Integrand size = 19, antiderivative size = 86 \[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{4 b^{5/2}} \]
3/4*a^2*arctanh(x^2*b^(1/2)/(b*x^4+a*x^3)^(1/2))/b^(5/2)+1/2*(b*x^4+a*x^3) ^(1/2)/b-3/4*a*(b*x^4+a*x^3)^(1/2)/b^2/x
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\frac {\sqrt {b} x^2 \left (-3 a^2-a b x+2 b^2 x^2\right )+6 a^2 x^{3/2} \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{5/2} \sqrt {x^3 (a+b x)}} \]
(Sqrt[b]*x^2*(-3*a^2 - a*b*x + 2*b^2*x^2) + 6*a^2*x^(3/2)*Sqrt[a + b*x]*Ar cTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(4*b^(5/2)*Sqrt[x^3*( a + b*x)])
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1930, 1930, 1935, 219}
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.
\(\displaystyle \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a x^3}}dx}{4 b}\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \int \frac {x}{\sqrt {b x^4+a x^3}}dx}{2 b}\right )}{4 b}\) |
\(\Big \downarrow \) 1935 |
\(\displaystyle \frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a x^3}}d\frac {x^2}{\sqrt {b x^4+a x^3}}}{b}\right )}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{b^{3/2}}\right )}{4 b}\) |
Sqrt[a*x^3 + b*x^4]/(2*b) - (3*a*(Sqrt[a*x^3 + b*x^4]/(b*x) - (a*ArcTanh[( Sqrt[b]*x^2)/Sqrt[a*x^3 + b*x^4]])/b^(3/2)))/(4*b)
3.4.11.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))) I nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp [-2/(n - j) Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
Time = 2.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{3} \left (b x +a \right )}}{x^{2} \sqrt {b}}\right ) a^{2} x +2 b^{\frac {3}{2}} x \sqrt {x^{3} \left (b x +a \right )}-3 a \sqrt {b}\, \sqrt {x^{3} \left (b x +a \right )}}{4 b^{\frac {5}{2}} x}\) | \(69\) |
risch | \(-\frac {\left (-2 b x +3 a \right ) x^{2} \left (b x +a \right )}{4 b^{2} \sqrt {x^{3} \left (b x +a \right )}}+\frac {3 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) x \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}} \sqrt {x^{3} \left (b x +a \right )}}\) | \(87\) |
default | \(\frac {x \sqrt {x \left (b x +a \right )}\, \left (4 x \sqrt {b \,x^{2}+a x}\, b^{\frac {5}{2}}-6 \sqrt {b \,x^{2}+a x}\, b^{\frac {3}{2}} a +3 \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b \right )}{8 \sqrt {b \,x^{4}+a \,x^{3}}\, b^{\frac {7}{2}}}\) | \(98\) |
1/4/b^(5/2)*(3*arctanh((x^3*(b*x+a))^(1/2)/x^2/b^(1/2))*a^2*x+2*b^(3/2)*x* (x^3*(b*x+a))^(1/2)-3*a*b^(1/2)*(x^3*(b*x+a))^(1/2))/x
Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.74 \[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x + 2 \, \sqrt {b x^{4} + a x^{3}} \sqrt {b}}{x}\right ) + 2 \, \sqrt {b x^{4} + a x^{3}} {\left (2 \, b^{2} x - 3 \, a b\right )}}{8 \, b^{3} x}, -\frac {3 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{4} + a x^{3}} \sqrt {-b}}{b x^{2}}\right ) - \sqrt {b x^{4} + a x^{3}} {\left (2 \, b^{2} x - 3 \, a b\right )}}{4 \, b^{3} x}\right ] \]
[1/8*(3*a^2*sqrt(b)*x*log((2*b*x^2 + a*x + 2*sqrt(b*x^4 + a*x^3)*sqrt(b))/ x) + 2*sqrt(b*x^4 + a*x^3)*(2*b^2*x - 3*a*b))/(b^3*x), -1/4*(3*a^2*sqrt(-b )*x*arctan(sqrt(b*x^4 + a*x^3)*sqrt(-b)/(b*x^2)) - sqrt(b*x^4 + a*x^3)*(2* b^2*x - 3*a*b))/(b^3*x)]
\[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\int \frac {x^{3}}{\sqrt {x^{3} \left (a + b x\right )}}\, dx \]
\[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\int { \frac {x^{3}}{\sqrt {b x^{4} + a x^{3}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} {\left (\frac {2 \, x}{b \mathrm {sgn}\left (x\right )} - \frac {3 \, a}{b^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {3 \, a^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, b^{\frac {5}{2}}} - \frac {3 \, a^{2} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{8 \, b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]
1/4*sqrt(b*x^2 + a*x)*(2*x/(b*sgn(x)) - 3*a/(b^2*sgn(x))) + 3/8*a^2*log(ab s(a))*sgn(x)/b^(5/2) - 3/8*a^2*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*s qrt(b) + a))/(b^(5/2)*sgn(x))
Timed out. \[ \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx=\int \frac {x^3}{\sqrt {b\,x^4+a\,x^3}} \,d x \]