Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}} \]
-15/4*b^2*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(7/2)+2/a/x/(b*x^3+a*x^ 2)^(1/2)-5/2*(b*x^3+a*x^2)^(1/2)/a^2/x^3+15/4*b*(b*x^3+a*x^2)^(1/2)/a^3/x^ 2
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-2 a^2+5 a b x+15 b^2 x^2\right )-15 b^2 x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2} x \sqrt {x^2 (a+b x)}} \]
(Sqrt[a]*(-2*a^2 + 5*a*b*x + 15*b^2*x^2) - 15*b^2*x^2*Sqrt[a + b*x]*ArcTan h[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(7/2)*x*Sqrt[x^2*(a + b*x)])
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1912, 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.
\(\displaystyle \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1912 |
\(\displaystyle \frac {5 \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {5 \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {5 \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {5 \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\) |
2/(a*x*Sqrt[a*x^2 + b*x^3]) + (5*(-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)))/a
3.3.66.3.1 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[((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}, x] && !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]
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 = 1.83 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.12
method | result | size |
pseudoelliptic | \(-\frac {2}{b \sqrt {b x +a}}\) | \(13\) |
default | \(-\frac {x \left (b x +a \right ) \left (15 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-5 a^{\frac {3}{2}} b x -15 \sqrt {a}\, b^{2} x^{2}+2 a^{\frac {5}{2}}\right )}{4 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {7}{2}}}\) | \(76\) |
risch | \(-\frac {\left (b x +a \right ) \left (-7 b x +2 a \right )}{4 a^{3} x \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{2} \left (\frac {16}{\sqrt {b x +a}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{8 a^{3} \sqrt {x^{2} \left (b x +a \right )}}\) | \(88\) |
Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.99 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{8 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{4 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \]
[1/8*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4 + a^5*x^3), 1/4*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(-a)*arct an(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3 )*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4 + a^5*x^3)]
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (x\right )} \]
15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3*sgn(x)) + 2*b^2/(sqr t(b*x + a)*a^3*sgn(x)) + 1/4*(7*(b*x + a)^(3/2)*b^2 - 9*sqrt(b*x + a)*a*b^ 2)/(a^3*b^2*x^2*sgn(x))
Time = 9.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2\,x\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a}{b\,x}\right )}{7\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \]