Integrand size = 15, antiderivative size = 105 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=\frac {3}{2 a x \left (a x^2+b x^3\right )^{2/3}}-\frac {27 \sqrt [3]{a x^2+b x^3}}{14 a^2 x^3}+\frac {81 b \sqrt [3]{a x^2+b x^3}}{28 a^3 x^2}-\frac {243 b^2 \sqrt [3]{a x^2+b x^3}}{28 a^4 x} \] Output:
3/2/a/x/(b*x^3+a*x^2)^(2/3)-27/14*(b*x^3+a*x^2)^(1/3)/a^2/x^3+81/28*b*(b*x ^3+a*x^2)^(1/3)/a^3/x^2-243/28*b^2*(b*x^3+a*x^2)^(1/3)/a^4/x
Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=-\frac {3 \left (4 a^3-9 a^2 b x+54 a b^2 x^2+81 b^3 x^3\right )}{28 a^4 x \left (x^2 (a+b x)\right )^{2/3}} \] Input:
Integrate[(a*x^2 + b*x^3)^(-5/3),x]
Output:
(-3*(4*a^3 - 9*a^2*b*x + 54*a*b^2*x^2 + 81*b^3*x^3))/(28*a^4*x*(x^2*(a + b *x))^(2/3))
Time = 0.45 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1907, 1922, 1922, 1906}
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.
\(\displaystyle \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx\) |
\(\Big \downarrow \) 1907 |
\(\displaystyle \frac {9 \int \frac {1}{x^2 \left (b x^3+a x^2\right )^{2/3}}dx}{2 a}+\frac {3}{2 a x \left (a x^2+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {9 \left (-\frac {6 b \int \frac {1}{x \left (b x^3+a x^2\right )^{2/3}}dx}{7 a}-\frac {3 \sqrt [3]{a x^2+b x^3}}{7 a x^3}\right )}{2 a}+\frac {3}{2 a x \left (a x^2+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {9 \left (-\frac {6 b \left (-\frac {3 b \int \frac {1}{\left (b x^3+a x^2\right )^{2/3}}dx}{4 a}-\frac {3 \sqrt [3]{a x^2+b x^3}}{4 a x^2}\right )}{7 a}-\frac {3 \sqrt [3]{a x^2+b x^3}}{7 a x^3}\right )}{2 a}+\frac {3}{2 a x \left (a x^2+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 1906 |
\(\displaystyle \frac {9 \left (-\frac {6 b \left (\frac {9 b \sqrt [3]{a x^2+b x^3}}{4 a^2 x}-\frac {3 \sqrt [3]{a x^2+b x^3}}{4 a x^2}\right )}{7 a}-\frac {3 \sqrt [3]{a x^2+b x^3}}{7 a x^3}\right )}{2 a}+\frac {3}{2 a x \left (a x^2+b x^3\right )^{2/3}}\) |
Input:
Int[(a*x^2 + b*x^3)^(-5/3),x]
Output:
3/(2*a*x*(a*x^2 + b*x^3)^(2/3)) + (9*((-3*(a*x^2 + b*x^3)^(1/3))/(7*a*x^3) - (6*b*((-3*(a*x^2 + b*x^3)^(1/3))/(4*a*x^2) + (9*b*(a*x^2 + b*x^3)^(1/3) )/(4*a^2*x)))/(7*a)))/(2*a)
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]
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*(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 = 0.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.48
method | result | size |
pseudoelliptic | \(\frac {-\frac {243}{28} b^{3} x^{3}-\frac {81}{14} a \,b^{2} x^{2}+\frac {27}{28} a^{2} b x -\frac {3}{7} a^{3}}{x \left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}} a^{4}}\) | \(50\) |
gosper | \(-\frac {3 x \left (b x +a \right ) \left (81 b^{3} x^{3}+54 a \,b^{2} x^{2}-9 a^{2} b x +4 a^{3}\right )}{28 a^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{3}}}\) | \(55\) |
orering | \(-\frac {3 x \left (b x +a \right ) \left (81 b^{3} x^{3}+54 a \,b^{2} x^{2}-9 a^{2} b x +4 a^{3}\right )}{28 a^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{3}}}\) | \(55\) |
trager | \(-\frac {3 \left (81 b^{3} x^{3}+54 a \,b^{2} x^{2}-9 a^{2} b x +4 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {1}{3}}}{28 \left (b x +a \right ) a^{4} x^{3}}\) | \(59\) |
risch | \(-\frac {3 \left (b x +a \right ) \left (67 b^{2} x^{2}-13 a b x +4 a^{2}\right )}{28 a^{4} x \left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}-\frac {3 b^{3} x^{2}}{2 a^{4} \left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}\) | \(67\) |
Input:
int(1/(b*x^3+a*x^2)^(5/3),x,method=_RETURNVERBOSE)
Output:
3/28*(-81*b^3*x^3-54*a*b^2*x^2+9*a^2*b*x-4*a^3)/x/(x^2*(b*x+a))^(2/3)/a^4
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=-\frac {3 \, {\left (81 \, b^{3} x^{3} + 54 \, a b^{2} x^{2} - 9 \, a^{2} b x + 4 \, a^{3}\right )} {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{28 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \] Input:
integrate(1/(b*x^3+a*x^2)^(5/3),x, algorithm="fricas")
Output:
-3/28*(81*b^3*x^3 + 54*a*b^2*x^2 - 9*a^2*b*x + 4*a^3)*(b*x^3 + a*x^2)^(1/3 )/(a^4*b*x^4 + a^5*x^3)
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {5}{3}}}\, dx \] Input:
integrate(1/(b*x**3+a*x**2)**(5/3),x)
Output:
Integral((a*x**2 + b*x**3)**(-5/3), x)
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate(1/(b*x^3+a*x^2)^(5/3),x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(-5/3), x)
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=-\frac {3 \, b^{3}}{2 \, a^{4} {\left (b + \frac {a}{x}\right )}^{\frac {2}{3}}} - \frac {3 \, {\left (4 \, a^{24} {\left (b + \frac {a}{x}\right )}^{\frac {7}{3}} - 21 \, a^{24} {\left (b + \frac {a}{x}\right )}^{\frac {4}{3}} b + 84 \, a^{24} {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b^{2}\right )}}{28 \, a^{28}} \] Input:
integrate(1/(b*x^3+a*x^2)^(5/3),x, algorithm="giac")
Output:
-3/2*b^3/(a^4*(b + a/x)^(2/3)) - 3/28*(4*a^24*(b + a/x)^(7/3) - 21*a^24*(b + a/x)^(4/3)*b + 84*a^24*(b + a/x)^(1/3)*b^2)/a^28
Time = 9.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=-\frac {3\,{\left (b\,x^3+a\,x^2\right )}^{1/3}\,\left (4\,a^3-9\,a^2\,b\,x+54\,a\,b^2\,x^2+81\,b^3\,x^3\right )}{28\,a^4\,x^3\,\left (a+b\,x\right )} \] Input:
int(1/(a*x^2 + b*x^3)^(5/3),x)
Output:
-(3*(a*x^2 + b*x^3)^(1/3)*(4*a^3 + 81*b^3*x^3 + 54*a*b^2*x^2 - 9*a^2*b*x)) /(28*a^4*x^3*(a + b*x))
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/3}} \, dx=\int \frac {1}{x^{\frac {10}{3}} \left (b x +a \right )^{\frac {2}{3}} a +x^{\frac {13}{3}} \left (b x +a \right )^{\frac {2}{3}} b}d x \] Input:
int(1/(b*x^3+a*x^2)^(5/3),x)
Output:
int(1/(x**(1/3)*(a + b*x)**(2/3)*a*x**3 + x**(1/3)*(a + b*x)**(2/3)*b*x**4 ),x)