Integrand size = 19, antiderivative size = 87 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{5/2}} \] Output:
-1/2*(b*x^3+a*x^2)^(1/2)/a/x^3+3/4*b*(b*x^3+a*x^2)^(1/2)/a^2/x^2-3/4*b^2*a rctanh(a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^(5/2)
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {a} \left (-2 a^2+a b x+3 b^2 x^2\right )-3 b^2 x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2} x \sqrt {x^2 (a+b x)}} \] Input:
Integrate[1/(x^2*Sqrt[a*x^2 + b*x^3]),x]
Output:
(Sqrt[a]*(-2*a^2 + a*b*x + 3*b^2*x^2) - 3*b^2*x^2*Sqrt[a + b*x]*ArcTanh[Sq rt[a + b*x]/Sqrt[a]])/(4*a^(5/2)*x*Sqrt[x^2*(a + b*x)])
Time = 0.37 (sec) , antiderivative size = 88, 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 = {1931, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle -\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle -\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle -\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\) |
Input:
Int[1/(x^2*Sqrt[a*x^2 + b*x^3]),x]
Output:
-1/2*Sqrt[a*x^2 + b*x^3]/(a*x^3) - (3*b*(-(Sqrt[a*x^2 + b*x^3]/(a*x^2)) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)))/(4*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*(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 = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41
method | result | size |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x -\sqrt {b x +a}\, \sqrt {a}}{a^{\frac {3}{2}} x}\) | \(36\) |
risch | \(-\frac {\left (b x +a \right ) \left (-3 b x +2 a \right )}{4 a^{2} x \sqrt {x^{2} \left (b x +a \right )}}-\frac {3 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{4 a^{\frac {5}{2}} \sqrt {x^{2} \left (b x +a \right )}}\) | \(73\) |
default | \(-\frac {\sqrt {b x +a}\, \left (-3 a^{\frac {3}{2}} b x \sqrt {b x +a}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} x^{2}+2 \sqrt {b x +a}\, a^{\frac {5}{2}}\right )}{4 x \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {7}{2}}}\) | \(77\) |
Input:
int(1/x^2/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
(arctanh((b*x+a)^(1/2)/a^(1/2))*b*x-(b*x+a)^(1/2)*a^(1/2))/a^(3/2)/x
Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.82 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{8 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{4 \, a^{3} x^{3}}\right ] \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
[1/8*(3*sqrt(a)*b^2*x^3*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 - 2*a^2))/(a^3*x^3), 1/4*(3*sqrt(- a)*b^2*x^3*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 - 2*a^2))/(a^3*x^3)]
\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{2} \sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:
integrate(1/x**2/(b*x**3+a*x**2)**(1/2),x)
Output:
Integral(1/(x**2*sqrt(x**2*(a + b*x))), x)
\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*x^3 + a*x^2)*x^2), x)
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b x + a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
1/4*(3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(3 /2)*b^3 - 5*sqrt(b*x + a)*a*b^3)/(a^2*b^2*x^2))/(b*sgn(x))
Time = 9.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=-\frac {2\,\sqrt {\frac {a}{b\,x}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {a}{b\,x}\right )}{5\,x\,\sqrt {b\,x^3+a\,x^2}} \] Input:
int(1/(x^2*(a*x^2 + b*x^3)^(1/2)),x)
Output:
-(2*(a/(b*x) + 1)^(1/2)*hypergeom([1/2, 5/2], 7/2, -a/(b*x)))/(5*x*(a*x^2 + b*x^3)^(1/2))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2}+6 \sqrt {b x +a}\, a b x +3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}}{8 a^{3} x^{2}} \] Input:
int(1/x^2/(b*x^3+a*x^2)^(1/2),x)
Output:
( - 4*sqrt(a + b*x)*a**2 + 6*sqrt(a + b*x)*a*b*x + 3*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**2*x**2 - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*x* *2)/(8*a**3*x**2)