Integrand size = 21, antiderivative size = 54 \[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=-\frac {x^{-3 p} \left (1+\frac {b x}{a}\right )^{-p} \left (a x^2+b x^3\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {b x}{a}\right )}{p} \] Output:
-(b*x^3+a*x^2)^p*hypergeom([-p, -p],[1-p],-b*x/a)/p/(x^(3*p))/((1+b*x/a)^p )
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=-\frac {x^{-3 p} \left (x^2 (a+b x)\right )^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {b x}{a}\right )}{p} \] Input:
Integrate[x^(-1 - 3*p)*(a*x^2 + b*x^3)^p,x]
Output:
-(((x^2*(a + b*x))^p*Hypergeometric2F1[-p, -p, 1 - p, -((b*x)/a)])/(p*x^(3 *p)*(1 + (b*x)/a)^p))
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, 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.
\(\displaystyle \int x^{-3 p-1} \left (a x^2+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle x^{-2 p} (a+b x)^{-p} \left (a x^2+b x^3\right )^p \int x^{-p-1} (a+b x)^pdx\) |
\(\Big \downarrow \) 76 |
\(\displaystyle x^{-2 p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \int x^{-p-1} \left (\frac {b x}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 74 |
\(\displaystyle -\frac {x^{-3 p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {b x}{a}\right )}{p}\) |
Input:
Int[x^(-1 - 3*p)*(a*x^2 + b*x^3)^p,x]
Output:
-(((a*x^2 + b*x^3)^p*Hypergeometric2F1[-p, -p, 1 - p, -((b*x)/a)])/(p*x^(3 *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]
\[\int x^{-1-3 p} \left (b \,x^{3}+a \,x^{2}\right )^{p}d x\]
Input:
int(x^(-1-3*p)*(b*x^3+a*x^2)^p,x)
Output:
int(x^(-1-3*p)*(b*x^3+a*x^2)^p,x)
\[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 1} \,d x } \] Input:
integrate(x^(-1-3*p)*(b*x^3+a*x^2)^p,x, algorithm="fricas")
Output:
integral((b*x^3 + a*x^2)^p*x^(-3*p - 1), x)
\[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int x^{- 3 p - 1} \left (x^{2} \left (a + b x\right )\right )^{p}\, dx \] Input:
integrate(x**(-1-3*p)*(b*x**3+a*x**2)**p,x)
Output:
Integral(x**(-3*p - 1)*(x**2*(a + b*x))**p, x)
\[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 1} \,d x } \] Input:
integrate(x^(-1-3*p)*(b*x^3+a*x^2)^p,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^p*x^(-3*p - 1), x)
\[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 1} \,d x } \] Input:
integrate(x^(-1-3*p)*(b*x^3+a*x^2)^p,x, algorithm="giac")
Output:
integrate((b*x^3 + a*x^2)^p*x^(-3*p - 1), x)
Timed out. \[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^p}{x^{3\,p+1}} \,d x \] Input:
int((a*x^2 + b*x^3)^p/x^(3*p + 1),x)
Output:
int((a*x^2 + b*x^3)^p/x^(3*p + 1), x)
\[ \int x^{-1-3 p} \left (a x^2+b x^3\right )^p \, dx=\int \frac {\left (b \,x^{3}+a \,x^{2}\right )^{p}}{x^{3 p} x}d x \] Input:
int(x^(-1-3*p)*(b*x^3+a*x^2)^p,x)
Output:
int((a*x**2 + b*x**3)**p/(x**(3*p)*x),x)