Integrand size = 15, antiderivative size = 75 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\frac {3}{a x \sqrt [3]{a x^2+b x^3}}-\frac {18 \left (a x^2+b x^3\right )^{2/3}}{5 a^2 x^3}+\frac {27 b \left (a x^2+b x^3\right )^{2/3}}{5 a^3 x^2} \] Output:
3/a/x/(b*x^3+a*x^2)^(1/3)-18/5*(b*x^3+a*x^2)^(2/3)/a^2/x^3+27/5*b*(b*x^3+a *x^2)^(2/3)/a^3/x^2
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=-\frac {3 x (a+b x) \left (a^2-3 a b x-9 b^2 x^2\right )}{5 a^3 \left (x^2 (a+b x)\right )^{4/3}} \] Input:
Integrate[(a*x^2 + b*x^3)^(-4/3),x]
Output:
(-3*x*(a + b*x)*(a^2 - 3*a*b*x - 9*b^2*x^2))/(5*a^3*(x^2*(a + b*x))^(4/3))
Time = 0.38 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1907, 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.
\(\displaystyle \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 1907 |
\(\displaystyle \frac {6 \int \frac {1}{x^2 \sqrt [3]{b x^3+a x^2}}dx}{a}+\frac {3}{a x \sqrt [3]{a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {6 \left (-\frac {3 b \int \frac {1}{x \sqrt [3]{b x^3+a x^2}}dx}{5 a}-\frac {3 \left (a x^2+b x^3\right )^{2/3}}{5 a x^3}\right )}{a}+\frac {3}{a x \sqrt [3]{a x^2+b x^3}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {6 \left (\frac {9 b \left (a x^2+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {3 \left (a x^2+b x^3\right )^{2/3}}{5 a x^3}\right )}{a}+\frac {3}{a x \sqrt [3]{a x^2+b x^3}}\) |
Input:
Int[(a*x^2 + b*x^3)^(-4/3),x]
Output:
3/(a*x*(a*x^2 + b*x^3)^(1/3)) + (6*((-3*(a*x^2 + b*x^3)^(2/3))/(5*a*x^3) + (9*b*(a*x^2 + b*x^3)^(2/3))/(10*a^2*x^2)))/a
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ (a*(n - j)*(p + 1)) Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, b, j, n}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1]
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 = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49
method | result | size |
pseudoelliptic | \(-\frac {3 \left (-9 b^{2} x^{2}-3 a b x +a^{2}\right )}{5 x \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} a^{3}}\) | \(37\) |
gosper | \(-\frac {3 x \left (b x +a \right ) \left (-9 b^{2} x^{2}-3 a b x +a^{2}\right )}{5 a^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {4}{3}}}\) | \(42\) |
orering | \(-\frac {3 x \left (b x +a \right ) \left (-9 b^{2} x^{2}-3 a b x +a^{2}\right )}{5 a^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {4}{3}}}\) | \(42\) |
trager | \(-\frac {3 \left (-9 b^{2} x^{2}-3 a b x +a^{2}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {2}{3}}}{5 \left (b x +a \right ) a^{3} x^{3}}\) | \(46\) |
risch | \(-\frac {3 \left (b x +a \right ) \left (-4 b x +a \right )}{5 a^{3} x \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}+\frac {3 b^{2} x}{\left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} a^{3}}\) | \(52\) |
Input:
int(1/(b*x^3+a*x^2)^(4/3),x,method=_RETURNVERBOSE)
Output:
-3/5/x*(-9*b^2*x^2-3*a*b*x+a^2)/(x^2*(b*x+a))^(1/3)/a^3
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\frac {3 \, {\left (9 \, b^{2} x^{2} + 3 \, a b x - a^{2}\right )} {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}}}{5 \, {\left (a^{3} b x^{4} + a^{4} x^{3}\right )}} \] Input:
integrate(1/(b*x^3+a*x^2)^(4/3),x, algorithm="fricas")
Output:
3/5*(9*b^2*x^2 + 3*a*b*x - a^2)*(b*x^3 + a*x^2)^(2/3)/(a^3*b*x^4 + a^4*x^3 )
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(1/(b*x**3+a*x**2)**(4/3),x)
Output:
Integral((a*x**2 + b*x**3)**(-4/3), x)
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(1/(b*x^3+a*x^2)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(-4/3), x)
Time = 3.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\frac {3 \, {\left (\frac {5 \, b^{2}}{a {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}}} - \frac {a^{4} {\left (b + \frac {a}{x}\right )}^{\frac {5}{3}} - 5 \, a^{4} {\left (b + \frac {a}{x}\right )}^{\frac {2}{3}} b}{a^{5}}\right )}}{5 \, a^{2}} \] Input:
integrate(1/(b*x^3+a*x^2)^(4/3),x, algorithm="giac")
Output:
3/5*(5*b^2/(a*(b + a/x)^(1/3)) - (a^4*(b + a/x)^(5/3) - 5*a^4*(b + a/x)^(2 /3)*b)/a^5)/a^2
Time = 9.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\frac {3\,{\left (b\,x^3+a\,x^2\right )}^{2/3}\,\left (-a^2+3\,a\,b\,x+9\,b^2\,x^2\right )}{5\,a^3\,x^3\,\left (a+b\,x\right )} \] Input:
int(1/(a*x^2 + b*x^3)^(4/3),x)
Output:
(3*(a*x^2 + b*x^3)^(2/3)*(9*b^2*x^2 - a^2 + 3*a*b*x))/(5*a^3*x^3*(a + b*x) )
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\int \frac {1}{x^{\frac {8}{3}} \left (b x +a \right )^{\frac {1}{3}} a +x^{\frac {11}{3}} \left (b x +a \right )^{\frac {1}{3}} b}d x \] Input:
int(1/(b*x^3+a*x^2)^(4/3),x)
Output:
int(1/(x**(2/3)*(a + b*x)**(1/3)*a*x**2 + x**(2/3)*(a + b*x)**(1/3)*b*x**3 ),x)