\(\int \frac {1}{x (a x^2+b x^3)^{3/2}} \, dx\) [246]

Optimal result

Integrand size = 19, antiderivative size = 138 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {35 b^3 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{9/2}} \] Output:

2/a/x^2/(b*x^3+a*x^2)^(1/2)-7/3*(b*x^3+a*x^2)^(1/2)/a^2/x^4+35/12*b*(b*x^3 
+a*x^2)^(1/2)/a^3/x^3-35/8*b^2*(b*x^3+a*x^2)^(1/2)/a^4/x^2+35/8*b^3*arctan 
h(a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^(9/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-\sqrt {a} \left (8 a^3-14 a^2 b x+35 a b^2 x^2+105 b^3 x^3\right )+105 b^3 x^3 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{9/2} x^2 \sqrt {x^2 (a+b x)}} \] Input:

Integrate[1/(x*(a*x^2 + b*x^3)^(3/2)),x]
 

Output:

(-(Sqrt[a]*(8*a^3 - 14*a^2*b*x + 35*a*b^2*x^2 + 105*b^3*x^3)) + 105*b^3*x^ 
3*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(24*a^(9/2)*x^2*Sqrt[x^2*( 
a + b*x)])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1929, 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.

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

\(\Big \downarrow \) 1929

\(\displaystyle \frac {7 \int \frac {1}{x^3 \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {7 \left (-\frac {5 b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {7 \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {7 \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {7 \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7 \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{a}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}}\)

Input:

Int[1/(x*(a*x^2 + b*x^3)^(3/2)),x]
 

Output:

2/(a*x^2*Sqrt[a*x^2 + b*x^3]) + (7*(-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)))/ 
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 = 0.46 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.22

method	result	size
pseudoelliptic	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {b x +a}}\)	\(31\)
default	\(\frac {\left (b x +a \right ) \left (105 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}+14 a^{\frac {5}{2}} b x -35 a^{\frac {3}{2}} b^{2} x^{2}-105 \sqrt {a}\, b^{3} x^{3}-8 a^{\frac {7}{2}}\right )}{24 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {9}{2}}}\)	\(86\)
risch	\(-\frac {\left (b x +a \right ) \left (57 b^{2} x^{2}-22 a b x +8 a^{2}\right )}{24 a^{4} x^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b^{3} \left (-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {32}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{16 a^{4} \sqrt {x^{2} \left (b x +a \right )}}\)	\(99\)

Input:

int(1/x/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2/a/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.78 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{5} + a b^{3} x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}, -\frac {105 \, {\left (b^{4} x^{5} + a b^{3} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}\right ] \] Input:

integrate(1/x/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/48*(105*(b^4*x^5 + a*b^3*x^4)*sqrt(a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 
 + a*x^2)*sqrt(a))/x^2) - 2*(105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 
 8*a^4)*sqrt(b*x^3 + a*x^2))/(a^5*b*x^5 + a^6*x^4), -1/24*(105*(b^4*x^5 + 
a*b^3*x^4)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + ( 
105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x^3 + a*x^2))/ 
(a^5*b*x^5 + a^6*x^4)]
 

Sympy [F]

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

integrate(1/x/(b*x**3+a*x**2)**(3/2),x)
 

Output:

Integral(1/(x*(x**2*(a + b*x))**(3/2)), x)
 

Maxima [F]

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

integrate(1/x/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^3 + a*x^2)^(3/2)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {35 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4} \mathrm {sgn}\left (x\right )} - \frac {2 \, b^{3}}{\sqrt {b x + a} a^{4} \mathrm {sgn}\left (x\right )} - \frac {57 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 136 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 87 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3} \mathrm {sgn}\left (x\right )} \] Input:

integrate(1/x/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
 

Output:

-35/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4*sgn(x)) - 2*b^3/(sq 
rt(b*x + a)*a^4*sgn(x)) - 1/24*(57*(b*x + a)^(5/2)*b^3 - 136*(b*x + a)^(3/ 
2)*a*b^3 + 87*sqrt(b*x + a)*a^2*b^3)/(a^4*b^3*x^3*sgn(x))
 

Mupad [F(-1)]

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

int(1/(x*(a*x^2 + b*x^3)^(3/2)),x)
 

Output:

int(1/(x*(a*x^2 + b*x^3)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}+105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}-16 a^{4}+28 a^{3} b x -70 a^{2} b^{2} x^{2}-210 a \,b^{3} x^{3}}{48 \sqrt {b x +a}\, a^{5} x^{3}} \] Input:

int(1/x/(b*x^3+a*x^2)^(3/2),x)
 

Output:

( - 105*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**3*x**3 + 105 
*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**3*x**3 - 16*a**4 + 
28*a**3*b*x - 70*a**2*b**2*x**2 - 210*a*b**3*x**3)/(48*sqrt(a + b*x)*a**5* 
x**3)