Integrand size = 18, antiderivative size = 92 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=-\frac {2 c \left (b c^2+a d^2\right ) (c+d x)^{3/2}}{3 d^4}+\frac {2 \left (3 b c^2+a d^2\right ) (c+d x)^{5/2}}{5 d^4}-\frac {6 b c (c+d x)^{7/2}}{7 d^4}+\frac {2 b (c+d x)^{9/2}}{9 d^4} \] Output:
-2/3*c*(a*d^2+b*c^2)*(d*x+c)^(3/2)/d^4+2/5*(a*d^2+3*b*c^2)*(d*x+c)^(5/2)/d ^4-6/7*b*c*(d*x+c)^(7/2)/d^4+2/9*b*(d*x+c)^(9/2)/d^4
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 (c+d x)^{3/2} \left (21 a d^2 (-2 c+3 d x)+b \left (-16 c^3+24 c^2 d x-30 c d^2 x^2+35 d^3 x^3\right )\right )}{315 d^4} \] Input:
Integrate[x*Sqrt[c + d*x]*(a + b*x^2),x]
Output:
(2*(c + d*x)^(3/2)*(21*a*d^2*(-2*c + 3*d*x) + b*(-16*c^3 + 24*c^2*d*x - 30 *c*d^2*x^2 + 35*d^3*x^3)))/(315*d^4)
Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {522, 2009}
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.
\(\displaystyle \int x \left (a+b x^2\right ) \sqrt {c+d x} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {(c+d x)^{3/2} \left (a d^2+3 b c^2\right )}{d^3}+\frac {c \sqrt {c+d x} \left (-a d^2-b c^2\right )}{d^3}+\frac {b (c+d x)^{7/2}}{d^3}-\frac {3 b c (c+d x)^{5/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{5/2} \left (a d^2+3 b c^2\right )}{5 d^4}-\frac {2 c (c+d x)^{3/2} \left (a d^2+b c^2\right )}{3 d^4}+\frac {2 b (c+d x)^{9/2}}{9 d^4}-\frac {6 b c (c+d x)^{7/2}}{7 d^4}\) |
Input:
Int[x*Sqrt[c + d*x]*(a + b*x^2),x]
Output:
(-2*c*(b*c^2 + a*d^2)*(c + d*x)^(3/2))/(3*d^4) + (2*(3*b*c^2 + a*d^2)*(c + d*x)^(5/2))/(5*d^4) - (6*b*c*(c + d*x)^(7/2))/(7*d^4) + (2*b*(c + d*x)^(9 /2))/(9*d^4)
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60
method | result | size |
pseudoelliptic | \(-\frac {4 \left (-\frac {3 x \left (\frac {5 b \,x^{2}}{9}+a \right ) d^{3}}{2}+c \left (\frac {5 b \,x^{2}}{7}+a \right ) d^{2}-\frac {4 b \,c^{2} d x}{7}+\frac {8 b \,c^{3}}{21}\right ) \left (d x +c \right )^{\frac {3}{2}}}{15 d^{4}}\) | \(55\) |
gosper | \(-\frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (-35 b \,d^{3} x^{3}+30 b c \,d^{2} x^{2}-63 a x \,d^{3}-24 b \,c^{2} d x +42 a \,d^{2} c +16 b \,c^{3}\right )}{315 d^{4}}\) | \(61\) |
orering | \(-\frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (-35 b \,d^{3} x^{3}+30 b c \,d^{2} x^{2}-63 a x \,d^{3}-24 b \,c^{2} d x +42 a \,d^{2} c +16 b \,c^{3}\right )}{315 d^{4}}\) | \(61\) |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {6 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,d^{2}+3 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 c \left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}}{d^{4}}\) | \(70\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {6 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,d^{2}+3 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 c \left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}}{d^{4}}\) | \(70\) |
trager | \(-\frac {2 \left (-35 b \,x^{4} d^{4}-5 c b \,d^{3} x^{3}-63 a \,d^{4} x^{2}+6 b \,c^{2} d^{2} x^{2}-21 a c \,d^{3} x -8 b \,c^{3} d x +42 a \,c^{2} d^{2}+16 b \,c^{4}\right ) \sqrt {d x +c}}{315 d^{4}}\) | \(85\) |
risch | \(-\frac {2 \left (-35 b \,x^{4} d^{4}-5 c b \,d^{3} x^{3}-63 a \,d^{4} x^{2}+6 b \,c^{2} d^{2} x^{2}-21 a c \,d^{3} x -8 b \,c^{3} d x +42 a \,c^{2} d^{2}+16 b \,c^{4}\right ) \sqrt {d x +c}}{315 d^{4}}\) | \(85\) |
Input:
int(x*(d*x+c)^(1/2)*(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
-4/15*(-3/2*x*(5/9*b*x^2+a)*d^3+c*(5/7*b*x^2+a)*d^2-4/7*b*c^2*d*x+8/21*b*c ^3)*(d*x+c)^(3/2)/d^4
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.92 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 \, {\left (35 \, b d^{4} x^{4} + 5 \, b c d^{3} x^{3} - 16 \, b c^{4} - 42 \, a c^{2} d^{2} - 3 \, {\left (2 \, b c^{2} d^{2} - 21 \, a d^{4}\right )} x^{2} + {\left (8 \, b c^{3} d + 21 \, a c d^{3}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{4}} \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a),x, algorithm="fricas")
Output:
2/315*(35*b*d^4*x^4 + 5*b*c*d^3*x^3 - 16*b*c^4 - 42*a*c^2*d^2 - 3*(2*b*c^2 *d^2 - 21*a*d^4)*x^2 + (8*b*c^3*d + 21*a*c*d^3)*x)*sqrt(d*x + c)/d^4
Time = 0.69 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\begin {cases} \frac {2 \left (- \frac {3 b c \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {b \left (c + d x\right )^{\frac {9}{2}}}{9 d^{2}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (a d^{2} + 3 b c^{2}\right )}{5 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- a c d^{2} - b c^{3}\right )}{3 d^{2}}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\begin {cases} \frac {a x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{2}}{4 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x*(d*x+c)**(1/2)*(b*x**2+a),x)
Output:
Piecewise((2*(-3*b*c*(c + d*x)**(7/2)/(7*d**2) + b*(c + d*x)**(9/2)/(9*d** 2) + (c + d*x)**(5/2)*(a*d**2 + 3*b*c**2)/(5*d**2) + (c + d*x)**(3/2)*(-a* c*d**2 - b*c**3)/(3*d**2))/d**2, Ne(d, 0)), (sqrt(c)*Piecewise((a*x**2/2, Eq(b, 0)), ((a + b*x**2)**2/(4*b), True)), True))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b - 135 \, {\left (d x + c\right )}^{\frac {7}{2}} b c + 63 \, {\left (3 \, b c^{2} + a d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 105 \, {\left (b c^{3} + a c d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{315 \, d^{4}} \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a),x, algorithm="maxima")
Output:
2/315*(35*(d*x + c)^(9/2)*b - 135*(d*x + c)^(7/2)*b*c + 63*(3*b*c^2 + a*d^ 2)*(d*x + c)^(5/2) - 105*(b*c^3 + a*c*d^2)*(d*x + c)^(3/2))/d^4
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (76) = 152\).
Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.97 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 \, {\left (\frac {105 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a c}{d} + \frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a}{d} + \frac {9 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b c}{d^{3}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b}{d^{3}}\right )}}{315 \, d} \] Input:
integrate(x*(d*x+c)^(1/2)*(b*x^2+a),x, algorithm="giac")
Output:
2/315*(105*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*c/d + 21*(3*(d*x + c)^( 5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a/d + 9*(5*(d*x + c)^( 7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^ 3)*b*c/d^3 + (35*(d*x + c)^(9/2) - 180*(d*x + c)^(7/2)*c + 378*(d*x + c)^( 5/2)*c^2 - 420*(d*x + c)^(3/2)*c^3 + 315*sqrt(d*x + c)*c^4)*b/d^3)/d
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.86 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2\,b\,{\left (c+d\,x\right )}^{9/2}}{9\,d^4}-\frac {\left (2\,b\,c^3+2\,a\,c\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}+\frac {\left (6\,b\,c^2+2\,a\,d^2\right )\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}-\frac {6\,b\,c\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4} \] Input:
int(x*(a + b*x^2)*(c + d*x)^(1/2),x)
Output:
(2*b*(c + d*x)^(9/2))/(9*d^4) - ((2*b*c^3 + 2*a*c*d^2)*(c + d*x)^(3/2))/(3 *d^4) + ((2*a*d^2 + 6*b*c^2)*(c + d*x)^(5/2))/(5*d^4) - (6*b*c*(c + d*x)^( 7/2))/(7*d^4)
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int x \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 \sqrt {d x +c}\, \left (35 b \,d^{4} x^{4}+5 b c \,d^{3} x^{3}+63 a \,d^{4} x^{2}-6 b \,c^{2} d^{2} x^{2}+21 a c \,d^{3} x +8 b \,c^{3} d x -42 a \,c^{2} d^{2}-16 b \,c^{4}\right )}{315 d^{4}} \] Input:
int(x*(d*x+c)^(1/2)*(b*x^2+a),x)
Output:
(2*sqrt(c + d*x)*( - 42*a*c**2*d**2 + 21*a*c*d**3*x + 63*a*d**4*x**2 - 16* b*c**4 + 8*b*c**3*d*x - 6*b*c**2*d**2*x**2 + 5*b*c*d**3*x**3 + 35*b*d**4*x **4))/(315*d**4)