output:
Integral((x**2*(a + b*x))**p/x**(3*p), x)
\[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{p}}{x^{3 \, p}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p/(x^(3*p)),x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^p/x^(3*p), x)
\[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{p}}{x^{3 \, p}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p/(x^(3*p)),x, algorithm="giac")
Output:
integrate((b*x^3 + a*x^2)^p/x^(3*p), x)
Timed out. \[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^p}{x^{3\,p}} \,d x \] Input:
int((a*x^2 + b*x^3)^p/x^(3*p),x)
Output:
int((a*x^2 + b*x^3)^p/x^(3*p), x)
\[ \int x^{-3 p} \left (a x^2+b x^3\right )^p \, dx=\frac {\left (b \,x^{3}+a \,x^{2}\right )^{p} x +x^{3 p} \left (\int \frac {\left (b \,x^{3}+a \,x^{2}\right )^{p}}{x^{3 p} a +x^{3 p} b x}d x \right ) a p}{x^{3 p}} \] Input:
int((b*x^3+a*x^2)^p/(x^(3*p)),x)
Output:
((a*x**2 + b*x**3)**p*x + x**(3*p)*int((a*x**2 + b*x**3)**p/(x**(3*p)*a + x**(3*p)*b*x),x)*a*p)/x**(3*p)