Integrand size = 13, antiderivative size = 48 \[ \int \left (a x^2+b x^3\right )^p \, dx=\frac {\left (a x^2+b x^3\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,2+3 p,2 (1+p),-\frac {b x}{a}\right )}{a (1+2 p) x} \] Output:
(b*x^3+a*x^2)^(p+1)*hypergeom([1, 2+3*p],[2*p+2],-b*x/a)/a/(1+2*p)/x
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \left (a x^2+b x^3\right )^p \, dx=\frac {x \left (x^2 (a+b x)\right )^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+2 p,2+2 p,-\frac {b x}{a}\right )}{1+2 p} \] Input:
Integrate[(a*x^2 + b*x^3)^p,x]
Output:
(x*(x^2*(a + b*x))^p*Hypergeometric2F1[-p, 1 + 2*p, 2 + 2*p, -((b*x)/a)])/ ((1 + 2*p)*(1 + (b*x)/a)^p)
Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1917, 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.
\(\displaystyle \int \left (a x^2+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle x^{-2 p} (a+b x)^{-p} \left (a x^2+b x^3\right )^p \int x^{2 p} (a+b x)^pdx\) |
\(\Big \downarrow \) 76 |
\(\displaystyle x^{-2 p} \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \int x^{2 p} \left (\frac {b x}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {x \left (\frac {b x}{a}+1\right )^{-p} \left (a x^2+b x^3\right )^p \operatorname {Hypergeometric2F1}\left (-p,2 p+1,2 (p+1),-\frac {b x}{a}\right )}{2 p+1}\) |
Input:
Int[(a*x^2 + b*x^3)^p,x]
Output:
(x*(a*x^2 + b*x^3)^p*Hypergeometric2F1[-p, 1 + 2*p, 2*(1 + p), -((b*x)/a)] )/((1 + 2*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[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
\[\int \left (b \,x^{3}+a \,x^{2}\right )^{p}d x\]
Input:
int((b*x^3+a*x^2)^p,x)
Output:
int((b*x^3+a*x^2)^p,x)
\[ \int \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p,x, algorithm="fricas")
Output:
integral((b*x^3 + a*x^2)^p, x)
\[ \int \left (a x^2+b x^3\right )^p \, dx=\int \left (a x^{2} + b x^{3}\right )^{p}\, dx \] Input:
integrate((b*x**3+a*x**2)**p,x)
Output:
Integral((a*x**2 + b*x**3)**p, x)
\[ \int \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^p, x)
\[ \int \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} \,d x } \] Input:
integrate((b*x^3+a*x^2)^p,x, algorithm="giac")
Output:
integrate((b*x^3 + a*x^2)^p, x)
Time = 8.74 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \left (a x^2+b x^3\right )^p \, dx=\frac {x\,{\left (b\,x^3+a\,x^2\right )}^p\,{{}}_2{\mathrm {F}}_1\left (2\,p+1,-p;\ 2\,p+2;\ -\frac {b\,x}{a}\right )}{\left (2\,p+1\right )\,{\left (\frac {b\,x}{a}+1\right )}^p} \] Input:
int((a*x^2 + b*x^3)^p,x)
Output:
(x*(a*x^2 + b*x^3)^p*hypergeom([2*p + 1, -p], 2*p + 2, -(b*x)/a))/((2*p + 1)*((b*x)/a + 1)^p)
\[ \int \left (a x^2+b x^3\right )^p \, dx=\frac {\left (b \,x^{3}+a \,x^{2}\right )^{p} a +3 \left (b \,x^{3}+a \,x^{2}\right )^{p} b x -6 \left (\int \frac {\left (b \,x^{3}+a \,x^{2}\right )^{p}}{3 b p \,x^{2}+3 a p x +b \,x^{2}+a x}d x \right ) a^{2} p^{2}-2 \left (\int \frac {\left (b \,x^{3}+a \,x^{2}\right )^{p}}{3 b p \,x^{2}+3 a p x +b \,x^{2}+a x}d x \right ) a^{2} p}{3 b \left (3 p +1\right )} \] Input:
int((b*x^3+a*x^2)^p,x)
Output:
((a*x**2 + b*x**3)**p*a + 3*(a*x**2 + b*x**3)**p*b*x - 6*int((a*x**2 + b*x **3)**p/(3*a*p*x + a*x + 3*b*p*x**2 + b*x**2),x)*a**2*p**2 - 2*int((a*x**2 + b*x**3)**p/(3*a*p*x + a*x + 3*b*p*x**2 + b*x**2),x)*a**2*p)/(3*b*(3*p + 1))