\(\int \frac {x^3}{(a x^2+b x^3)^{5/2}} \, dx\) [322]

Optimal result

Integrand size = 19, antiderivative size = 105 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {x^2}{a \left (a x^2+b x^3\right )^{3/2}}-\frac {5 b x^3}{3 a^2 \left (a x^2+b x^3\right )^{3/2}}-\frac {5 b x}{a^3 \sqrt {a x^2+b x^3}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{a^{7/2}} \] Output:

-x^2/a/(b*x^3+a*x^2)^(3/2)-5/3*b*x^3/a^2/(b*x^3+a*x^2)^(3/2)-5*b*x/a^3/(b* 
x^3+a*x^2)^(1/2)+5*b*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {x^2 \left (-\sqrt {a} \left (3 a^2+20 a b x+15 b^2 x^2\right )+15 b x (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{3 a^{7/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(x^2*(-(Sqrt[a]*(3*a^2 + 20*a*b*x + 15*b^2*x^2)) + 15*b*x*(a + b*x)^(3/2)* 
ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(3*a^(7/2)*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1929, 1929, 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.

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

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*x^2)/(3*a*(a*x^2 + b*x^3)^(3/2)) + (5*(2/(a*Sqrt[a*x^2 + b*x^3]) + (3*( 
-(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)))/a))/(3*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 = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.40

method	result	size
pseudoelliptic	\(\frac {\frac {2}{3} b^{3} x^{3}-4 a \,b^{2} x^{2}-16 a^{2} b x -\frac {32}{3} a^{3}}{b^{4} \left (b x +a \right )^{\frac {3}{2}}}\)	\(42\)
default	\(\frac {x^{4} \left (b x +a \right ) \left (15 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x -20 a^{\frac {3}{2}} b x -15 \sqrt {a}\, b^{2} x^{2}-3 a^{\frac {5}{2}}\right )}{3 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {7}{2}}}\)	\(74\)
risch	\(-\frac {b x +a}{a^{3} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b \left (-\frac {10 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {8}{\sqrt {b x +a}}+\frac {4 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{2 a^{3} \sqrt {x^{2} \left (b x +a \right )}}\)	\(85\)

Input:

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

Output:

2/3*(b^3*x^3-6*a*b^2*x^2-24*a^2*b*x-16*a^3)/b^4/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.51 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{6 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, -\frac {15 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/6*(15*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*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 + 20*a^2*b*x + 3*a^3 
)*sqrt(b*x^3 + a*x^2))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), -1/3*(15*(b^ 
3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt( 
-a)/(b*x^2 + a*x)) + (15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x^3 + a*x^ 
2))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {5 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} b + a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\left (x\right )} - \frac {\sqrt {b x + a}}{a^{3} x \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3*sgn(x)) - 2/3*(6*(b*x + 
a)*b + a*b)/((b*x + a)^(3/2)*a^3*sgn(x)) - sqrt(b*x + a)/(a^3*x*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.40 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {-15 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b x -15 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}+15 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b x +15 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}-6 a^{3}-40 a^{2} b x -30 a \,b^{2} x^{2}}{6 \sqrt {b x +a}\, a^{4} x \left (b x +a \right )} \] Input:

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

Output:

( - 15*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b*x - 15*sqrt( 
a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**2*x**2 + 15*sqrt(a)*sqrt( 
a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b*x + 15*sqrt(a)*sqrt(a + b*x)*log 
(sqrt(a + b*x) + sqrt(a))*b**2*x**2 - 6*a**3 - 40*a**2*b*x - 30*a*b**2*x** 
2)/(6*sqrt(a + b*x)*a**4*x*(a + b*x))