Integrand size = 24, antiderivative size = 92 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 a (b c-a d) \sqrt {a x^2+b x^3}}{b^3 x}+\frac {2 (b c-2 a d) \left (a x^2+b x^3\right )^{3/2}}{3 b^3 x^3}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{5 b^3 x^5} \] Output:
-2*a*(-a*d+b*c)*(b*x^3+a*x^2)^(1/2)/b^3/x+2/3*(-2*a*d+b*c)*(b*x^3+a*x^2)^( 3/2)/b^3/x^3+2/5*d*(b*x^3+a*x^2)^(5/2)/b^3/x^5
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {x^2 (a+b x)} \left (8 a^2 d-2 a b (5 c+2 d x)+b^2 x (5 c+3 d x)\right )}{15 b^3 x} \] Input:
Integrate[(x^2*(c + d*x))/Sqrt[a*x^2 + b*x^3],x]
Output:
(2*Sqrt[x^2*(a + b*x)]*(8*a^2*d - 2*a*b*(5*c + 2*d*x) + b^2*x*(5*c + 3*d*x )))/(15*b^3*x)
Time = 0.42 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1945, 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.
\(\displaystyle \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(5 b c-4 a d) \int \frac {x^2}{\sqrt {b x^3+a x^2}}dx}{5 b}+\frac {2 d x \sqrt {a x^2+b x^3}}{5 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(5 b c-4 a d) \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{3 b}\right )}{5 b}+\frac {2 d x \sqrt {a x^2+b x^3}}{5 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {4 a \sqrt {a x^2+b x^3}}{3 b^2 x}\right ) (5 b c-4 a d)}{5 b}+\frac {2 d x \sqrt {a x^2+b x^3}}{5 b}\) |
Input:
Int[(x^2*(c + d*x))/Sqrt[a*x^2 + b*x^3],x]
Output:
(2*d*x*Sqrt[a*x^2 + b*x^3])/(5*b) + ((5*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)))/(5*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[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.64 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61
method | result | size |
trager | \(\frac {2 \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 b^{3} x}\) | \(56\) |
risch | \(\frac {2 \left (b x +a \right ) x \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right )}{15 \sqrt {x^{2} \left (b x +a \right )}\, b^{3}}\) | \(57\) |
pseudoelliptic | \(-\frac {32 \sqrt {b x +a}\, \left (-\frac {7 \left (\frac {5 d x}{7}+c \right ) x^{2} b^{3}}{16}+\frac {7 \left (\frac {9 d x}{14}+c \right ) x a \,b^{2}}{12}-\frac {7 \left (\frac {3 d x}{7}+c \right ) a^{2} b}{6}+a^{3} d \right )}{35 b^{4}}\) | \(58\) |
gosper | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right ) x}{15 b^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(59\) |
default | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right ) x}{15 b^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(59\) |
orering | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right ) x}{15 b^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(59\) |
Input:
int(x^2*(d*x+c)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/15*(3*b^2*d*x^2-4*a*b*d*x+5*b^2*c*x+8*a^2*d-10*a*b*c)/b^3/x*(b*x^3+a*x^2 )^(1/2)
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (3 \, b^{2} d x^{2} - 10 \, a b c + 8 \, a^{2} d + {\left (5 \, b^{2} c - 4 \, a b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, b^{3} x} \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
2/15*(3*b^2*d*x^2 - 10*a*b*c + 8*a^2*d + (5*b^2*c - 4*a*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^3*x)
\[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{2} \left (c + d x\right )}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:
integrate(x**2*(d*x+c)/(b*x**3+a*x**2)**(1/2),x)
Output:
Integral(x**2*(c + d*x)/sqrt(x**2*(a + b*x)), x)
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (b^{2} x^{2} - a b x - 2 \, a^{2}\right )} c}{3 \, \sqrt {b x + a} b^{2}} + \frac {2 \, {\left (3 \, b^{3} x^{3} - a b^{2} x^{2} + 4 \, a^{2} b x + 8 \, a^{3}\right )} d}{15 \, \sqrt {b x + a} b^{3}} \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
2/3*(b^2*x^2 - a*b*x - 2*a^2)*c/(sqrt(b*x + a)*b^2) + 2/15*(3*b^3*x^3 - a* b^2*x^2 + 4*a^2*b*x + 8*a^3)*d/(sqrt(b*x + a)*b^3)
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} c}{b} + \frac {{\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} d}{b^{2}}\right )}}{15 \, b \mathrm {sgn}\left (x\right )} + \frac {4 \, {\left (5 \, a^{\frac {3}{2}} b c - 4 \, a^{\frac {5}{2}} d\right )} \mathrm {sgn}\left (x\right )}{15 \, b^{3}} \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
2/15*(5*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*c/b + (3*(b*x + a)^(5/2) - 1 0*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*d/b^2)/(b*sgn(x)) + 4/15*(5*a^ (3/2)*b*c - 4*a^(5/2)*d)*sgn(x)/b^3
Time = 9.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {16\,a^2\,d-20\,a\,b\,c}{15\,b^3}+\frac {x\,\left (10\,b^2\,c-8\,a\,b\,d\right )}{15\,b^3}+\frac {2\,d\,x^2}{5\,b}\right )}{x} \] Input:
int((x^2*(c + d*x))/(a*x^2 + b*x^3)^(1/2),x)
Output:
((a*x^2 + b*x^3)^(1/2)*((16*a^2*d - 20*a*b*c)/(15*b^3) + (x*(10*b^2*c - 8* a*b*d))/(15*b^3) + (2*d*x^2)/(5*b)))/x
Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {b x +a}\, \left (3 b^{2} d \,x^{2}-4 a b d x +5 b^{2} c x +8 a^{2} d -10 a b c \right )}{15 b^{3}} \] Input:
int(x^2*(d*x+c)/(b*x^3+a*x^2)^(1/2),x)
Output:
(2*sqrt(a + b*x)*(8*a**2*d - 10*a*b*c - 4*a*b*d*x + 5*b**2*c*x + 3*b**2*d* x**2))/(15*b**3)