\(\int \frac {x^8 (c+d x)}{(a x^2+b x^3)^{5/2}} \, dx\) [291]

Optimal result

Integrand size = 24, antiderivative size = 161 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 a^3 (b c-a d) x^3}{3 b^5 \left (a x^2+b x^3\right )^{3/2}}-\frac {2 a^2 (3 b c-4 a d) x}{b^5 \sqrt {a x^2+b x^3}}-\frac {6 a (b c-2 a d) \sqrt {a x^2+b x^3}}{b^5 x}+\frac {2 (b c-4 a d) \left (a x^2+b x^3\right )^{3/2}}{3 b^5 x^3}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{5 b^5 x^5} \] Output:

2/3*a^3*(-a*d+b*c)*x^3/b^5/(b*x^3+a*x^2)^(3/2)-2*a^2*(-4*a*d+3*b*c)*x/b^5/ 
(b*x^3+a*x^2)^(1/2)-6*a*(-2*a*d+b*c)*(b*x^3+a*x^2)^(1/2)/b^5/x+2/3*(-4*a*d 
+b*c)*(b*x^3+a*x^2)^(3/2)/b^5/x^3+2/5*d*(b*x^3+a*x^2)^(5/2)/b^5/x^5
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.57 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 x^3 \left (128 a^4 d+24 a^2 b^2 x (-5 c+2 d x)+b^4 x^3 (5 c+3 d x)-2 a b^3 x^2 (15 c+4 d x)+a^3 b (-80 c+192 d x)\right )}{15 b^5 \left (x^2 (a+b x)\right )^{3/2}} \] Input:

Integrate[(x^8*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

(2*x^3*(128*a^4*d + 24*a^2*b^2*x*(-5*c + 2*d*x) + b^4*x^3*(5*c + 3*d*x) - 
2*a*b^3*x^2*(15*c + 4*d*x) + a^3*b*(-80*c + 192*d*x)))/(15*b^5*(x^2*(a + b 
*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1943, 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.

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

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1921

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

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1920

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

Input:

Int[(x^8*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

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

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 = 0.70 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (3 d \,x^{4} b^{4}-8 a \,b^{3} d \,x^{3}+5 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-30 a \,b^{3} c \,x^{2}+192 a^{3} b d x -120 a^{2} b^{2} c x +128 a^{4} d -80 a^{3} b c \right ) x^{5}}{15 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(109\)
default	\(\frac {2 \left (b x +a \right ) \left (3 d \,x^{4} b^{4}-8 a \,b^{3} d \,x^{3}+5 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-30 a \,b^{3} c \,x^{2}+192 a^{3} b d x -120 a^{2} b^{2} c x +128 a^{4} d -80 a^{3} b c \right ) x^{5}}{15 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(109\)
orering	\(\frac {2 \left (b x +a \right ) \left (3 d \,x^{4} b^{4}-8 a \,b^{3} d \,x^{3}+5 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-30 a \,b^{3} c \,x^{2}+192 a^{3} b d x -120 a^{2} b^{2} c x +128 a^{4} d -80 a^{3} b c \right ) x^{5}}{15 b^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(109\)
risch	\(\frac {2 \left (3 b^{2} d \,x^{2}-14 a b d x +5 b^{2} c x +73 a^{2} d -40 a b c \right ) \left (b x +a \right ) x}{15 b^{5} \sqrt {x^{2} \left (b x +a \right )}}+\frac {2 a^{2} \left (12 a b d x -9 b^{2} c x +11 a^{2} d -8 a b c \right ) x}{3 b^{5} \left (b x +a \right ) \sqrt {x^{2} \left (b x +a \right )}}\)	\(110\)
trager	\(\frac {2 \left (3 d \,x^{4} b^{4}-8 a \,b^{3} d \,x^{3}+5 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-30 a \,b^{3} c \,x^{2}+192 a^{3} b d x -120 a^{2} b^{2} c x +128 a^{4} d -80 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 x \,b^{5} \left (b x +a \right )^{2}}\)	\(111\)
pseudoelliptic	\(\frac {\left (858 d \,x^{9}+990 c \,x^{8}\right ) b^{9}-1440 x^{7} \left (\frac {33 d x}{40}+c \right ) a \,b^{8}+2240 \left (\frac {27 d x}{35}+c \right ) x^{6} a^{2} b^{7}-3840 x^{5} \left (\frac {7 d x}{10}+c \right ) a^{3} b^{6}+7680 x^{4} \left (\frac {3 d x}{5}+c \right ) a^{4} b^{5}-20480 x^{3} a^{5} \left (\frac {9 d x}{20}+c \right ) b^{4}+122880 \left (\frac {d x}{5}+c \right ) x^{2} a^{6} b^{3}+491520 x \left (-\frac {3 d x}{10}+c \right ) a^{7} b^{2}+327680 \left (-\frac {9 d x}{5}+c \right ) a^{8} b -393216 a^{9} d}{6435 \left (b x +a \right )^{\frac {3}{2}} b^{10}}\)	\(164\)

Input:

int(x^8*(d*x+c)/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/15*(b*x+a)*(3*b^4*d*x^4-8*a*b^3*d*x^3+5*b^4*c*x^3+48*a^2*b^2*d*x^2-30*a* 
b^3*c*x^2+192*a^3*b*d*x-120*a^2*b^2*c*x+128*a^4*d-80*a^3*b*c)*x^5/b^5/(b*x 
^3+a*x^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{4} d x^{4} - 80 \, a^{3} b c + 128 \, a^{4} d + {\left (5 \, b^{4} c - 8 \, a b^{3} d\right )} x^{3} - 6 \, {\left (5 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{2} - 24 \, {\left (5 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, {\left (b^{7} x^{3} + 2 \, a b^{6} x^{2} + a^{2} b^{5} x\right )}} \] Input:

integrate(x^8*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
 

Output:

2/15*(3*b^4*d*x^4 - 80*a^3*b*c + 128*a^4*d + (5*b^4*c - 8*a*b^3*d)*x^3 - 6 
*(5*a*b^3*c - 8*a^2*b^2*d)*x^2 - 24*(5*a^2*b^2*c - 8*a^3*b*d)*x)*sqrt(b*x^ 
3 + a*x^2)/(b^7*x^3 + 2*a*b^6*x^2 + a^2*b^5*x)
 

Sympy [F]

\[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {x^{8} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**8*(d*x+c)/(b*x**3+a*x**2)**(5/2),x)
 

Output:

Integral(x**8*(c + d*x)/(x**2*(a + b*x))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.73 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} c}{3 \, {\left (b^{5} x + a b^{4}\right )} \sqrt {b x + a}} + \frac {2 \, {\left (3 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} + 192 \, a^{3} b x + 128 \, a^{4}\right )} d}{15 \, {\left (b^{6} x + a b^{5}\right )} \sqrt {b x + a}} \] Input:

integrate(x^8*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
 

Output:

2/3*(b^3*x^3 - 6*a*b^2*x^2 - 24*a^2*b*x - 16*a^3)*c/((b^5*x + a*b^4)*sqrt( 
b*x + a)) + 2/15*(3*b^4*x^4 - 8*a*b^3*x^3 + 48*a^2*b^2*x^2 + 192*a^3*b*x + 
 128*a^4)*d/((b^6*x + a*b^5)*sqrt(b*x + a))
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {32 \, {\left (5 \, a^{2} b c - 8 \, a^{3} d\right )} \mathrm {sgn}\left (x\right )}{15 \, \sqrt {a} b^{5}} - \frac {2 \, {\left (9 \, {\left (b x + a\right )} a^{2} b c - a^{3} b c - 12 \, {\left (b x + a\right )} a^{3} d + a^{4} d\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{21} c - 45 \, \sqrt {b x + a} a b^{21} c + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{20} d - 20 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{20} d + 90 \, \sqrt {b x + a} a^{2} b^{20} d\right )}}{15 \, b^{25} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^8*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
 

Output:

32/15*(5*a^2*b*c - 8*a^3*d)*sgn(x)/(sqrt(a)*b^5) - 2/3*(9*(b*x + a)*a^2*b* 
c - a^3*b*c - 12*(b*x + a)*a^3*d + a^4*d)/((b*x + a)^(3/2)*b^5*sgn(x)) + 2 
/15*(5*(b*x + a)^(3/2)*b^21*c - 45*sqrt(b*x + a)*a*b^21*c + 3*(b*x + a)^(5 
/2)*b^20*d - 20*(b*x + a)^(3/2)*a*b^20*d + 90*sqrt(b*x + a)*a^2*b^20*d)/(b 
^25*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.68 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (128\,d\,a^4+192\,d\,a^3\,b\,x-80\,c\,a^3\,b+48\,d\,a^2\,b^2\,x^2-120\,c\,a^2\,b^2\,x-8\,d\,a\,b^3\,x^3-30\,c\,a\,b^3\,x^2+3\,d\,b^4\,x^4+5\,c\,b^4\,x^3\right )}{15\,b^5\,x\,{\left (a+b\,x\right )}^2} \] Input:

int((x^8*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x)
 

Output:

(2*(a*x^2 + b*x^3)^(1/2)*(128*a^4*d + 5*b^4*c*x^3 + 3*b^4*d*x^4 - 80*a^3*b 
*c + 48*a^2*b^2*d*x^2 + 192*a^3*b*d*x - 120*a^2*b^2*c*x - 30*a*b^3*c*x^2 - 
 8*a*b^3*d*x^3))/(15*b^5*x*(a + b*x)^2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.63 \[ \int \frac {x^8 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\frac {2}{5} b^{4} d \,x^{4}-\frac {16}{15} a \,b^{3} d \,x^{3}+\frac {2}{3} b^{4} c \,x^{3}+\frac {32}{5} a^{2} b^{2} d \,x^{2}-4 a \,b^{3} c \,x^{2}+\frac {128}{5} a^{3} b d x -16 a^{2} b^{2} c x +\frac {256}{15} a^{4} d -\frac {32}{3} a^{3} b c}{\sqrt {b x +a}\, b^{5} \left (b x +a \right )} \] Input:

int(x^8*(d*x+c)/(b*x^3+a*x^2)^(5/2),x)
 

Output:

(2*(128*a**4*d - 80*a**3*b*c + 192*a**3*b*d*x - 120*a**2*b**2*c*x + 48*a** 
2*b**2*d*x**2 - 30*a*b**3*c*x**2 - 8*a*b**3*d*x**3 + 5*b**4*c*x**3 + 3*b** 
4*d*x**4))/(15*sqrt(a + b*x)*b**5*(a + b*x))