Integrand size = 17, antiderivative size = 75 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a \sqrt {a x^2+b x^3}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}} \] Output:
2/a/(b*x^3+a*x^2)^(1/2)-3*(b*x^3+a*x^2)^(1/2)/a^2/x^2+3*b*arctanh(a^(1/2)* x/(b*x^3+a*x^2)^(1/2))/a^(5/2)
Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-\sqrt {a} (a+3 b x)+3 b x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[x/(a*x^2 + b*x^3)^(3/2),x]
Output:
(-(Sqrt[a]*(a + 3*b*x)) + 3*b*x*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a ]])/(a^(5/2)*Sqrt[x^2*(a + b*x)])
Time = 0.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1929, 1931, 1914, 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.
\(\displaystyle \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1929 |
\(\displaystyle \frac {3 \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {3 \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{a}+\frac {2}{a \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {3 \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{a}+\frac {2}{a \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{a}+\frac {2}{a \sqrt {a x^2+b x^3}}\) |
Input:
Int[x/(a*x^2 + b*x^3)^(3/2),x]
Output:
2/(a*Sqrt[a*x^2 + b*x^3]) + (3*(-(Sqrt[a*x^2 + b*x^3]/(a*x^2)) + (b*ArcTan h[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)))/a
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[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]
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] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] & & !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 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, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ m + j*p + 1, 0]
Time = 0.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.27
method | result | size |
pseudoelliptic | \(\frac {2 b x +4 a}{b^{2} \sqrt {b x +a}}\) | \(20\) |
default | \(-\frac {x^{2} \left (b x +a \right ) \left (-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, b x +3 b x \sqrt {a}+a^{\frac {3}{2}}\right )}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {5}{2}}}\) | \(61\) |
risch | \(-\frac {b x +a}{a^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b \left (-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {4}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{2 a^{2} \sqrt {x^{2} \left (b x +a \right )}}\) | \(75\) |
Input:
int(x/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
(2*b*x+4*a)/b^2/(b*x+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.59 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x + a^{2}\right )}}{2 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x + a^{2}\right )}}{a^{3} b x^{3} + a^{4} x^{2}}\right ] \] Input:
integrate(x/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/2*(3*(b^2*x^3 + a*b*x^2)*sqrt(a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a* x^2)*sqrt(a))/x^2) - 2*sqrt(b*x^3 + a*x^2)*(3*a*b*x + a^2))/(a^3*b*x^3 + a ^4*x^2), -(3*(b^2*x^3 + a*b*x^2)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt( -a)/(b*x^2 + a*x)) + sqrt(b*x^3 + a*x^2)*(3*a*b*x + a^2))/(a^3*b*x^3 + a^4 *x^2)]
\[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x/(x**2*(a + b*x))**(3/2), x)
\[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate(x/(b*x^3 + a*x^2)^(3/2), x)
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (x\right )} - \frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
-3*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(x)) - (3*(b*x + a)*b - 2*a*b)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)*a^2*sgn(x))
Timed out. \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x}{{\left (b\,x^3+a\,x^2\right )}^{3/2}} \,d x \] Input:
int(x/(a*x^2 + b*x^3)^(3/2),x)
Output:
int(x/(a*x^2 + b*x^3)^(3/2), x)
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b x +3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b x -2 a^{2}-6 a b x}{2 \sqrt {b x +a}\, a^{3} x} \] Input:
int(x/(b*x^3+a*x^2)^(3/2),x)
Output:
( - 3*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b*x + 3*sqrt(a)*s qrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b*x - 2*a**2 - 6*a*b*x)/(2*sqrt( a + b*x)*a**3*x)