Integrand size = 24, antiderivative size = 125 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 a^2 (b c-a d) x}{b^4 \sqrt {a x^2+b x^3}}-\frac {2 a (2 b c-3 a d) \sqrt {a x^2+b x^3}}{b^4 x}+\frac {2 (b c-3 a d) \left (a x^2+b x^3\right )^{3/2}}{3 b^4 x^3}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{5 b^4 x^5} \] Output:
-2*a^2*(-a*d+b*c)*x/b^4/(b*x^3+a*x^2)^(1/2)-2*a*(-3*a*d+2*b*c)*(b*x^3+a*x^ 2)^(1/2)/b^4/x+2/3*(-3*a*d+b*c)*(b*x^3+a*x^2)^(3/2)/b^4/x^3+2/5*d*(b*x^3+a *x^2)^(5/2)/b^4/x^5
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.58 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (48 a^3 d-8 a^2 b (5 c-3 d x)+b^3 x^2 (5 c+3 d x)-2 a b^2 x (10 c+3 d x)\right )}{15 b^4 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(x^5*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*x*(48*a^3*d - 8*a^2*b*(5*c - 3*d*x) + b^3*x^2*(5*c + 3*d*x) - 2*a*b^2*x *(10*c + 3*d*x)))/(15*b^4*Sqrt[x^2*(a + b*x)])
Time = 0.53 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1943, 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.
\(\displaystyle \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {2 x^4 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(5 b c-6 a d) \int \frac {x^3}{\sqrt {b x^3+a x^2}}dx}{a b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 x^4 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(5 b c-6 a d) \left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \int \frac {x^2}{\sqrt {b x^3+a x^2}}dx}{5 b}\right )}{a b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 x^4 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(5 b c-6 a d) \left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{3 b}\right )}{5 b}\right )}{a b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 x^4 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {\left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {4 a \sqrt {a x^2+b x^3}}{3 b^2 x}\right )}{5 b}\right ) (5 b c-6 a d)}{a b}\) |
Input:
Int[(x^5*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*(b*c - a*d)*x^4)/(a*b*Sqrt[a*x^2 + b*x^3]) - ((5*b*c - 6*a*d)*((2*x*Sqr t[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)))/(a*b)
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])
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])
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])
Time = 0.54 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (3 b^{3} d \,x^{3}-6 a \,b^{2} d \,x^{2}+5 b^{3} c \,x^{2}+24 a^{2} b d x -20 a \,b^{2} c x +48 a^{3} d -40 c \,a^{2} b \right ) x^{3}}{15 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(85\) |
default | \(\frac {2 \left (b x +a \right ) \left (3 b^{3} d \,x^{3}-6 a \,b^{2} d \,x^{2}+5 b^{3} c \,x^{2}+24 a^{2} b d x -20 a \,b^{2} c x +48 a^{3} d -40 c \,a^{2} b \right ) x^{3}}{15 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(85\) |
orering | \(\frac {2 \left (b x +a \right ) \left (3 b^{3} d \,x^{3}-6 a \,b^{2} d \,x^{2}+5 b^{3} c \,x^{2}+24 a^{2} b d x -20 a \,b^{2} c x +48 a^{3} d -40 c \,a^{2} b \right ) x^{3}}{15 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(85\) |
risch | \(\frac {2 \left (3 b^{2} d \,x^{2}-9 a b d x +5 b^{2} c x +33 a^{2} d -25 a b c \right ) \left (b x +a \right ) x}{15 b^{4} \sqrt {x^{2} \left (b x +a \right )}}+\frac {2 a^{2} \left (a d -b c \right ) x}{b^{4} \sqrt {x^{2} \left (b x +a \right )}}\) | \(86\) |
trager | \(\frac {2 \left (3 b^{3} d \,x^{3}-6 a \,b^{2} d \,x^{2}+5 b^{3} c \,x^{2}+24 a^{2} b d x -20 a \,b^{2} c x +48 a^{3} d -40 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 \left (b x +a \right ) b^{4} x}\) | \(87\) |
pseudoelliptic | \(-\frac {2048 \left (-\frac {77 x^{5} \left (\frac {9 d x}{11}+c \right ) b^{6}}{3072}+\frac {55 \left (\frac {42 d x}{55}+c \right ) x^{4} a \,b^{5}}{1536}-\frac {11 x^{3} a^{2} \left (\frac {15 d x}{22}+c \right ) b^{4}}{192}+\frac {11 \left (\frac {6 d x}{11}+c \right ) x^{2} a^{3} b^{3}}{96}-\frac {11 x \left (\frac {3 d x}{11}+c \right ) a^{4} b^{2}}{24}-\frac {11 \left (-\frac {6 d x}{11}+c \right ) a^{5} b}{12}+a^{6} d \right )}{231 \sqrt {b x +a}\, b^{7}}\) | \(109\) |
Input:
int(x^5*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/15*(b*x+a)*(3*b^3*d*x^3-6*a*b^2*d*x^2+5*b^3*c*x^2+24*a^2*b*d*x-20*a*b^2* c*x+48*a^3*d-40*a^2*b*c)*x^3/b^4/(b*x^3+a*x^2)^(3/2)
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.72 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (3 \, b^{3} d x^{3} - 40 \, a^{2} b c + 48 \, a^{3} d + {\left (5 \, b^{3} c - 6 \, a b^{2} d\right )} x^{2} - 4 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, {\left (b^{5} x^{2} + a b^{4} x\right )}} \] Input:
integrate(x^5*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/15*(3*b^3*d*x^3 - 40*a^2*b*c + 48*a^3*d + (5*b^3*c - 6*a*b^2*d)*x^2 - 4* (5*a*b^2*c - 6*a^2*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^5*x^2 + a*b^4*x)
\[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{5} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**5*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x**5*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.59 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )} c}{3 \, \sqrt {b x + a} b^{3}} + \frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} d}{5 \, \sqrt {b x + a} b^{4}} \] Input:
integrate(x^5*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
2/3*(b^2*x^2 - 4*a*b*x - 8*a^2)*c/(sqrt(b*x + a)*b^3) + 2/5*(b^3*x^3 - 2*a *b^2*x^2 + 8*a^2*b*x + 16*a^3)*d/(sqrt(b*x + a)*b^4)
Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {16 \, {\left (5 \, a^{2} b c - 6 \, a^{3} d\right )} \mathrm {sgn}\left (x\right )}{15 \, \sqrt {a} b^{4}} - \frac {2 \, {\left (a^{2} b c - a^{3} d\right )}}{\sqrt {b x + a} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{17} c - 30 \, \sqrt {b x + a} a b^{17} c + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{16} d - 15 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{16} d + 45 \, \sqrt {b x + a} a^{2} b^{16} d\right )}}{15 \, b^{20} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^5*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
16/15*(5*a^2*b*c - 6*a^3*d)*sgn(x)/(sqrt(a)*b^4) - 2*(a^2*b*c - a^3*d)/(sq rt(b*x + a)*b^4*sgn(x)) + 2/15*(5*(b*x + a)^(3/2)*b^17*c - 30*sqrt(b*x + a )*a*b^17*c + 3*(b*x + a)^(5/2)*b^16*d - 15*(b*x + a)^(3/2)*a*b^16*d + 45*s qrt(b*x + a)*a^2*b^16*d)/(b^20*sgn(x))
Time = 9.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.69 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (48\,d\,a^3+24\,d\,a^2\,b\,x-40\,c\,a^2\,b-6\,d\,a\,b^2\,x^2-20\,c\,a\,b^2\,x+3\,d\,b^3\,x^3+5\,c\,b^3\,x^2\right )}{15\,b^4\,x\,\left (a+b\,x\right )} \] Input:
int((x^5*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(48*a^3*d + 5*b^3*c*x^2 + 3*b^3*d*x^3 - 40*a^2*b* c - 20*a*b^2*c*x + 24*a^2*b*d*x - 6*a*b^2*d*x^2))/(15*b^4*x*(a + b*x))
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.57 \[ \int \frac {x^5 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\frac {2}{5} b^{3} d \,x^{3}-\frac {4}{5} a \,b^{2} d \,x^{2}+\frac {2}{3} b^{3} c \,x^{2}+\frac {16}{5} a^{2} b d x -\frac {8}{3} a \,b^{2} c x +\frac {32}{5} a^{3} d -\frac {16}{3} a^{2} b c}{\sqrt {b x +a}\, b^{4}} \] Input:
int(x^5*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*(48*a**3*d - 40*a**2*b*c + 24*a**2*b*d*x - 20*a*b**2*c*x - 6*a*b**2*d*x **2 + 5*b**3*c*x**2 + 3*b**3*d*x**3))/(15*sqrt(a + b*x)*b**4)