Integrand size = 22, antiderivative size = 131 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 a^2 (b c-a d) \left (a x^2+b x^3\right )^{3/2}}{3 b^4 x^3}-\frac {2 a (2 b c-3 a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^4 x^5}+\frac {2 (b c-3 a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^4 x^7}+\frac {2 d \left (a x^2+b x^3\right )^{9/2}}{9 b^4 x^9} \] Output:
2/3*a^2*(-a*d+b*c)*(b*x^3+a*x^2)^(3/2)/b^4/x^3-2/5*a*(-3*a*d+2*b*c)*(b*x^3 +a*x^2)^(5/2)/b^4/x^5+2/7*(-3*a*d+b*c)*(b*x^3+a*x^2)^(7/2)/b^4/x^7+2/9*d*( b*x^3+a*x^2)^(9/2)/b^4/x^9
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.55 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 \left (x^2 (a+b x)\right )^{3/2} \left (-16 a^3 d+24 a^2 b (c+d x)-6 a b^2 x (6 c+5 d x)+5 b^3 x^2 (9 c+7 d x)\right )}{315 b^4 x^3} \] Input:
Integrate[x*(c + d*x)*Sqrt[a*x^2 + b*x^3],x]
Output:
(2*(x^2*(a + b*x))^(3/2)*(-16*a^3*d + 24*a^2*b*(c + d*x) - 6*a*b^2*x*(6*c + 5*d*x) + 5*b^3*x^2*(9*c + 7*d*x)))/(315*b^4*x^3)
Time = 0.49 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1945, 1922, 1908, 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.
\(\displaystyle \int x \sqrt {a x^2+b x^3} (c+d x) \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(3 b c-2 a d) \int x \sqrt {b x^3+a x^2}dx}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{9 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \int \sqrt {b x^3+a x^2}dx}{7 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{9 b}\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {2 a \int \frac {\sqrt {b x^3+a x^2}}{x}dx}{5 b}\right )}{7 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{9 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3}\right )}{7 b}\right ) (3 b c-2 a d)}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{9 b}\) |
Input:
Int[x*(c + d*x)*Sqrt[a*x^2 + b*x^3],x]
Output:
(2*d*(a*x^2 + b*x^3)^(3/2))/(9*b) + ((3*b*c - 2*a*d)*((2*(a*x^2 + b*x^3)^( 3/2))/(7*b*x) - (4*a*((-4*a*(a*x^2 + b*x^3)^(3/2))/(15*b^2*x^3) + (2*(a*x^ 2 + b*x^3)^(3/2))/(5*b*x^2)))/(7*b)))/(3*b)
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j - 1)), x] - Simp[b*((n*p + n - j + 1)/(a*( j*p + 1))) Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 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*(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 = 1.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.31
method | result | size |
pseudoelliptic | \(\frac {16 \left (b x +a \right )^{\frac {3}{2}} \left (\frac {21 x \left (\frac {5 d x}{7}+c \right ) b^{2}}{8}-\frac {7 \left (\frac {6 d x}{7}+c \right ) a b}{4}+a^{2} d \right )}{105 b^{3}}\) | \(41\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} d \,x^{3}+30 a \,b^{2} d \,x^{2}-45 b^{3} c \,x^{2}-24 a^{2} b d x +36 a \,b^{2} c x +16 a^{3} d -24 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(85\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} d \,x^{3}+30 a \,b^{2} d \,x^{2}-45 b^{3} c \,x^{2}-24 a^{2} b d x +36 a \,b^{2} c x +16 a^{3} d -24 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(85\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} d \,x^{3}+30 a \,b^{2} d \,x^{2}-45 b^{3} c \,x^{2}-24 a^{2} b d x +36 a \,b^{2} c x +16 a^{3} d -24 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(85\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-35 d \,x^{4} b^{4}-5 a \,b^{3} d \,x^{3}-45 b^{4} c \,x^{3}+6 a^{2} b^{2} d \,x^{2}-9 a \,b^{3} c \,x^{2}-8 a^{3} b d x +12 a^{2} b^{2} c x +16 a^{4} d -24 a^{3} b c \right )}{315 x \,b^{4}}\) | \(102\) |
trager | \(-\frac {2 \left (-35 d \,x^{4} b^{4}-5 a \,b^{3} d \,x^{3}-45 b^{4} c \,x^{3}+6 a^{2} b^{2} d \,x^{2}-9 a \,b^{3} c \,x^{2}-8 a^{3} b d x +12 a^{2} b^{2} c x +16 a^{4} d -24 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(104\) |
Input:
int(x*(d*x+c)*(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
16/105*(b*x+a)^(3/2)*(21/8*x*(5/7*d*x+c)*b^2-7/4*(6/7*d*x+c)*a*b+a^2*d)/b^ 3
Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.79 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 \, {\left (35 \, b^{4} d x^{4} + 24 \, a^{3} b c - 16 \, a^{4} d + 5 \, {\left (9 \, b^{4} c + a b^{3} d\right )} x^{3} + 3 \, {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d\right )} x^{2} - 4 \, {\left (3 \, a^{2} b^{2} c - 2 \, a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{315 \, b^{4} x} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
2/315*(35*b^4*d*x^4 + 24*a^3*b*c - 16*a^4*d + 5*(9*b^4*c + a*b^3*d)*x^3 + 3*(3*a*b^3*c - 2*a^2*b^2*d)*x^2 - 4*(3*a^2*b^2*c - 2*a^3*b*d)*x)*sqrt(b*x^ 3 + a*x^2)/(b^4*x)
\[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\int x \sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )\, dx \] Input:
integrate(x*(d*x+c)*(b*x**3+a*x**2)**(1/2),x)
Output:
Integral(x*sqrt(x**2*(a + b*x))*(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 \, {\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a} c}{105 \, b^{3}} + \frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a} d}{315 \, b^{4}} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)*c/b^3 + 2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqr t(b*x + a)*d/b^4
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (115) = 230\).
Time = 0.11 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.81 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a c \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} c \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a d \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} d \mathrm {sgn}\left (x\right )}{b^{3}}\right )}}{315 \, b} - \frac {16 \, {\left (3 \, a^{\frac {7}{2}} b c - 2 \, a^{\frac {9}{2}} d\right )} \mathrm {sgn}\left (x\right )}{315 \, b^{4}} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
2/315*(21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2 )*a*c*sgn(x)/b^2 + 9*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*c*sgn(x)/b^2 + 9*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a* d*sgn(x)/b^3 + (35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a) ^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*d*sgn(x)/b^3 )/b - 16/315*(3*a^(7/2)*b*c - 2*a^(9/2)*d)*sgn(x)/b^4
Time = 8.78 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,d\,x^4}{9}-\frac {32\,a^4\,d-48\,a^3\,b\,c}{315\,b^4}+\frac {x^3\,\left (90\,c\,b^4+10\,a\,d\,b^3\right )}{315\,b^4}-\frac {2\,a\,x^2\,\left (2\,a\,d-3\,b\,c\right )}{105\,b^2}+\frac {8\,a^2\,x\,\left (2\,a\,d-3\,b\,c\right )}{315\,b^3}\right )}{x} \] Input:
int(x*(a*x^2 + b*x^3)^(1/2)*(c + d*x),x)
Output:
((a*x^2 + b*x^3)^(1/2)*((2*d*x^4)/9 - (32*a^4*d - 48*a^3*b*c)/(315*b^4) + (x^3*(90*b^4*c + 10*a*b^3*d))/(315*b^4) - (2*a*x^2*(2*a*d - 3*b*c))/(105*b ^2) + (8*a^2*x*(2*a*d - 3*b*c))/(315*b^3)))/x
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.71 \[ \int x (c+d x) \sqrt {a x^2+b x^3} \, dx=\frac {2 \sqrt {b x +a}\, \left (35 b^{4} d \,x^{4}+5 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}-6 a^{2} b^{2} d \,x^{2}+9 a \,b^{3} c \,x^{2}+8 a^{3} b d x -12 a^{2} b^{2} c x -16 a^{4} d +24 a^{3} b c \right )}{315 b^{4}} \] Input:
int(x*(d*x+c)*(b*x^3+a*x^2)^(1/2),x)
Output:
(2*sqrt(a + b*x)*( - 16*a**4*d + 24*a**3*b*c + 8*a**3*b*d*x - 12*a**2*b**2 *c*x - 6*a**2*b**2*d*x**2 + 9*a*b**3*c*x**2 + 5*a*b**3*d*x**3 + 45*b**4*c* x**3 + 35*b**4*d*x**4))/(315*b**4)