Integrand size = 20, antiderivative size = 121 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 c^2 \left (b c^2+a d^2\right ) \sqrt {c+d x}}{d^5}-\frac {4 c \left (2 b c^2+a d^2\right ) (c+d x)^{3/2}}{3 d^5}+\frac {2 \left (6 b c^2+a d^2\right ) (c+d x)^{5/2}}{5 d^5}-\frac {8 b c (c+d x)^{7/2}}{7 d^5}+\frac {2 b (c+d x)^{9/2}}{9 d^5} \] Output:
2*c^2*(a*d^2+b*c^2)*(d*x+c)^(1/2)/d^5-4/3*c*(a*d^2+2*b*c^2)*(d*x+c)^(3/2)/ d^5+2/5*(a*d^2+6*b*c^2)*(d*x+c)^(5/2)/d^5-8/7*b*c*(d*x+c)^(7/2)/d^5+2/9*b* (d*x+c)^(9/2)/d^5
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (21 a d^2 \left (8 c^2-4 c d x+3 d^2 x^2\right )+b \left (128 c^4-64 c^3 d x+48 c^2 d^2 x^2-40 c d^3 x^3+35 d^4 x^4\right )\right )}{315 d^5} \] Input:
Integrate[(x^2*(a + b*x^2))/Sqrt[c + d*x],x]
Output:
(2*Sqrt[c + d*x]*(21*a*d^2*(8*c^2 - 4*c*d*x + 3*d^2*x^2) + b*(128*c^4 - 64 *c^3*d*x + 48*c^2*d^2*x^2 - 40*c*d^3*x^3 + 35*d^4*x^4)))/(315*d^5)
Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (-\frac {2 \sqrt {c+d x} \left (a c d^2+2 b c^3\right )}{d^4}+\frac {(c+d x)^{3/2} \left (a d^2+6 b c^2\right )}{d^4}+\frac {a c^2 d^2+b c^4}{d^4 \sqrt {c+d x}}+\frac {b (c+d x)^{7/2}}{d^4}-\frac {4 b c (c+d x)^{5/2}}{d^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{5/2} \left (a d^2+6 b c^2\right )}{5 d^5}-\frac {4 c (c+d x)^{3/2} \left (a d^2+2 b c^2\right )}{3 d^5}+\frac {2 c^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}{d^5}+\frac {2 b (c+d x)^{9/2}}{9 d^5}-\frac {8 b c (c+d x)^{7/2}}{7 d^5}\) |
Input:
Int[(x^2*(a + b*x^2))/Sqrt[c + d*x],x]
Output:
(2*c^2*(b*c^2 + a*d^2)*Sqrt[c + d*x])/d^5 - (4*c*(2*b*c^2 + a*d^2)*(c + d* x)^(3/2))/(3*d^5) + (2*(6*b*c^2 + a*d^2)*(c + d*x)^(5/2))/(5*d^5) - (8*b*c *(c + d*x)^(7/2))/(7*d^5) + (2*b*(c + d*x)^(9/2))/(9*d^5)
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 = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {16 \sqrt {d x +c}\, \left (\frac {3 x^{2} \left (\frac {5 b \,x^{2}}{9}+a \right ) d^{4}}{8}-\frac {x c \left (\frac {10 b \,x^{2}}{21}+a \right ) d^{3}}{2}+c^{2} \left (\frac {2 b \,x^{2}}{7}+a \right ) d^{2}-\frac {8 b \,c^{3} d x}{21}+\frac {16 b \,c^{4}}{21}\right )}{15 d^{5}}\) | \(74\) |
gosper | \(\frac {2 \sqrt {d x +c}\, \left (35 b \,x^{4} d^{4}-40 c b \,d^{3} x^{3}+63 a \,d^{4} x^{2}+48 b \,c^{2} d^{2} x^{2}-84 a c \,d^{3} x -64 b \,c^{3} d x +168 a \,c^{2} d^{2}+128 b \,c^{4}\right )}{315 d^{5}}\) | \(85\) |
trager | \(\frac {2 \sqrt {d x +c}\, \left (35 b \,x^{4} d^{4}-40 c b \,d^{3} x^{3}+63 a \,d^{4} x^{2}+48 b \,c^{2} d^{2} x^{2}-84 a c \,d^{3} x -64 b \,c^{3} d x +168 a \,c^{2} d^{2}+128 b \,c^{4}\right )}{315 d^{5}}\) | \(85\) |
risch | \(\frac {2 \sqrt {d x +c}\, \left (35 b \,x^{4} d^{4}-40 c b \,d^{3} x^{3}+63 a \,d^{4} x^{2}+48 b \,c^{2} d^{2} x^{2}-84 a c \,d^{3} x -64 b \,c^{3} d x +168 a \,c^{2} d^{2}+128 b \,c^{4}\right )}{315 d^{5}}\) | \(85\) |
orering | \(\frac {2 \sqrt {d x +c}\, \left (35 b \,x^{4} d^{4}-40 c b \,d^{3} x^{3}+63 a \,d^{4} x^{2}+48 b \,c^{2} d^{2} x^{2}-84 a c \,d^{3} x -64 b \,c^{3} d x +168 a \,c^{2} d^{2}+128 b \,c^{4}\right )}{315 d^{5}}\) | \(85\) |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {8 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,d^{2}+6 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 b \,c^{3}-2 c \left (a \,d^{2}+b \,c^{2}\right )\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}+2 c^{2} \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{d^{5}}\) | \(101\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {8 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,d^{2}+6 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 b \,c^{3}-2 c \left (a \,d^{2}+b \,c^{2}\right )\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}+2 c^{2} \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{d^{5}}\) | \(101\) |
Input:
int(x^2*(b*x^2+a)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
16/15*(d*x+c)^(1/2)*(3/8*x^2*(5/9*b*x^2+a)*d^4-1/2*x*c*(10/21*b*x^2+a)*d^3 +c^2*(2/7*b*x^2+a)*d^2-8/21*b*c^3*d*x+16/21*b*c^4)/d^5
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.71 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, b d^{4} x^{4} - 40 \, b c d^{3} x^{3} + 128 \, b c^{4} + 168 \, a c^{2} d^{2} + 3 \, {\left (16 \, b c^{2} d^{2} + 21 \, a d^{4}\right )} x^{2} - 4 \, {\left (16 \, b c^{3} d + 21 \, a c d^{3}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{5}} \] Input:
integrate(x^2*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
2/315*(35*b*d^4*x^4 - 40*b*c*d^3*x^3 + 128*b*c^4 + 168*a*c^2*d^2 + 3*(16*b *c^2*d^2 + 21*a*d^4)*x^2 - 4*(16*b*c^3*d + 21*a*c*d^3)*x)*sqrt(d*x + c)/d^ 5
Time = 0.51 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (- \frac {4 b c \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {b \left (c + d x\right )^{\frac {9}{2}}}{9 d^{2}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (a d^{2} + 6 b c^{2}\right )}{5 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- 2 a c d^{2} - 4 b c^{3}\right )}{3 d^{2}} + \frac {\sqrt {c + d x} \left (a c^{2} d^{2} + b c^{4}\right )}{d^{2}}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\frac {a x^{3}}{3} + \frac {b x^{5}}{5}}{\sqrt {c}} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(b*x**2+a)/(d*x+c)**(1/2),x)
Output:
Piecewise((2*(-4*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 + 6*b*c**2)/(5*d**2) + (c + d*x)**(3/2)*(-2* a*c*d**2 - 4*b*c**3)/(3*d**2) + sqrt(c + d*x)*(a*c**2*d**2 + b*c**4)/d**2) /d**3, Ne(d, 0)), ((a*x**3/3 + b*x**5/5)/sqrt(c), True))
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} b c + 63 \, {\left (6 \, b c^{2} + a d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 210 \, {\left (2 \, b c^{3} + a c d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 315 \, {\left (b c^{4} + a c^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{315 \, d^{5}} \] Input:
integrate(x^2*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
2/315*(35*(d*x + c)^(9/2)*b - 180*(d*x + c)^(7/2)*b*c + 63*(6*b*c^2 + a*d^ 2)*(d*x + c)^(5/2) - 210*(2*b*c^3 + a*c*d^2)*(d*x + c)^(3/2) + 315*(b*c^4 + a*c^2*d^2)*sqrt(d*x + c))/d^5
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a}{d^{2}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b}{d^{4}}\right )}}{315 \, d} \] Input:
integrate(x^2*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="giac")
Output:
2/315*(21*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2 )*a/d^2 + (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^4)/d
Time = 7.94 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {\left (2\,b\,c^4+2\,a\,c^2\,d^2\right )\,\sqrt {c+d\,x}}{d^5}-\frac {\left (8\,b\,c^3+4\,a\,c\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^5}+\frac {2\,b\,{\left (c+d\,x\right )}^{9/2}}{9\,d^5}+\frac {\left (12\,b\,c^2+2\,a\,d^2\right )\,{\left (c+d\,x\right )}^{5/2}}{5\,d^5}-\frac {8\,b\,c\,{\left (c+d\,x\right )}^{7/2}}{7\,d^5} \] Input:
int((x^2*(a + b*x^2))/(c + d*x)^(1/2),x)
Output:
((2*b*c^4 + 2*a*c^2*d^2)*(c + d*x)^(1/2))/d^5 - ((8*b*c^3 + 4*a*c*d^2)*(c + d*x)^(3/2))/(3*d^5) + (2*b*(c + d*x)^(9/2))/(9*d^5) + ((2*a*d^2 + 12*b*c ^2)*(c + d*x)^(5/2))/(5*d^5) - (8*b*c*(c + d*x)^(7/2))/(7*d^5)
Time = 0.18 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {d x +c}\, \left (35 b \,d^{4} x^{4}-40 b c \,d^{3} x^{3}+63 a \,d^{4} x^{2}+48 b \,c^{2} d^{2} x^{2}-84 a c \,d^{3} x -64 b \,c^{3} d x +168 a \,c^{2} d^{2}+128 b \,c^{4}\right )}{315 d^{5}} \] Input:
int(x^2*(b*x^2+a)/(d*x+c)^(1/2),x)
Output:
(2*sqrt(c + d*x)*(168*a*c**2*d**2 - 84*a*c*d**3*x + 63*a*d**4*x**2 + 128*b *c**4 - 64*b*c**3*d*x + 48*b*c**2*d**2*x**2 - 40*b*c*d**3*x**3 + 35*b*d**4 *x**4))/(315*d**5)