Integrand size = 25, antiderivative size = 116 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (a+b x)^{3/2}}{3 b^4}+\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right ) (a+b x)^{5/2}}{5 b^4}+\frac {2 (b C-3 a D) (a+b x)^{7/2}}{7 b^4}+\frac {2 D (a+b x)^{9/2}}{9 b^4} \] Output:
2/3*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*(b*x+a)^(3/2)/b^4+2/5*(B*b^2-2*C*a*b+3*D *a^2)*(b*x+a)^(5/2)/b^4+2/7*(C*b-3*D*a)*(b*x+a)^(7/2)/b^4+2/9*D*(b*x+a)^(9 /2)/b^4
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 (a+b x)^{3/2} \left (105 A b^3-16 a^3 D+24 a^2 b (C+D x)-6 a b^2 (7 B+x (6 C+5 D x))+b^3 x (63 B+5 x (9 C+7 D x))\right )}{315 b^4} \] Input:
Integrate[Sqrt[a + b*x]*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*(a + b*x)^(3/2)*(105*A*b^3 - 16*a^3*D + 24*a^2*b*(C + D*x) - 6*a*b^2*(7 *B + x*(6*C + 5*D*x)) + b^3*x*(63*B + 5*x*(9*C + 7*D*x))))/(315*b^4)
Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2389, 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.
\(\displaystyle \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {\sqrt {a+b x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{b^3}+\frac {(a+b x)^{3/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^3}+\frac {(a+b x)^{5/2} (b C-3 a D)}{b^3}+\frac {D (a+b x)^{7/2}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^4}+\frac {2 (a+b x)^{5/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{5 b^4}+\frac {2 (a+b x)^{7/2} (b C-3 a D)}{7 b^4}+\frac {2 D (a+b x)^{9/2}}{9 b^4}\) |
Input:
Int[Sqrt[a + b*x]*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*(a + b*x)^(3/2))/(3*b^4) + (2*(b^2* B - 2*a*b*C + 3*a^2*D)*(a + b*x)^(5/2))/(5*b^4) + (2*(b*C - 3*a*D)*(a + b* x)^(7/2))/(7*b^4) + (2*D*(a + b*x)^(9/2))/(9*b^4)
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {1}{3} D x^{3}+\frac {3}{7} C \,x^{2}+\frac {3}{5} B x +A \right ) b^{3}-\frac {2 \left (\frac {5}{7} D x^{2}+\frac {6}{7} C x +B \right ) a \,b^{2}}{5}+\frac {8 a^{2} \left (D x +C \right ) b}{35}-\frac {16 a^{3} D}{105}\right ) \left (b x +a \right )^{\frac {3}{2}}}{3 b^{4}}\) | \(71\) |
gosper | \(\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (35 D x^{3} b^{3}+45 C \,x^{2} b^{3}-30 D x^{2} a \,b^{2}+63 b^{3} B x -36 C x a \,b^{2}+24 D x \,a^{2} b +105 b^{3} A -42 a \,b^{2} B +24 a^{2} b C -16 a^{3} D\right )}{315 b^{4}}\) | \(91\) |
orering | \(\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (35 D x^{3} b^{3}+45 C \,x^{2} b^{3}-30 D x^{2} a \,b^{2}+63 b^{3} B x -36 C x a \,b^{2}+24 D x \,a^{2} b +105 b^{3} A -42 a \,b^{2} B +24 a^{2} b C -16 a^{3} D\right )}{315 b^{4}}\) | \(91\) |
derivativedivides | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (C b -3 D a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (B \,b^{2}-2 C a b +3 D a^{2}\right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{4}}\) | \(94\) |
default | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (C b -3 D a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (B \,b^{2}-2 C a b +3 D a^{2}\right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{4}}\) | \(94\) |
trager | \(\frac {2 \left (35 b^{4} D x^{4}+45 b^{4} C \,x^{3}+5 a \,b^{3} D x^{3}+63 B \,b^{4} x^{2}+9 C a \,b^{3} x^{2}-6 D a^{2} b^{2} x^{2}+105 A \,b^{4} x +21 B a \,b^{3} x -12 a^{2} b^{2} C x +8 a^{3} b D x +105 a \,b^{3} A -42 a^{2} b^{2} B +24 a^{3} b C -16 a^{4} D\right ) \sqrt {b x +a}}{315 b^{4}}\) | \(135\) |
Input:
int((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
2/3*((1/3*D*x^3+3/7*C*x^2+3/5*B*x+A)*b^3-2/5*(5/7*D*x^2+6/7*C*x+B)*a*b^2+8 /35*a^2*(D*x+C)*b-16/105*a^3*D)*(b*x+a)^(3/2)/b^4
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (35 \, D b^{4} x^{4} - 16 \, D a^{4} + 24 \, C a^{3} b - 42 \, B a^{2} b^{2} + 105 \, A a b^{3} + 5 \, {\left (D a b^{3} + 9 \, C b^{4}\right )} x^{3} - 3 \, {\left (2 \, D a^{2} b^{2} - 3 \, C a b^{3} - 21 \, B b^{4}\right )} x^{2} + {\left (8 \, D a^{3} b - 12 \, C a^{2} b^{2} + 21 \, B a b^{3} + 105 \, A b^{4}\right )} x\right )} \sqrt {b x + a}}{315 \, b^{4}} \] Input:
integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
2/315*(35*D*b^4*x^4 - 16*D*a^4 + 24*C*a^3*b - 42*B*a^2*b^2 + 105*A*a*b^3 + 5*(D*a*b^3 + 9*C*b^4)*x^3 - 3*(2*D*a^2*b^2 - 3*C*a*b^3 - 21*B*b^4)*x^2 + (8*D*a^3*b - 12*C*a^2*b^2 + 21*B*a*b^3 + 105*A*b^4)*x)*sqrt(b*x + a)/b^4
Time = 0.79 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.23 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \frac {2 \left (\frac {D \left (a + b x\right )^{\frac {9}{2}}}{9 b^{3}} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (C b - 3 D a\right )}{7 b^{3}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (B b^{2} - 2 C a b + 3 D a^{2}\right )}{5 b^{3}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b^{3} - B a b^{2} + C a^{2} b - D a^{3}\right )}{3 b^{3}}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**(1/2)*(D*x**3+C*x**2+B*x+A),x)
Output:
Piecewise((2*(D*(a + b*x)**(9/2)/(9*b**3) + (a + b*x)**(7/2)*(C*b - 3*D*a) /(7*b**3) + (a + b*x)**(5/2)*(B*b**2 - 2*C*a*b + 3*D*a**2)/(5*b**3) + (a + b*x)**(3/2)*(A*b**3 - B*a*b**2 + C*a**2*b - D*a**3)/(3*b**3))/b, Ne(b, 0) ), (sqrt(a)*(A*x + B*x**2/2 + C*x**3/3 + D*x**4/4), True))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} D - 45 \, {\left (3 \, D a - C b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 63 \, {\left (3 \, D a^{2} - 2 \, C a b + B b^{2}\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 105 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} {\left (b x + a\right )}^{\frac {3}{2}}\right )}}{315 \, b^{4}} \] Input:
integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
2/315*(35*(b*x + a)^(9/2)*D - 45*(3*D*a - C*b)*(b*x + a)^(7/2) + 63*(3*D*a ^2 - 2*C*a*b + B*b^2)*(b*x + a)^(5/2) - 105*(D*a^3 - C*a^2*b + B*a*b^2 - A *b^3)*(b*x + a)^(3/2))/b^4
Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (101) = 202\).
Time = 0.12 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.60 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {b x + a} A a + 105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A + \frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} B a}{b} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} C a}{b^{2}} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B}{b} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} D a}{b^{3}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} C}{b^{2}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} D}{b^{3}}\right )}}{315 \, b} \] Input:
integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
2/315*(315*sqrt(b*x + a)*A*a + 105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A + 105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*B*a/b + 21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*C*a/b^2 + 21*(3*(b*x + a)^ (5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*B/b + 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)*D*a/b^3 + 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/b^2 + (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/b^3)/b
Time = 3.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.16 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {70\,{\left (a+b\,x\right )}^{9/2}\,D-210\,a^3\,{\left (a+b\,x\right )}^{3/2}\,D+378\,a^2\,{\left (a+b\,x\right )}^{5/2}\,D-270\,a\,{\left (a+b\,x\right )}^{7/2}\,D}{315\,b^4}+\frac {30\,C\,{\left (a+b\,x\right )}^{7/2}-84\,C\,a\,{\left (a+b\,x\right )}^{5/2}+70\,C\,a^2\,{\left (a+b\,x\right )}^{3/2}}{105\,b^3}+\frac {6\,B\,{\left (a+b\,x\right )}^{5/2}-10\,B\,a\,{\left (a+b\,x\right )}^{3/2}}{15\,b^2}+\frac {2\,A\,{\left (a+b\,x\right )}^{3/2}}{3\,b} \] Input:
int((a + b*x)^(1/2)*(A + B*x + C*x^2 + x^3*D),x)
Output:
(70*(a + b*x)^(9/2)*D - 210*a^3*(a + b*x)^(3/2)*D + 378*a^2*(a + b*x)^(5/2 )*D - 270*a*(a + b*x)^(7/2)*D)/(315*b^4) + (30*C*(a + b*x)^(7/2) - 84*C*a* (a + b*x)^(5/2) + 70*C*a^2*(a + b*x)^(3/2))/(105*b^3) + (6*B*(a + b*x)^(5/ 2) - 10*B*a*(a + b*x)^(3/2))/(15*b^2) + (2*A*(a + b*x)^(3/2))/(3*b)
Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00 \[ \int \sqrt {a+b x} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {b x +a}\, \left (35 b^{4} d \,x^{4}+5 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}-6 a^{2} b^{2} d \,x^{2}+9 a \,b^{3} c \,x^{2}+63 b^{5} x^{2}+8 a^{3} b d x -12 a^{2} b^{2} c x +126 a \,b^{4} x -16 a^{4} d +24 a^{3} b c +63 a^{2} b^{3}\right )}{315 b^{4}} \] Input:
int((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x)
Output:
(2*sqrt(a + b*x)*( - 16*a**4*d + 24*a**3*b*c + 8*a**3*b*d*x + 63*a**2*b**3 - 12*a**2*b**2*c*x - 6*a**2*b**2*d*x**2 + 126*a*b**4*x + 9*a*b**3*c*x**2 + 5*a*b**3*d*x**3 + 63*b**5*x**2 + 45*b**4*c*x**3 + 35*b**4*d*x**4))/(315* b**4)