\(\int \frac {(c x)^{3/2}}{(a x^2+b x^3)^{5/2}} \, dx\) [369]

Optimal result

Integrand size = 23, antiderivative size = 163 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}+\frac {16 c^3}{3 a^2 (c x)^{3/2} \sqrt {a x^2+b x^3}}-\frac {32 c^5 \sqrt {a x^2+b x^3}}{5 a^3 (c x)^{7/2}}+\frac {128 b c^4 \sqrt {a x^2+b x^3}}{15 a^4 (c x)^{5/2}}-\frac {256 b^2 c^3 \sqrt {a x^2+b x^3}}{15 a^5 (c x)^{3/2}} \] Output:

2/3*c*(c*x)^(1/2)/a/(b*x^3+a*x^2)^(3/2)+16/3*c^3/a^2/(c*x)^(3/2)/(b*x^3+a* 
x^2)^(1/2)-32/5*c^5*(b*x^3+a*x^2)^(1/2)/a^3/(c*x)^(7/2)+128/15*b*c^4*(b*x^ 
3+a*x^2)^(1/2)/a^4/(c*x)^(5/2)-256/15*b^2*c^3*(b*x^3+a*x^2)^(1/2)/a^5/(c*x 
)^(3/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.42 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 c \sqrt {c x} \left (3 a^4-8 a^3 b x+48 a^2 b^2 x^2+192 a b^3 x^3+128 b^4 x^4\right )}{15 a^5 \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(-2*c*Sqrt[c*x]*(3*a^4 - 8*a^3*b*x + 48*a^2*b^2*x^2 + 192*a*b^3*x^3 + 128* 
b^4*x^4))/(15*a^5*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1921, 1921, 1922, 1922, 1920}

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.

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

\(\Big \downarrow \) 1921

\(\displaystyle \frac {8 c^2 \int \frac {1}{\sqrt {c x} \left (b x^3+a x^2\right )^{3/2}}dx}{3 a}+\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1921

\(\displaystyle \frac {8 c^2 \left (\frac {6 c^2 \int \frac {1}{(c x)^{5/2} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 c}{a (c x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {8 c^2 \left (\frac {6 c^2 \left (-\frac {4 b \int \frac {1}{(c x)^{3/2} \sqrt {b x^3+a x^2}}dx}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{a}+\frac {2 c}{a (c x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {8 c^2 \left (\frac {6 c^2 \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{\sqrt {c x} \sqrt {b x^3+a x^2}}dx}{3 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{3 a (c x)^{5/2}}\right )}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{a}+\frac {2 c}{a (c x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1920

\(\displaystyle \frac {8 c^2 \left (\frac {6 c^2 \left (-\frac {4 b \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 (c x)^{3/2}}-\frac {2 c \sqrt {a x^2+b x^3}}{3 a (c x)^{5/2}}\right )}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{a}+\frac {2 c}{a (c x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2 c \sqrt {c x}}{3 a \left (a x^2+b x^3\right )^{3/2}}\)

Input:

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

Output:

(2*c*Sqrt[c*x])/(3*a*(a*x^2 + b*x^3)^(3/2)) + (8*c^2*((2*c)/(a*(c*x)^(3/2) 
*Sqrt[a*x^2 + b*x^3]) + (6*c^2*((-2*c*Sqrt[a*x^2 + b*x^3])/(5*a*(c*x)^(7/2 
)) - (4*b*((-2*c*Sqrt[a*x^2 + b*x^3])/(3*a*(c*x)^(5/2)) + (4*b*Sqrt[a*x^2 
+ b*x^3])/(3*a^2*(c*x)^(3/2))))/(5*a*c)))/a))/(3*a)
 

Defintions of rubi rules used

rule 1920

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] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[ 
n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1921

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, j, m, n} 
, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( 
n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1922

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, j, m, n, 
p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) 
/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.44

method	result	size
gosper	\(-\frac {2 x \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right ) \left (c x \right )^{\frac {3}{2}}}{15 a^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(71\)
orering	\(-\frac {2 x \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right ) \left (c x \right )^{\frac {3}{2}}}{15 a^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(71\)
default	\(-\frac {2 x^{2} \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right ) \sqrt {c x}\, c}{15 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{5}}\)	\(74\)
risch	\(-\frac {2 \left (b x +a \right ) \left (73 b^{2} x^{2}-14 a b x +3 a^{2}\right ) c^{2}}{15 a^{5} x \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {c x}}-\frac {2 b^{3} \left (11 b x +12 a \right ) x^{2} c^{2}}{3 \left (b x +a \right ) a^{5} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {c x}}\)	\(98\)

Input:

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

Output:

-2/15*x*(b*x+a)*(128*b^4*x^4+192*a*b^3*x^3+48*a^2*b^2*x^2-8*a^3*b*x+3*a^4) 
*(c*x)^(3/2)/a^5/(b*x^3+a*x^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.58 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (128 \, b^{4} c x^{4} + 192 \, a b^{3} c x^{3} + 48 \, a^{2} b^{2} c x^{2} - 8 \, a^{3} b c x + 3 \, a^{4} c\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {c x}}{15 \, {\left (a^{5} b^{2} x^{6} + 2 \, a^{6} b x^{5} + a^{7} x^{4}\right )}} \] Input:

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

Output:

-2/15*(128*b^4*c*x^4 + 192*a*b^3*c*x^3 + 48*a^2*b^2*c*x^2 - 8*a^3*b*c*x + 
3*a^4*c)*sqrt(b*x^3 + a*x^2)*sqrt(c*x)/(a^5*b^2*x^6 + 2*a^6*b*x^5 + a^7*x^ 
4)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (133) = 266\).

Time = 104.84 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.90 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2}{15} \, c {\left (\frac {5 \, \sqrt {c x} {\left (\frac {11 \, b^{4} c^{2} x {\left | c \right |}}{a^{5} \mathrm {sgn}\left (x\right )} + \frac {12 \, b^{3} c^{2} {\left | c \right |}}{a^{4} \mathrm {sgn}\left (x\right )}\right )}}{{\left (b c^{2} x + a c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (73 \, \sqrt {b c} a^{4} b^{2} c^{11} - 320 \, \sqrt {b c} {\left (\sqrt {b c} \sqrt {c x} - \sqrt {b c^{2} x + a c^{2}}\right )}^{2} a^{3} b^{2} c^{9} + 490 \, \sqrt {b c} {\left (\sqrt {b c} \sqrt {c x} - \sqrt {b c^{2} x + a c^{2}}\right )}^{4} a^{2} b^{2} c^{7} - 240 \, \sqrt {b c} {\left (\sqrt {b c} \sqrt {c x} - \sqrt {b c^{2} x + a c^{2}}\right )}^{6} a b^{2} c^{5} + 45 \, \sqrt {b c} {\left (\sqrt {b c} \sqrt {c x} - \sqrt {b c^{2} x + a c^{2}}\right )}^{8} b^{2} c^{3}\right )}}{{\left (a c^{2} - {\left (\sqrt {b c} \sqrt {c x} - \sqrt {b c^{2} x + a c^{2}}\right )}^{2}\right )}^{5} a^{4} {\left | c \right |} \mathrm {sgn}\left (x\right )}\right )} \] Input:

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

Output:

-2/15*c*(5*sqrt(c*x)*(11*b^4*c^2*x*abs(c)/(a^5*sgn(x)) + 12*b^3*c^2*abs(c) 
/(a^4*sgn(x)))/(b*c^2*x + a*c^2)^(3/2) + 2*(73*sqrt(b*c)*a^4*b^2*c^11 - 32 
0*sqrt(b*c)*(sqrt(b*c)*sqrt(c*x) - sqrt(b*c^2*x + a*c^2))^2*a^3*b^2*c^9 + 
490*sqrt(b*c)*(sqrt(b*c)*sqrt(c*x) - sqrt(b*c^2*x + a*c^2))^4*a^2*b^2*c^7 
- 240*sqrt(b*c)*(sqrt(b*c)*sqrt(c*x) - sqrt(b*c^2*x + a*c^2))^6*a*b^2*c^5 
+ 45*sqrt(b*c)*(sqrt(b*c)*sqrt(c*x) - sqrt(b*c^2*x + a*c^2))^8*b^2*c^3)/(( 
a*c^2 - (sqrt(b*c)*sqrt(c*x) - sqrt(b*c^2*x + a*c^2))^2)^5*a^4*abs(c)*sgn( 
x)))
 

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,c\,\sqrt {c\,x}}{5\,a\,b^2}+\frac {32\,c\,x^2\,\sqrt {c\,x}}{5\,a^3}-\frac {16\,c\,x\,\sqrt {c\,x}}{15\,a^2\,b}+\frac {128\,b\,c\,x^3\,\sqrt {c\,x}}{5\,a^4}+\frac {256\,b^2\,c\,x^4\,\sqrt {c\,x}}{15\,a^5}\right )}{x^6+\frac {2\,a\,x^5}{b}+\frac {a^2\,x^4}{b^2}} \] Input:

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

Output:

-((a*x^2 + b*x^3)^(1/2)*((2*c*(c*x)^(1/2))/(5*a*b^2) + (32*c*x^2*(c*x)^(1/ 
2))/(5*a^3) - (16*c*x*(c*x)^(1/2))/(15*a^2*b) + (128*b*c*x^3*(c*x)^(1/2))/ 
(5*a^4) + (256*b^2*c*x^4*(c*x)^(1/2))/(15*a^5)))/(x^6 + (2*a*x^5)/b + (a^2 
*x^4)/b^2)
 

Reduce [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67 \[ \int \frac {(c x)^{3/2}}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {c}\, c \left (128 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} x^{3}+128 \sqrt {b}\, \sqrt {b x +a}\, b^{3} x^{4}-3 \sqrt {x}\, a^{4}+8 \sqrt {x}\, a^{3} b x -48 \sqrt {x}\, a^{2} b^{2} x^{2}-192 \sqrt {x}\, a \,b^{3} x^{3}-128 \sqrt {x}\, b^{4} x^{4}\right )}{15 \sqrt {b x +a}\, a^{5} x^{3} \left (b x +a \right )} \] Input:

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

Output:

(2*sqrt(c)*c*(128*sqrt(b)*sqrt(a + b*x)*a*b**2*x**3 + 128*sqrt(b)*sqrt(a + 
 b*x)*b**3*x**4 - 3*sqrt(x)*a**4 + 8*sqrt(x)*a**3*b*x - 48*sqrt(x)*a**2*b* 
*2*x**2 - 192*sqrt(x)*a*b**3*x**3 - 128*sqrt(x)*b**4*x**4))/(15*sqrt(a + b 
*x)*a**5*x**3*(a + b*x))