\(\int \frac {x^2}{(a x^2+b x^3)^{5/2}} \, dx\) [323]

Optimal result

Integrand size = 19, antiderivative size = 141 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {x}{2 a \left (a x^2+b x^3\right )^{3/2}}+\frac {7 b x^2}{4 a^2 \left (a x^2+b x^3\right )^{3/2}}+\frac {35 b^2 x^3}{12 a^3 \left (a x^2+b x^3\right )^{3/2}}+\frac {35 b^2 x}{4 a^4 \sqrt {a x^2+b x^3}}-\frac {35 b^2 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{4 a^{9/2}} \] Output:

-1/2*x/a/(b*x^3+a*x^2)^(3/2)+7/4*b*x^2/a^2/(b*x^3+a*x^2)^(3/2)+35/12*b^2*x 
^3/a^3/(b*x^3+a*x^2)^(3/2)+35/4*b^2*x/a^4/(b*x^3+a*x^2)^(1/2)-35/4*b^2*arc 
tanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(9/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {x \left (\sqrt {a} \left (-6 a^3+21 a^2 b x+140 a b^2 x^2+105 b^3 x^3\right )-105 b^2 x^2 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{12 a^{9/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(x*(Sqrt[a]*(-6*a^3 + 21*a^2*b*x + 140*a*b^2*x^2 + 105*b^3*x^3) - 105*b^2* 
x^2*(a + b*x)^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(12*a^(9/2)*(x^2*(a + 
 b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1929, 1912, 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.

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

\(\Big \downarrow \) 1929

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

\(\Big \downarrow \) 1912

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*x)/(3*a*(a*x^2 + b*x^3)^(3/2)) + (7*(2/(a*x*Sqrt[a*x^2 + b*x^3]) + (5*( 
-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)))/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 1912

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + 
 b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ 
(a*(n - j)*(p + 1))   Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, 
b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]
 

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 = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.23

method	result	size
pseudoelliptic	\(\frac {2 b^{2} x^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\)	\(32\)
default	\(-\frac {x^{3} \left (b x +a \right ) \left (105 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-140 a^{\frac {3}{2}} b^{2} x^{2}-105 \sqrt {a}\, b^{3} x^{3}-21 a^{\frac {5}{2}} b x +6 a^{\frac {7}{2}}\right )}{12 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {9}{2}}}\)	\(89\)
risch	\(-\frac {\left (b x +a \right ) \left (-11 b x +2 a \right )}{4 a^{4} x \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{2} \left (-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {48}{\sqrt {b x +a}}+\frac {16 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{8 a^{4} \sqrt {x^{2} \left (b x +a \right )}}\)	\(98\)

Input:

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

Output:

2/3*(3*b^2*x^2+12*a*b*x+8*a^2)/b^3/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.06 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac {105 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{12 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \] Input:

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

Output:

[1/24*(105*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*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 + 140*a^2*b^2*x 
^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x^3 + a*x^2))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + 
 a^7*x^3), 1/12*(105*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(-a)*arctan 
(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (105*a*b^3*x^3 + 140*a^2*b^ 
2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x^3 + a*x^2))/(a^5*b^2*x^5 + 2*a^6*b*x^ 
4 + a^7*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (9 \, {\left (b x + a\right )} b^{2} + a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} \mathrm {sgn}\left (x\right )} + \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 13 \, \sqrt {b x + a} a b^{2}}{4 \, a^{4} b^{2} x^{2} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

35/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4*sgn(x)) + 2/3*(9*(b* 
x + a)*b^2 + a*b^2)/((b*x + a)^(3/2)*a^4*sgn(x)) + 1/4*(11*(b*x + a)^(3/2) 
*b^2 - 13*sqrt(b*x + a)*a*b^2)/(a^4*b^2*x^2*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} x^{2}+105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} x^{2}-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}-12 a^{4}+42 a^{3} b x +280 a^{2} b^{2} x^{2}+210 a \,b^{3} x^{3}}{24 \sqrt {b x +a}\, a^{5} x^{2} \left (b x +a \right )} \] Input:

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

Output:

(105*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*x**2 + 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))*a*b**2*x**2 - 105*sqrt(a)*sqrt 
(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**3*x**3 - 12*a**4 + 42*a**3*b*x + 
 280*a**2*b**2*x**2 + 210*a*b**3*x**3)/(24*sqrt(a + b*x)*a**5*x**2*(a + b* 
x))