Integrand size = 18, antiderivative size = 90 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=-\frac {2 c \left (b c^2+a d^2\right ) \sqrt {c+d x}}{d^4}+\frac {2 \left (3 b c^2+a d^2\right ) (c+d x)^{3/2}}{3 d^4}-\frac {6 b c (c+d x)^{5/2}}{5 d^4}+\frac {2 b (c+d x)^{7/2}}{7 d^4} \] Output:
-2*c*(a*d^2+b*c^2)*(d*x+c)^(1/2)/d^4+2/3*(a*d^2+3*b*c^2)*(d*x+c)^(3/2)/d^4 -6/5*b*c*(d*x+c)^(5/2)/d^4+2/7*b*(d*x+c)^(7/2)/d^4
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.70 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (35 a d^2 (-2 c+d x)-3 b \left (16 c^3-8 c^2 d x+6 c d^2 x^2-5 d^3 x^3\right )\right )}{105 d^4} \] Input:
Integrate[(x*(a + b*x^2))/Sqrt[c + d*x],x]
Output:
(2*Sqrt[c + d*x]*(35*a*d^2*(-2*c + d*x) - 3*b*(16*c^3 - 8*c^2*d*x + 6*c*d^ 2*x^2 - 5*d^3*x^3)))/(105*d^4)
Time = 0.37 (sec) , antiderivative size = 90, 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.
\(\displaystyle \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {\sqrt {c+d x} \left (a d^2+3 b c^2\right )}{d^3}+\frac {c \left (-a d^2-b c^2\right )}{d^3 \sqrt {c+d x}}+\frac {b (c+d x)^{5/2}}{d^3}-\frac {3 b c (c+d x)^{3/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a d^2+3 b c^2\right )}{3 d^4}-\frac {2 c \sqrt {c+d x} \left (a d^2+b c^2\right )}{d^4}+\frac {2 b (c+d x)^{7/2}}{7 d^4}-\frac {6 b c (c+d x)^{5/2}}{5 d^4}\) |
Input:
Int[(x*(a + b*x^2))/Sqrt[c + d*x],x]
Output:
(-2*c*(b*c^2 + a*d^2)*Sqrt[c + d*x])/d^4 + (2*(3*b*c^2 + a*d^2)*(c + d*x)^ (3/2))/(3*d^4) - (6*b*c*(c + d*x)^(5/2))/(5*d^4) + (2*b*(c + d*x)^(7/2))/( 7*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 = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {4 \sqrt {d x +c}\, \left (-\frac {x \left (\frac {3 b \,x^{2}}{7}+a \right ) d^{3}}{2}+c \left (\frac {9 b \,x^{2}}{35}+a \right ) d^{2}-\frac {12 b \,c^{2} d x}{35}+\frac {24 b \,c^{3}}{35}\right )}{3 d^{4}}\) | \(55\) |
gosper | \(-\frac {2 \sqrt {d x +c}\, \left (-15 b \,d^{3} x^{3}+18 b c \,d^{2} x^{2}-35 a x \,d^{3}-24 b \,c^{2} d x +70 a \,d^{2} c +48 b \,c^{3}\right )}{105 d^{4}}\) | \(61\) |
trager | \(-\frac {2 \sqrt {d x +c}\, \left (-15 b \,d^{3} x^{3}+18 b c \,d^{2} x^{2}-35 a x \,d^{3}-24 b \,c^{2} d x +70 a \,d^{2} c +48 b \,c^{3}\right )}{105 d^{4}}\) | \(61\) |
risch | \(-\frac {2 \sqrt {d x +c}\, \left (-15 b \,d^{3} x^{3}+18 b c \,d^{2} x^{2}-35 a x \,d^{3}-24 b \,c^{2} d x +70 a \,d^{2} c +48 b \,c^{3}\right )}{105 d^{4}}\) | \(61\) |
orering | \(-\frac {2 \sqrt {d x +c}\, \left (-15 b \,d^{3} x^{3}+18 b c \,d^{2} x^{2}-35 a x \,d^{3}-24 b \,c^{2} d x +70 a \,d^{2} c +48 b \,c^{3}\right )}{105 d^{4}}\) | \(61\) |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {6 b c \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a \,d^{2}+3 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}-2 c \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{d^{4}}\) | \(70\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {6 b c \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a \,d^{2}+3 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}-2 c \left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d x +c}}{d^{4}}\) | \(70\) |
Input:
int(x*(b*x^2+a)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-4/3*(d*x+c)^(1/2)*(-1/2*x*(3/7*b*x^2+a)*d^3+c*(9/35*b*x^2+a)*d^2-12/35*b* c^2*d*x+24/35*b*c^3)/d^4
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.68 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (15 \, b d^{3} x^{3} - 18 \, b c d^{2} x^{2} - 48 \, b c^{3} - 70 \, a c d^{2} + {\left (24 \, b c^{2} d + 35 \, a d^{3}\right )} x\right )} \sqrt {d x + c}}{105 \, d^{4}} \] Input:
integrate(x*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
2/105*(15*b*d^3*x^3 - 18*b*c*d^2*x^2 - 48*b*c^3 - 70*a*c*d^2 + (24*b*c^2*d + 35*a*d^3)*x)*sqrt(d*x + c)/d^4
Time = 0.58 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.29 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (- \frac {3 b c \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {b \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a d^{2} + 3 b c^{2}\right )}{3 d^{2}} + \frac {\sqrt {c + d x} \left (- a c d^{2} - b c^{3}\right )}{d^{2}}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\begin {cases} \frac {a x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{2}}{4 b} & \text {otherwise} \end {cases}}{\sqrt {c}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(b*x**2+a)/(d*x+c)**(1/2),x)
Output:
Piecewise((2*(-3*b*c*(c + d*x)**(5/2)/(5*d**2) + b*(c + d*x)**(7/2)/(7*d** 2) + (c + d*x)**(3/2)*(a*d**2 + 3*b*c**2)/(3*d**2) + sqrt(c + d*x)*(-a*c*d **2 - b*c**3)/d**2)/d**2, Ne(d, 0)), (Piecewise((a*x**2/2, Eq(b, 0)), ((a + b*x**2)**2/(4*b), True))/sqrt(c), True))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (15 \, {\left (d x + c\right )}^{\frac {7}{2}} b - 63 \, {\left (d x + c\right )}^{\frac {5}{2}} b c + 35 \, {\left (3 \, b c^{2} + a d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 105 \, {\left (b c^{3} + a c d^{2}\right )} \sqrt {d x + c}\right )}}{105 \, d^{4}} \] Input:
integrate(x*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
2/105*(15*(d*x + c)^(7/2)*b - 63*(d*x + c)^(5/2)*b*c + 35*(3*b*c^2 + a*d^2 )*(d*x + c)^(3/2) - 105*(b*c^3 + a*c*d^2)*sqrt(d*x + c))/d^4
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (\frac {35 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a}{d} + \frac {3 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b}{d^{3}}\right )}}{105 \, d} \] Input:
integrate(x*(b*x^2+a)/(d*x+c)^(1/2),x, algorithm="giac")
Output:
2/105*(35*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a/d + 3*(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 /d^3)/d
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2\,b\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4}-\frac {\left (2\,b\,c^3+2\,a\,c\,d^2\right )\,\sqrt {c+d\,x}}{d^4}+\frac {\left (6\,b\,c^2+2\,a\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}-\frac {6\,b\,c\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4} \] Input:
int((x*(a + b*x^2))/(c + d*x)^(1/2),x)
Output:
(2*b*(c + d*x)^(7/2))/(7*d^4) - ((2*b*c^3 + 2*a*c*d^2)*(c + d*x)^(1/2))/d^ 4 + ((2*a*d^2 + 6*b*c^2)*(c + d*x)^(3/2))/(3*d^4) - (6*b*c*(c + d*x)^(5/2) )/(5*d^4)
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int \frac {x \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {d x +c}\, \left (15 b \,d^{3} x^{3}-18 b c \,d^{2} x^{2}+35 a \,d^{3} x +24 b \,c^{2} d x -70 a c \,d^{2}-48 b \,c^{3}\right )}{105 d^{4}} \] Input:
int(x*(b*x^2+a)/(d*x+c)^(1/2),x)
Output:
(2*sqrt(c + d*x)*( - 70*a*c*d**2 + 35*a*d**3*x - 48*b*c**3 + 24*b*c**2*d*x - 18*b*c*d**2*x**2 + 15*b*d**3*x**3))/(105*d**4)