\(\int \frac {x}{(a x^2+b x^3)^{5/2}} \, dx\) [324]

Optimal result

Integrand size = 17, antiderivative size = 166 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {1}{3 a \left (a x^2+b x^3\right )^{3/2}}+\frac {3 b x}{4 a^2 \left (a x^2+b x^3\right )^{3/2}}-\frac {21 b^2 x^2}{8 a^3 \left (a x^2+b x^3\right )^{3/2}}-\frac {35 b^3 x^3}{8 a^4 \left (a x^2+b x^3\right )^{3/2}}-\frac {105 b^3 x}{8 a^5 \sqrt {a x^2+b x^3}}+\frac {105 b^3 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 a^{11/2}} \] Output:

-1/3/a/(b*x^3+a*x^2)^(3/2)+3/4*b*x/a^2/(b*x^3+a*x^2)^(3/2)-21/8*b^2*x^2/a^ 
3/(b*x^3+a*x^2)^(3/2)-35/8*b^3*x^3/a^4/(b*x^3+a*x^2)^(3/2)-105/8*b^3*x/a^5 
/(b*x^3+a*x^2)^(1/2)+105/8*b^3*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(1 
1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {-\sqrt {a} \left (8 a^4-18 a^3 b x+63 a^2 b^2 x^2+420 a b^3 x^3+315 b^4 x^4\right )+315 b^3 x^3 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{11/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

Integrate[x/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

(-(Sqrt[a]*(8*a^4 - 18*a^3*b*x + 63*a^2*b^2*x^2 + 420*a*b^3*x^3 + 315*b^4* 
x^4)) + 315*b^3*x^3*(a + b*x)^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(24*a^ 
(11/2)*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1929, 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.

\(\displaystyle \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

Int[x/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

2/(3*a*(a*x^2 + b*x^3)^(3/2)) + (3*(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))/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 = 0.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.13

method	result	size
pseudoelliptic	\(\frac {-6 b x -4 a}{3 b^{2} \left (b x +a \right )^{\frac {3}{2}}}\)	\(21\)
default	\(\frac {x^{2} \left (b x +a \right ) \left (315 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-63 a^{\frac {5}{2}} b^{2} x^{2}-420 a^{\frac {3}{2}} b^{3} x^{3}-315 \sqrt {a}\, b^{4} x^{4}+18 a^{\frac {7}{2}} b x -8 a^{\frac {9}{2}}\right )}{24 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {11}{2}}}\)	\(100\)
risch	\(-\frac {\left (b x +a \right ) \left (123 b^{2} x^{2}-34 a b x +8 a^{2}\right )}{24 a^{5} x^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b^{3} \left (-\frac {210 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {128}{\sqrt {b x +a}}+\frac {32 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{16 a^{5} \sqrt {x^{2} \left (b x +a \right )}}\)	\(109\)

Input:

int(x/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*(-6*b*x-4*a)/b^2/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.88 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [\frac {315 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{5} + a^{2} b^{3} x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}}, -\frac {315 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{5} + a^{2} b^{3} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}}\right ] \] Input:

integrate(x/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(315*(b^5*x^6 + 2*a*b^4*x^5 + a^2*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*(315*a*b^4*x^4 + 420*a^2*b^3*x 
^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x^3 + a*x^2))/(a^6*b^2*x^ 
6 + 2*a^7*b*x^5 + a^8*x^4), -1/24*(315*(b^5*x^6 + 2*a*b^4*x^5 + a^2*b^3*x^ 
4)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (315*a*b^ 
4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x^3 
+ a*x^2))/(a^6*b^2*x^6 + 2*a^7*b*x^5 + a^8*x^4)]
 

Sympy [F]

\[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {x}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x/(b*x**3+a*x**2)**(5/2),x)
 

Output:

Integral(x/(x**2*(a + b*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int { \frac {x}{{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x/(b*x^3 + a*x^2)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {105 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (x\right )} - \frac {315 \, {\left (b x + a\right )}^{4} b^{3} - 840 \, {\left (b x + a\right )}^{3} a b^{3} + 693 \, {\left (b x + a\right )}^{2} a^{2} b^{3} - 144 \, {\left (b x + a\right )} a^{3} b^{3} - 16 \, a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
 

Output:

-105/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5*sgn(x)) - 1/24*(31 
5*(b*x + a)^4*b^3 - 840*(b*x + a)^3*a*b^3 + 693*(b*x + a)^2*a^2*b^3 - 144* 
(b*x + a)*a^3*b^3 - 16*a^4*b^3)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^5 
*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {x}{{\left (b\,x^3+a\,x^2\right )}^{5/2}} \,d x \] Input:

int(x/(a*x^2 + b*x^3)^(5/2),x)
 

Output:

int(x/(a*x^2 + b*x^3)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} x^{3}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} x^{3}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}-16 a^{5}+36 a^{4} b x -126 a^{3} b^{2} x^{2}-840 a^{2} b^{3} x^{3}-630 a \,b^{4} x^{4}}{48 \sqrt {b x +a}\, a^{6} x^{3} \left (b x +a \right )} \] Input:

int(x/(b*x^3+a*x^2)^(5/2),x)
 

Output:

( - 315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**3*x**3 - 3 
15*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**4*x**4 + 315*sqrt 
(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**3*x**3 + 315*sqrt(a)*s 
qrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**4*x**4 - 16*a**5 + 36*a**4*b* 
x - 126*a**3*b**2*x**2 - 840*a**2*b**3*x**3 - 630*a*b**4*x**4)/(48*sqrt(a 
+ b*x)*a**6*x**3*(a + b*x))