Integrand size = 21, antiderivative size = 116 \[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=-\frac {x^{-5-3 p} \left (a x^2+b x^3\right )^{1+p}}{a (3+p)}+\frac {2 b x^{-4-3 p} \left (a x^2+b x^3\right )^{1+p}}{a^2 (2+p) (3+p)}-\frac {2 b^2 x^{-3 (1+p)} \left (a x^2+b x^3\right )^{1+p}}{a^3 (1+p) (2+p) (3+p)} \] Output:
-x^(-5-3*p)*(b*x^3+a*x^2)^(p+1)/a/(3+p)+2*b*x^(-4-3*p)*(b*x^3+a*x^2)^(p+1) /a^2/(2+p)/(3+p)-2*b^2*(b*x^3+a*x^2)^(p+1)/a^3/(p+1)/(2+p)/(3+p)/(x^(3*p+3 ))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62 \[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=-\frac {x^{-3 (1+p)} (a+b x) \left (x^2 (a+b x)\right )^p \left (a^2 \left (2+3 p+p^2\right )-2 a b (1+p) x+2 b^2 x^2\right )}{a^3 (1+p) (2+p) (3+p)} \] Input:
Integrate[x^(-4 - 3*p)*(a*x^2 + b*x^3)^p,x]
Output:
-(((a + b*x)*(x^2*(a + b*x))^p*(a^2*(2 + 3*p + p^2) - 2*a*b*(1 + p)*x + 2* b^2*x^2))/(a^3*(1 + p)*(2 + p)*(3 + p)*x^(3*(1 + p))))
Time = 0.46 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1922, 1922, 1920}
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.
\(\displaystyle \int x^{-3 p-4} \left (a x^2+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {2 b \int x^{-3 (p+1)} \left (b x^3+a x^2\right )^pdx}{a (p+3)}-\frac {x^{-3 p-5} \left (a x^2+b x^3\right )^{p+1}}{a (p+3)}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {2 b \left (-\frac {b \int x^{-3 p-2} \left (b x^3+a x^2\right )^pdx}{a (p+2)}-\frac {x^{-3 p-4} \left (a x^2+b x^3\right )^{p+1}}{a (p+2)}\right )}{a (p+3)}-\frac {x^{-3 p-5} \left (a x^2+b x^3\right )^{p+1}}{a (p+3)}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle -\frac {2 b \left (\frac {b x^{-3 (p+1)} \left (a x^2+b x^3\right )^{p+1}}{a^2 (p+1) (p+2)}-\frac {x^{-3 p-4} \left (a x^2+b x^3\right )^{p+1}}{a (p+2)}\right )}{a (p+3)}-\frac {x^{-3 p-5} \left (a x^2+b x^3\right )^{p+1}}{a (p+3)}\) |
Input:
Int[x^(-4 - 3*p)*(a*x^2 + b*x^3)^p,x]
Output:
-((x^(-5 - 3*p)*(a*x^2 + b*x^3)^(1 + p))/(a*(3 + p))) - (2*b*(-((x^(-4 - 3 *p)*(a*x^2 + b*x^3)^(1 + p))/(a*(2 + p))) + (b*(a*x^2 + b*x^3)^(1 + p))/(a ^2*(1 + p)*(2 + p)*x^(3*(1 + p)))))/(a*(3 + p))
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Time = 0.49 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72
method | result | size |
gosper | \(-\frac {x^{-3-3 p} \left (b x +a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{p} \left (a^{2} p^{2}-2 a b p x +2 b^{2} x^{2}+3 a^{2} p -2 a b x +2 a^{2}\right )}{a^{3} \left (2+p \right ) \left (3+p \right ) \left (p +1\right )}\) | \(84\) |
orering | \(-\frac {\left (b x +a \right ) x \left (a^{2} p^{2}-2 a b p x +2 b^{2} x^{2}+3 a^{2} p -2 a b x +2 a^{2}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{p} x^{-4-3 p}}{\left (3+p \right ) \left (2+p \right ) \left (p +1\right ) a^{3}}\) | \(85\) |
Input:
int(x^(-4-3*p)*(b*x^3+a*x^2)^p,x,method=_RETURNVERBOSE)
Output:
-x^(-3-3*p)/a^3/(2+p)/(3+p)/(p+1)*(b*x+a)*(b*x^3+a*x^2)^p*(a^2*p^2-2*a*b*p *x+2*b^2*x^2+3*a^2*p-2*a*b*x+2*a^2)
Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=\frac {{\left (2 \, a b^{2} p x^{3} - 2 \, b^{3} x^{4} - {\left (a^{2} b p^{2} + a^{2} b p\right )} x^{2} - {\left (a^{3} p^{2} + 3 \, a^{3} p + 2 \, a^{3}\right )} x\right )} {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 4}}{a^{3} p^{3} + 6 \, a^{3} p^{2} + 11 \, a^{3} p + 6 \, a^{3}} \] Input:
integrate(x^(-4-3*p)*(b*x^3+a*x^2)^p,x, algorithm="fricas")
Output:
(2*a*b^2*p*x^3 - 2*b^3*x^4 - (a^2*b*p^2 + a^2*b*p)*x^2 - (a^3*p^2 + 3*a^3* p + 2*a^3)*x)*(b*x^3 + a*x^2)^p*x^(-3*p - 4)/(a^3*p^3 + 6*a^3*p^2 + 11*a^3 *p + 6*a^3)
\[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=\int x^{- 3 p - 4} \left (x^{2} \left (a + b x\right )\right )^{p}\, dx \] Input:
integrate(x**(-4-3*p)*(b*x**3+a*x**2)**p,x)
Output:
Integral(x**(-3*p - 4)*(x**2*(a + b*x))**p, x)
\[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 4} \,d x } \] Input:
integrate(x^(-4-3*p)*(b*x^3+a*x^2)^p,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^p*x^(-3*p - 4), x)
\[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=\int { {\left (b x^{3} + a x^{2}\right )}^{p} x^{-3 \, p - 4} \,d x } \] Input:
integrate(x^(-4-3*p)*(b*x^3+a*x^2)^p,x, algorithm="giac")
Output:
integrate((b*x^3 + a*x^2)^p*x^(-3*p - 4), x)
Time = 9.35 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.35 \[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=-{\left (b\,x^3+a\,x^2\right )}^p\,\left (\frac {x\,\left (p^2+3\,p+2\right )}{x^{3\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {2\,b^3\,x^4}{a^3\,x^{3\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {2\,b^2\,p\,x^3}{a^2\,x^{3\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {b\,p\,x^2\,\left (p+1\right )}{a\,x^{3\,p+4}\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:
int((a*x^2 + b*x^3)^p/x^(3*p + 4),x)
Output:
-(a*x^2 + b*x^3)^p*((x*(3*p + p^2 + 2))/(x^(3*p + 4)*(11*p + 6*p^2 + p^3 + 6)) + (2*b^3*x^4)/(a^3*x^(3*p + 4)*(11*p + 6*p^2 + p^3 + 6)) - (2*b^2*p*x ^3)/(a^2*x^(3*p + 4)*(11*p + 6*p^2 + p^3 + 6)) + (b*p*x^2*(p + 1))/(a*x^(3 *p + 4)*(11*p + 6*p^2 + p^3 + 6)))
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84 \[ \int x^{-4-3 p} \left (a x^2+b x^3\right )^p \, dx=\frac {\left (b \,x^{3}+a \,x^{2}\right )^{p} \left (-a^{2} b \,p^{2} x +2 a \,b^{2} p \,x^{2}-2 b^{3} x^{3}-a^{3} p^{2}-a^{2} b p x -3 a^{3} p -2 a^{3}\right )}{x^{3 p} a^{3} x^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^(-4-3*p)*(b*x^3+a*x^2)^p,x)
Output:
((a*x**2 + b*x**3)**p*( - a**3*p**2 - 3*a**3*p - 2*a**3 - a**2*b*p**2*x - a**2*b*p*x + 2*a*b**2*p*x**2 - 2*b**3*x**3))/(x**(3*p)*a**3*x**3*(p**3 + 6 *p**2 + 11*p + 6))