Integrand size = 19, antiderivative size = 168 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {105 b^3}{64 a^4 \sqrt {a x^2+b x^3}}-\frac {1}{4 a x^3 \sqrt {a x^2+b x^3}}+\frac {3 b}{8 a^2 x^2 \sqrt {a x^2+b x^3}}-\frac {21 b^2}{32 a^3 x \sqrt {a x^2+b x^3}}+\frac {315 b^4 x}{64 a^5 \sqrt {a x^2+b x^3}}-\frac {315 b^4 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{11/2}} \] Output:
105/64*b^3/a^4/(b*x^3+a*x^2)^(1/2)-1/4/a/x^3/(b*x^3+a*x^2)^(1/2)+3/8*b/a^2 /x^2/(b*x^3+a*x^2)^(1/2)-21/32*b^2/a^3/x/(b*x^3+a*x^2)^(1/2)+315/64*b^4*x/ a^5/(b*x^3+a*x^2)^(1/2)-315/64*b^4*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/ a^(11/2)
Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-16 a^4+24 a^3 b x-42 a^2 b^2 x^2+105 a b^3 x^3+315 b^4 x^4\right )-315 b^4 x^4 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2} x^3 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[1/(x^2*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(Sqrt[a]*(-16*a^4 + 24*a^3*b*x - 42*a^2*b^2*x^2 + 105*a*b^3*x^3 + 315*b^4* x^4) - 315*b^4*x^4*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(11 /2)*x^3*Sqrt[x^2*(a + b*x)])
Time = 0.68 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1929, 1931, 1931, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1929 |
\(\displaystyle \frac {9 \int \frac {1}{x^4 \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {9 \left (-\frac {7 b \int \frac {1}{x^3 \sqrt {b x^3+a x^2}}dx}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {9 \left (-\frac {7 b \left (-\frac {5 b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\) |
Input:
Int[1/(x^2*(a*x^2 + b*x^3)^(3/2)),x]
Output:
2/(a*x^3*Sqrt[a*x^2 + b*x^3]) + (9*(-1/4*Sqrt[a*x^2 + b*x^3]/(a*x^5) - (7* b*(-1/3*Sqrt[a*x^2 + b*x^3]/(a*x^4) - (5*b*(-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)))/(6*a)))/(8*a)))/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 = 0.42 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.30
method | result | size |
pseudoelliptic | \(-\frac {-3 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x +\sqrt {a}\, \left (3 b x +a \right )}{\sqrt {b x +a}\, x \,a^{\frac {5}{2}}}\) | \(50\) |
default | \(-\frac {\left (b x +a \right ) \left (315 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}-24 a^{\frac {7}{2}} b x +42 a^{\frac {5}{2}} b^{2} x^{2}-105 a^{\frac {3}{2}} b^{3} x^{3}-315 \sqrt {a}\, b^{4} x^{4}+16 a^{\frac {9}{2}}\right )}{64 x \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {11}{2}}}\) | \(100\) |
risch | \(-\frac {\left (b x +a \right ) \left (-187 b^{3} x^{3}+82 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right )}{64 a^{5} x^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{4} \left (-\frac {630 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {256}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{128 a^{5} \sqrt {x^{2} \left (b x +a \right )}}\) | \(110\) |
Input:
int(1/x^2/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/(b*x+a)^(1/2)*(-3*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*b*x+a^(1 /2)*(3*b*x+a))/x/a^(5/2)
Time = 0.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.60 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{128 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}, \frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{64 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}\right ] \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
[1/128*(315*(b^5*x^6 + a*b^4*x^5)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^ 3 + a*x^2)*sqrt(a))/x^2) + 2*(315*a*b^4*x^4 + 105*a^2*b^3*x^3 - 42*a^3*b^2 *x^2 + 24*a^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2))/(a^6*b*x^6 + a^7*x^5), 1/ 64*(315*(b^5*x^6 + a*b^4*x^5)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a) /(b*x^2 + a*x)) + (315*a*b^4*x^4 + 105*a^2*b^3*x^3 - 42*a^3*b^2*x^2 + 24*a ^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2))/(a^6*b*x^6 + a^7*x^5)]
\[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**2/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(1/(x**2*(x**2*(a + b*x))**(3/2)), x)
\[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a*x^2)^(3/2)*x^2), x)
Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {315 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{4}}{\sqrt {b x + a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {187 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} - 643 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} + 765 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 325 \, \sqrt {b x + a} a^{3} b^{4}}{64 \, a^{5} b^{4} x^{4} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^2/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
315/64*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5*sgn(x)) + 2*b^4/(s qrt(b*x + a)*a^5*sgn(x)) + 1/64*(187*(b*x + a)^(7/2)*b^4 - 643*(b*x + a)^( 5/2)*a*b^4 + 765*(b*x + a)^(3/2)*a^2*b^4 - 325*sqrt(b*x + a)*a^3*b^4)/(a^5 *b^4*x^4*sgn(x))
Time = 9.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {11}{2};\ \frac {13}{2};\ -\frac {a}{b\,x}\right )}{11\,x\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \] Input:
int(1/(x^2*(a*x^2 + b*x^3)^(3/2)),x)
Output:
-(2*(a/(b*x) + 1)^(3/2)*hypergeom([3/2, 11/2], 13/2, -a/(b*x)))/(11*x*(a*x ^2 + b*x^3)^(3/2))
Time = 0.42 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}-32 a^{5}+48 a^{4} b x -84 a^{3} b^{2} x^{2}+210 a^{2} b^{3} x^{3}+630 a \,b^{4} x^{4}}{128 \sqrt {b x +a}\, a^{6} x^{4}} \] Input:
int(1/x^2/(b*x^3+a*x^2)^(3/2),x)
Output:
(315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**4*x**4 - 315*sq rt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**4*x**4 - 32*a**5 + 48* a**4*b*x - 84*a**3*b**2*x**2 + 210*a**2*b**3*x**3 + 630*a*b**4*x**4)/(128* sqrt(a + b*x)*a**6*x**4)