\(\int \frac {x^6 (c+d x)}{(a x^2+b x^3)^{3/2}} \, dx\) [282]

Optimal result

Integrand size = 24, antiderivative size = 159 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 a^3 (b c-a d) x}{b^5 \sqrt {a x^2+b x^3}}+\frac {2 a^2 (3 b c-4 a d) \sqrt {a x^2+b x^3}}{b^5 x}-\frac {2 a (b c-2 a d) \left (a x^2+b x^3\right )^{3/2}}{b^5 x^3}+\frac {2 (b c-4 a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^5 x^5}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{7 b^5 x^7} \] Output:

2*a^3*(-a*d+b*c)*x/b^5/(b*x^3+a*x^2)^(1/2)+2*a^2*(-4*a*d+3*b*c)*(b*x^3+a*x 
^2)^(1/2)/b^5/x-2*a*(-2*a*d+b*c)*(b*x^3+a*x^2)^(3/2)/b^5/x^3+2/5*(-4*a*d+b 
*c)*(b*x^3+a*x^2)^(5/2)/b^5/x^5+2/7*d*(b*x^3+a*x^2)^(7/2)/b^5/x^7
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.57 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (-128 a^4 d+16 a^3 b (7 c-4 d x)+8 a^2 b^2 x (7 c+2 d x)-2 a b^3 x^2 (7 c+4 d x)+b^4 x^3 (7 c+5 d x)\right )}{35 b^5 \sqrt {x^2 (a+b x)}} \] Input:

Integrate[(x^6*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
 

Output:

(2*x*(-128*a^4*d + 16*a^3*b*(7*c - 4*d*x) + 8*a^2*b^2*x*(7*c + 2*d*x) - 2* 
a*b^3*x^2*(7*c + 4*d*x) + b^4*x^3*(7*c + 5*d*x)))/(35*b^5*Sqrt[x^2*(a + b* 
x)])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1943, 1922, 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.

\(\displaystyle \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1943

\(\displaystyle \frac {2 x^5 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(7 b c-8 a d) \int \frac {x^4}{\sqrt {b x^3+a x^2}}dx}{a b}\)

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1920

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

Input:

Int[(x^6*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
 

Output:

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

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 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])
 

rule 1943

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j 
+ 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( 
m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1)))   Int[(e*x)^(m 
 - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, 
n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 
] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {2 \left (b x +a \right ) \left (-5 d \,x^{4} b^{4}+8 a \,b^{3} d \,x^{3}-7 b^{4} c \,x^{3}-16 a^{2} b^{2} d \,x^{2}+14 a \,b^{3} c \,x^{2}+64 a^{3} b d x -56 a^{2} b^{2} c x +128 a^{4} d -112 a^{3} b c \right ) x^{3}}{35 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(109\)
default	\(-\frac {2 \left (b x +a \right ) \left (-5 d \,x^{4} b^{4}+8 a \,b^{3} d \,x^{3}-7 b^{4} c \,x^{3}-16 a^{2} b^{2} d \,x^{2}+14 a \,b^{3} c \,x^{2}+64 a^{3} b d x -56 a^{2} b^{2} c x +128 a^{4} d -112 a^{3} b c \right ) x^{3}}{35 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(109\)
orering	\(-\frac {2 \left (b x +a \right ) \left (-5 d \,x^{4} b^{4}+8 a \,b^{3} d \,x^{3}-7 b^{4} c \,x^{3}-16 a^{2} b^{2} d \,x^{2}+14 a \,b^{3} c \,x^{2}+64 a^{3} b d x -56 a^{2} b^{2} c x +128 a^{4} d -112 a^{3} b c \right ) x^{3}}{35 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(109\)
risch	\(-\frac {2 \left (-5 b^{3} d \,x^{3}+13 a \,b^{2} d \,x^{2}-7 b^{3} c \,x^{2}-29 a^{2} b d x +21 a \,b^{2} c x +93 a^{3} d -77 c \,a^{2} b \right ) \left (b x +a \right ) x}{35 b^{5} \sqrt {x^{2} \left (b x +a \right )}}-\frac {2 a^{3} \left (a d -b c \right ) x}{b^{5} \sqrt {x^{2} \left (b x +a \right )}}\)	\(110\)
trager	\(-\frac {2 \left (-5 d \,x^{4} b^{4}+8 a \,b^{3} d \,x^{3}-7 b^{4} c \,x^{3}-16 a^{2} b^{2} d \,x^{2}+14 a \,b^{3} c \,x^{2}+64 a^{3} b d x -56 a^{2} b^{2} c x +128 a^{4} d -112 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{35 \left (b x +a \right ) b^{5} x}\)	\(111\)
pseudoelliptic	\(\frac {\frac {2 \left (\frac {11 d x}{13}+c \right ) x^{6} b^{7}}{11}-\frac {8 x^{5} \left (\frac {21 d x}{26}+c \right ) a \,b^{6}}{33}+\frac {80 x^{4} a^{2} \left (\frac {49 d x}{65}+c \right ) b^{5}}{231}-\frac {128 x^{3} a^{3} \left (\frac {35 d x}{52}+c \right ) b^{4}}{231}+\frac {256 x^{2} \left (\frac {7 d x}{13}+c \right ) a^{4} b^{3}}{231}-\frac {1024 \left (\frac {7 d x}{26}+c \right ) x \,a^{5} b^{2}}{231}-\frac {2048 a^{6} \left (-\frac {7 d x}{13}+c \right ) b}{231}+\frac {4096 a^{7} d}{429}}{b^{8} \sqrt {b x +a}}\)	\(126\)

Input:

int(x^6*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/35*(b*x+a)*(-5*b^4*d*x^4+8*a*b^3*d*x^3-7*b^4*c*x^3-16*a^2*b^2*d*x^2+14* 
a*b^3*c*x^2+64*a^3*b*d*x-56*a^2*b^2*c*x+128*a^4*d-112*a^3*b*c)*x^3/b^5/(b* 
x^3+a*x^2)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.72 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (5 \, b^{4} d x^{4} + 112 \, a^{3} b c - 128 \, a^{4} d + {\left (7 \, b^{4} c - 8 \, a b^{3} d\right )} x^{3} - 2 \, {\left (7 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{2} + 8 \, {\left (7 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, {\left (b^{6} x^{2} + a b^{5} x\right )}} \] Input:

integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
 

Output:

2/35*(5*b^4*d*x^4 + 112*a^3*b*c - 128*a^4*d + (7*b^4*c - 8*a*b^3*d)*x^3 - 
2*(7*a*b^3*c - 8*a^2*b^2*d)*x^2 + 8*(7*a^2*b^2*c - 8*a^3*b*d)*x)*sqrt(b*x^ 
3 + a*x^2)/(b^6*x^2 + a*b^5*x)
 

Sympy [F]

\[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{6} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**6*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
 

Output:

Integral(x**6*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.61 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} c}{5 \, \sqrt {b x + a} b^{4}} + \frac {2 \, {\left (5 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 16 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x - 128 \, a^{4}\right )} d}{35 \, \sqrt {b x + a} b^{5}} \] Input:

integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
 

Output:

2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)*c/(sqrt(b*x + a)*b^4) + 2 
/35*(5*b^4*x^4 - 8*a*b^3*x^3 + 16*a^2*b^2*x^2 - 64*a^3*b*x - 128*a^4)*d/(s 
qrt(b*x + a)*b^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {32 \, {\left (7 \, a^{3} b c - 8 \, a^{4} d\right )} \mathrm {sgn}\left (x\right )}{35 \, \sqrt {a} b^{5}} + \frac {2 \, {\left (a^{3} b c - a^{4} d\right )}}{\sqrt {b x + a} b^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (7 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{31} c - 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{31} c + 105 \, \sqrt {b x + a} a^{2} b^{31} c + 5 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{30} d - 28 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{30} d + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{30} d - 140 \, \sqrt {b x + a} a^{3} b^{30} d\right )}}{35 \, b^{35} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^6*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
 

Output:

-32/35*(7*a^3*b*c - 8*a^4*d)*sgn(x)/(sqrt(a)*b^5) + 2*(a^3*b*c - a^4*d)/(s 
qrt(b*x + a)*b^5*sgn(x)) + 2/35*(7*(b*x + a)^(5/2)*b^31*c - 35*(b*x + a)^( 
3/2)*a*b^31*c + 105*sqrt(b*x + a)*a^2*b^31*c + 5*(b*x + a)^(7/2)*b^30*d - 
28*(b*x + a)^(5/2)*a*b^30*d + 70*(b*x + a)^(3/2)*a^2*b^30*d - 140*sqrt(b*x 
 + a)*a^3*b^30*d)/(b^35*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-128\,d\,a^4-64\,d\,a^3\,b\,x+112\,c\,a^3\,b+16\,d\,a^2\,b^2\,x^2+56\,c\,a^2\,b^2\,x-8\,d\,a\,b^3\,x^3-14\,c\,a\,b^3\,x^2+5\,d\,b^4\,x^4+7\,c\,b^4\,x^3\right )}{35\,b^5\,x\,\left (a+b\,x\right )} \] Input:

int((x^6*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
 

Output:

(2*(a*x^2 + b*x^3)^(1/2)*(7*b^4*c*x^3 - 128*a^4*d + 5*b^4*d*x^4 + 112*a^3* 
b*c + 16*a^2*b^2*d*x^2 - 64*a^3*b*d*x + 56*a^2*b^2*c*x - 14*a*b^3*c*x^2 - 
8*a*b^3*d*x^3))/(35*b^5*x*(a + b*x))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60 \[ \int \frac {x^6 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\frac {2}{7} b^{4} d \,x^{4}-\frac {16}{35} a \,b^{3} d \,x^{3}+\frac {2}{5} b^{4} c \,x^{3}+\frac {32}{35} a^{2} b^{2} d \,x^{2}-\frac {4}{5} a \,b^{3} c \,x^{2}-\frac {128}{35} a^{3} b d x +\frac {16}{5} a^{2} b^{2} c x -\frac {256}{35} a^{4} d +\frac {32}{5} a^{3} b c}{\sqrt {b x +a}\, b^{5}} \] Input:

int(x^6*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
 

Output:

(2*( - 128*a**4*d + 112*a**3*b*c - 64*a**3*b*d*x + 56*a**2*b**2*c*x + 16*a 
**2*b**2*d*x**2 - 14*a*b**3*c*x**2 - 8*a*b**3*d*x**3 + 7*b**4*c*x**3 + 5*b 
**4*d*x**4))/(35*sqrt(a + b*x)*b**5)