\(\int \frac {1}{x^2 (a x^2+b x^3)^{3/2}} \, dx\) [247]

Optimal result

Integrand size = 19, antiderivative size = 166 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a x^3 \sqrt {a x^2+b x^3}}-\frac {9 \sqrt {a x^2+b x^3}}{4 a^2 x^5}+\frac {21 b \sqrt {a x^2+b x^3}}{8 a^3 x^4}-\frac {105 b^2 \sqrt {a x^2+b x^3}}{32 a^4 x^3}+\frac {315 b^3 \sqrt {a x^2+b x^3}}{64 a^5 x^2}-\frac {315 b^4 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{11/2}} \] Output:

2/a/x^3/(b*x^3+a*x^2)^(1/2)-9/4*(b*x^3+a*x^2)^(1/2)/a^2/x^5+21/8*b*(b*x^3+ 
a*x^2)^(1/2)/a^3/x^4-105/32*b^2*(b*x^3+a*x^2)^(1/2)/a^4/x^3+315/64*b^3*(b* 
x^3+a*x^2)^(1/2)/a^5/x^2-315/64*b^4*arctanh(a^(1/2)*x/(b*x^3+a*x^2)^(1/2)) 
/a^(11/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-16 a^4+24 a^3 b x-42 a^2 b^2 x^2+105 a b^3 x^3+315 b^4 x^4\right )-315 b^4 x^4 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2} x^3 \sqrt {x^2 (a+b x)}} \] 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)])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.11, 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
 

Defintions of rubi rules used

rule 219

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])
 

rule 1914

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]
 

rule 1929

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]
 

rule 1931

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]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.30

method	result	size
pseudoelliptic	\(-\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, b x +\left (3 b x +a \right ) \sqrt {a}}{\sqrt {b x +a}\, x \,a^{\frac {5}{2}}}\)	\(50\)
default	\(-\frac {\left (b x +a \right ) \left (315 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}-24 a^{\frac {7}{2}} b x +42 a^{\frac {5}{2}} b^{2} x^{2}-105 a^{\frac {3}{2}} b^{3} x^{3}-315 b^{4} x^{4} \sqrt {a}+16 a^{\frac {9}{2}}\right )}{64 x \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {11}{2}}}\)	\(100\)
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*arctanh((b*x+a)^(1/2)/a^(1/2))*(b*x+a)^(1/2)*b*x+(3*b 
*x+a)*a^(1/2))/x/a^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.61 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{128 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}, \frac {315 \, {\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{64 \, {\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}\right ] \] 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)]
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] 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)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x^{2}} \,d x } \] 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)
 

Giac [A] (verification not implemented)

Time = 0.26 (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))
 

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {11}{2};\ \frac {13}{2};\ -\frac {a}{b\,x}\right )}{11\,x\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \] 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))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}-32 a^{5}+48 a^{4} b x -84 a^{3} b^{2} x^{2}+210 a^{2} b^{3} x^{3}+630 a \,b^{4} x^{4}}{128 \sqrt {b x +a}\, a^{6} x^{4}} \] 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)