Integrand size = 19, antiderivative size = 108 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{35 a^3 x^3}+\frac {32 b^3 \sqrt {a x^3+b x^4}}{35 a^4 x^2} \] Output:
-2/7*(b*x^4+a*x^3)^(1/2)/a/x^5+12/35*b*(b*x^4+a*x^3)^(1/2)/a^2/x^4-16/35*b ^2*(b*x^4+a*x^3)^(1/2)/a^3/x^3+32/35*b^3*(b*x^4+a*x^3)^(1/2)/a^4/x^2
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\frac {2 \sqrt {x^3 (a+b x)} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^5} \] Input:
Integrate[1/(x^3*Sqrt[a*x^3 + b*x^4]),x]
Output:
(2*Sqrt[x^3*(a + b*x)]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(3 5*a^4*x^5)
Time = 0.45 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1922, 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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \int \frac {1}{x^2 \sqrt {b x^4+a x^3}}dx}{7 a}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {1}{x \sqrt {b x^4+a x^3}}dx}{5 a}-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}\right )}{7 a}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{\sqrt {b x^4+a x^3}}dx}{3 a}-\frac {2 \sqrt {a x^3+b x^4}}{3 a x^3}\right )}{5 a}-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}\right )}{7 a}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}\) |
\(\Big \downarrow \) 1906 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (\frac {4 b \sqrt {a x^3+b x^4}}{3 a^2 x^2}-\frac {2 \sqrt {a x^3+b x^4}}{3 a x^3}\right )}{5 a}-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}\right )}{7 a}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}\) |
Input:
Int[1/(x^3*Sqrt[a*x^3 + b*x^4]),x]
Output:
(-2*Sqrt[a*x^3 + b*x^4])/(7*a*x^5) - (6*b*((-2*Sqrt[a*x^3 + b*x^4])/(5*a*x ^4) - (4*b*((-2*Sqrt[a*x^3 + b*x^4])/(3*a*x^3) + (4*b*Sqrt[a*x^3 + b*x^4]) /(3*a^2*x^2)))/(5*a)))/(7*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[((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 = 0.70 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.46
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right ) \sqrt {x^{3} \left (b x +a \right )}}{35 a^{4} x^{5}}\) | \(50\) |
trager | \(-\frac {2 \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right ) \sqrt {b \,x^{4}+a \,x^{3}}}{35 a^{4} x^{5}}\) | \(52\) |
risch | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{2} \sqrt {x^{3} \left (b x +a \right )}\, a^{4}}\) | \(55\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{2} a^{4} \sqrt {b \,x^{4}+a \,x^{3}}}\) | \(57\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{2} a^{4} \sqrt {b \,x^{4}+a \,x^{3}}}\) | \(57\) |
default | \(-\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b \,x^{2}+a x}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{3} \sqrt {b \,x^{4}+a \,x^{3}}\, a^{4}}\) | \(72\) |
Input:
int(1/x^3/(b*x^4+a*x^3)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/35*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b*x+5*a^3)/a^4/x^5*(x^3*(b*x+a))^(1/2 )
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x^{4} + a x^{3}}}{35 \, a^{4} x^{5}} \] Input:
integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="fricas")
Output:
2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x^4 + a*x^3)/(a ^4*x^5)
\[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\int \frac {1}{x^{3} \sqrt {x^{3} \left (a + b x\right )}}\, dx \] Input:
integrate(1/x**3/(b*x**4+a*x**3)**(1/2),x)
Output:
Integral(1/(x**3*sqrt(x**3*(a + b*x))), x)
\[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\int { \frac {1}{\sqrt {b x^{4} + a x^{3}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*x^4 + a*x^3)*x^3), x)
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\frac {2 \, {\left (70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} b^{\frac {3}{2}} + 84 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a b + 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{2} \sqrt {b} + 5 \, a^{3}\right )}}{35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{7} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="giac")
Output:
2/35*(70*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*b^(3/2) + 84*(sqrt(b)*x - sqrt( b*x^2 + a*x))^2*a*b + 35*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^2*sqrt(b) + 5*a ^3)/((sqrt(b)*x - sqrt(b*x^2 + a*x))^7*sgn(x))
Time = 8.92 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\frac {12\,b\,\sqrt {b\,x^4+a\,x^3}}{35\,a^2\,x^4}-\frac {2\,\sqrt {b\,x^4+a\,x^3}}{7\,a\,x^5}-\frac {16\,b^2\,\sqrt {b\,x^4+a\,x^3}}{35\,a^3\,x^3}+\frac {32\,b^3\,\sqrt {b\,x^4+a\,x^3}}{35\,a^4\,x^2} \] Input:
int(1/(x^3*(a*x^3 + b*x^4)^(1/2)),x)
Output:
(12*b*(a*x^3 + b*x^4)^(1/2))/(35*a^2*x^4) - (2*(a*x^3 + b*x^4)^(1/2))/(7*a *x^5) - (16*b^2*(a*x^3 + b*x^4)^(1/2))/(35*a^3*x^3) + (32*b^3*(a*x^3 + b*x ^4)^(1/2))/(35*a^4*x^2)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{3}}{7}+\frac {12 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b x}{35}-\frac {16 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x^{2}}{35}+\frac {32 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{3}}{35}-\frac {32 \sqrt {b}\, b^{3} x^{4}}{35}}{a^{4} x^{4}} \] Input:
int(1/x^3/(b*x^4+a*x^3)^(1/2),x)
Output:
(2*( - 5*sqrt(x)*sqrt(a + b*x)*a**3 + 6*sqrt(x)*sqrt(a + b*x)*a**2*b*x - 8 *sqrt(x)*sqrt(a + b*x)*a*b**2*x**2 + 16*sqrt(x)*sqrt(a + b*x)*b**3*x**3 - 16*sqrt(b)*b**3*x**4))/(35*a**4*x**4)