Integrand size = 24, antiderivative size = 88 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 a (b c-a d) x}{b^3 \sqrt {a x^2+b x^3}}+\frac {2 (b c-2 a d) \sqrt {a x^2+b x^3}}{b^3 x}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b^3 x^3} \] Output:
2*a*(-a*d+b*c)*x/b^3/(b*x^3+a*x^2)^(1/2)+2*(-2*a*d+b*c)*(b*x^3+a*x^2)^(1/2 )/b^3/x+2/3*d*(b*x^3+a*x^2)^(3/2)/b^3/x^3
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.58 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (-8 a^2 d+a b (6 c-4 d x)+b^2 x (3 c+d x)\right )}{3 b^3 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(x^4*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*x*(-8*a^2*d + a*b*(6*c - 4*d*x) + b^2*x*(3*c + d*x)))/(3*b^3*Sqrt[x^2*( a + b*x)])
Time = 0.44 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1943, 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.
\(\displaystyle \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {2 x^3 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(3 b c-4 a d) \int \frac {x^2}{\sqrt {b x^3+a x^2}}dx}{a b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 x^3 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {(3 b c-4 a d) \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{3 b}\right )}{a b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 x^3 (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {\left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {4 a \sqrt {a x^2+b x^3}}{3 b^2 x}\right ) (3 b c-4 a d)}{a b}\) |
Input:
Int[(x^4*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*(b*c - a*d)*x^3)/(a*b*Sqrt[a*x^2 + b*x^3]) - ((3*b*c - 4*a*d)*((2*Sqrt[ a*x^2 + b*x^3])/(3*b) - (4*a*Sqrt[a*x^2 + b*x^3])/(3*b^2*x)))/(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 = 0.52 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-b^{2} d \,x^{2}+4 a b d x -3 b^{2} c x +8 a^{2} d -6 a b c \right ) x^{3}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(61\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-b^{2} d \,x^{2}+4 a b d x -3 b^{2} c x +8 a^{2} d -6 a b c \right ) x^{3}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(61\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-b^{2} d \,x^{2}+4 a b d x -3 b^{2} c x +8 a^{2} d -6 a b c \right ) x^{3}}{3 b^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(61\) |
trager | \(-\frac {2 \left (-b^{2} d \,x^{2}+4 a b d x -3 b^{2} c x +8 a^{2} d -6 a b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 \left (b x +a \right ) b^{3} x}\) | \(63\) |
risch | \(-\frac {2 \left (-b d x +5 a d -3 b c \right ) \left (b x +a \right ) x}{3 b^{3} \sqrt {x^{2} \left (b x +a \right )}}-\frac {2 a \left (a d -b c \right ) x}{b^{3} \sqrt {x^{2} \left (b x +a \right )}}\) | \(64\) |
pseudoelliptic | \(\frac {\left (70 d \,x^{5}+90 c \,x^{4}\right ) b^{5}-144 \left (\frac {25 d x}{36}+c \right ) x^{3} a \,b^{4}+288 \left (\frac {5 d x}{9}+c \right ) x^{2} a^{2} b^{3}-1152 \left (\frac {5 d x}{18}+c \right ) x \,a^{3} b^{2}-2304 \left (-\frac {5 d x}{9}+c \right ) a^{4} b +2560 a^{5} d}{315 \sqrt {b x +a}\, b^{6}}\) | \(96\) |
Input:
int(x^4*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(b*x+a)*(-b^2*d*x^2+4*a*b*d*x-3*b^2*c*x+8*a^2*d-6*a*b*c)*x^3/b^3/(b*x ^3+a*x^2)^(3/2)
Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{2} d x^{2} + 6 \, a b c - 8 \, a^{2} d + {\left (3 \, b^{2} c - 4 \, a b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, {\left (b^{4} x^{2} + a b^{3} x\right )}} \] Input:
integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/3*(b^2*d*x^2 + 6*a*b*c - 8*a^2*d + (3*b^2*c - 4*a*b*d)*x)*sqrt(b*x^3 + a *x^2)/(b^4*x^2 + a*b^3*x)
\[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{4} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**4*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x**4*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.59 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b x + 2 \, a\right )} c}{\sqrt {b x + a} b^{2}} + \frac {2 \, {\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )} d}{3 \, \sqrt {b x + a} b^{3}} \] Input:
integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
2*(b*x + 2*a)*c/(sqrt(b*x + a)*b^2) + 2/3*(b^2*x^2 - 4*a*b*x - 8*a^2)*d/(s qrt(b*x + a)*b^3)
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {4 \, {\left (3 \, a b c - 4 \, a^{2} d\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {a} b^{3}} + \frac {2 \, {\left (a b c - a^{2} d\right )}}{\sqrt {b x + a} b^{3} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (3 \, \sqrt {b x + a} b^{7} c + {\left (b x + a\right )}^{\frac {3}{2}} b^{6} d - 6 \, \sqrt {b x + a} a b^{6} d\right )}}{3 \, b^{9} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
-4/3*(3*a*b*c - 4*a^2*d)*sgn(x)/(sqrt(a)*b^3) + 2*(a*b*c - a^2*d)/(sqrt(b* x + a)*b^3*sgn(x)) + 2/3*(3*sqrt(b*x + a)*b^7*c + (b*x + a)^(3/2)*b^6*d - 6*sqrt(b*x + a)*a*b^6*d)/(b^9*sgn(x))
Time = 8.95 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-8\,d\,a^2-4\,d\,a\,b\,x+6\,c\,a\,b+d\,b^2\,x^2+3\,c\,b^2\,x\right )}{3\,b^3\,x\,\left (a+b\,x\right )} \] Input:
int((x^4*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(b^2*d*x^2 - 8*a^2*d + 6*a*b*c + 3*b^2*c*x - 4*a* b*d*x))/(3*b^3*x*(a + b*x))
Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52 \[ \int \frac {x^4 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\frac {2}{3} b^{2} d \,x^{2}-\frac {8}{3} a b d x +2 b^{2} c x -\frac {16}{3} a^{2} d +4 a b c}{\sqrt {b x +a}\, b^{3}} \] Input:
int(x^4*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*( - 8*a**2*d + 6*a*b*c - 4*a*b*d*x + 3*b**2*c*x + b**2*d*x**2))/(3*sqrt (a + b*x)*b**3)