Integrand size = 17, antiderivative size = 71 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 (b c-a d)^2 (c+d x)^{3/2}}{3 d^3}-\frac {4 b (b c-a d) (c+d x)^{5/2}}{5 d^3}+\frac {2 b^2 (c+d x)^{7/2}}{7 d^3} \] Output:
2/3*(-a*d+b*c)^2*(d*x+c)^(3/2)/d^3-4/5*b*(-a*d+b*c)*(d*x+c)^(5/2)/d^3+2/7* b^2*(d*x+c)^(7/2)/d^3
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 (c+d x)^{3/2} \left (35 a^2 d^2+14 a b d (-2 c+3 d x)+b^2 \left (8 c^2-12 c d x+15 d^2 x^2\right )\right )}{105 d^3} \] Input:
Integrate[(a + b*x)^2*Sqrt[c + d*x],x]
Output:
(2*(c + d*x)^(3/2)*(35*a^2*d^2 + 14*a*b*d*(-2*c + 3*d*x) + b^2*(8*c^2 - 12 *c*d*x + 15*d^2*x^2)))/(105*d^3)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {53, 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.
\(\displaystyle \int (a+b x)^2 \sqrt {c+d x} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {2 b (c+d x)^{3/2} (b c-a d)}{d^2}+\frac {\sqrt {c+d x} (a d-b c)^2}{d^2}+\frac {b^2 (c+d x)^{5/2}}{d^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 b (c+d x)^{5/2} (b c-a d)}{5 d^3}+\frac {2 (c+d x)^{3/2} (b c-a d)^2}{3 d^3}+\frac {2 b^2 (c+d x)^{7/2}}{7 d^3}\) |
Input:
Int[(a + b*x)^2*Sqrt[c + d*x],x]
Output:
(2*(b*c - a*d)^2*(c + d*x)^(3/2))/(3*d^3) - (4*b*(b*c - a*d)*(c + d*x)^(5/ 2))/(5*d^3) + (2*b^2*(c + d*x)^(7/2))/(7*d^3)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {3}{7} b^{2} x^{2}+\frac {6}{5} a b x +a^{2}\right ) d^{2}-\frac {4 c \left (\frac {3 b x}{7}+a \right ) b d}{5}+\frac {8 b^{2} c^{2}}{35}\right ) \left (x d +c \right )^{\frac {3}{2}}}{3 d^{3}}\) | \(54\) |
derivativedivides | \(\frac {\frac {2 b^{2} \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {4 b \left (a d -b c \right ) \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{2}}}{3}}{d^{3}}\) | \(56\) |
default | \(\frac {\frac {2 b^{2} \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {4 b \left (a d -b c \right ) \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{2}}}{3}}{d^{3}}\) | \(56\) |
gosper | \(\frac {2 \left (x d +c \right )^{\frac {3}{2}} \left (15 d^{2} x^{2} b^{2}+42 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-28 a b c d +8 b^{2} c^{2}\right )}{105 d^{3}}\) | \(63\) |
orering | \(\frac {2 \left (x d +c \right )^{\frac {3}{2}} \left (15 d^{2} x^{2} b^{2}+42 x a b \,d^{2}-12 x \,b^{2} c d +35 a^{2} d^{2}-28 a b c d +8 b^{2} c^{2}\right )}{105 d^{3}}\) | \(63\) |
trager | \(\frac {2 \left (15 b^{2} d^{3} x^{3}+42 a b \,d^{3} x^{2}+3 b^{2} c \,d^{2} x^{2}+35 a^{2} d^{3} x +14 a b c \,d^{2} x -4 b^{2} c^{2} d x +35 a^{2} c \,d^{2}-28 a b \,c^{2} d +8 b^{2} c^{3}\right ) \sqrt {x d +c}}{105 d^{3}}\) | \(100\) |
risch | \(\frac {2 \left (15 b^{2} d^{3} x^{3}+42 a b \,d^{3} x^{2}+3 b^{2} c \,d^{2} x^{2}+35 a^{2} d^{3} x +14 a b c \,d^{2} x -4 b^{2} c^{2} d x +35 a^{2} c \,d^{2}-28 a b \,c^{2} d +8 b^{2} c^{3}\right ) \sqrt {x d +c}}{105 d^{3}}\) | \(100\) |
Input:
int((b*x+a)^2*(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*((3/7*b^2*x^2+6/5*a*b*x+a^2)*d^2-4/5*c*(3/7*b*x+a)*b*d+8/35*b^2*c^2)*( d*x+c)^(3/2)/d^3
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 \, {\left (15 \, b^{2} d^{3} x^{3} + 8 \, b^{2} c^{3} - 28 \, a b c^{2} d + 35 \, a^{2} c d^{2} + 3 \, {\left (b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{2} - {\left (4 \, b^{2} c^{2} d - 14 \, a b c d^{2} - 35 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{105 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(1/2),x, algorithm="fricas")
Output:
2/105*(15*b^2*d^3*x^3 + 8*b^2*c^3 - 28*a*b*c^2*d + 35*a^2*c*d^2 + 3*(b^2*c *d^2 + 14*a*b*d^3)*x^2 - (4*b^2*c^2*d - 14*a*b*c*d^2 - 35*a^2*d^3)*x)*sqrt (d*x + c)/d^3
Time = 0.59 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \cdot \left (2 a b d - 2 b^{2} c\right )}{5 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{3 d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*(d*x+c)**(1/2),x)
Output:
Piecewise((2*(b**2*(c + d*x)**(7/2)/(7*d**2) + (c + d*x)**(5/2)*(2*a*b*d - 2*b**2*c)/(5*d**2) + (c + d*x)**(3/2)*(a**2*d**2 - 2*a*b*c*d + b**2*c**2) /(3*d**2))/d, Ne(d, 0)), (sqrt(c)*Piecewise((a**2*x, Eq(b, 0)), ((a + b*x) **3/(3*b), True)), True))
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 \, {\left (15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{2} - 42 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {5}{2}} + 35 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{105 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(1/2),x, algorithm="maxima")
Output:
2/105*(15*(d*x + c)^(7/2)*b^2 - 42*(b^2*c - a*b*d)*(d*x + c)^(5/2) + 35*(b ^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x + c)^(3/2))/d^3
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (59) = 118\).
Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.82 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 \, {\left (105 \, \sqrt {d x + c} a^{2} c + 35 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{2} + \frac {70 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b c}{d} + \frac {7 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b^{2} c}{d^{2}} + \frac {14 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b}{d} + \frac {3 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b^{2}}{d^{2}}\right )}}{105 \, d} \] Input:
integrate((b*x+a)^2*(d*x+c)^(1/2),x, algorithm="giac")
Output:
2/105*(105*sqrt(d*x + c)*a^2*c + 35*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)* a^2 + 70*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b*c/d + 7*(3*(d*x + c)^(5 /2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b^2*c/d^2 + 14*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a*b/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^2/d^2)/d
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2\,{\left (c+d\,x\right )}^{3/2}\,\left (15\,b^2\,{\left (c+d\,x\right )}^2+35\,a^2\,d^2+35\,b^2\,c^2-42\,b^2\,c\,\left (c+d\,x\right )+42\,a\,b\,d\,\left (c+d\,x\right )-70\,a\,b\,c\,d\right )}{105\,d^3} \] Input:
int((a + b*x)^2*(c + d*x)^(1/2),x)
Output:
(2*(c + d*x)^(3/2)*(15*b^2*(c + d*x)^2 + 35*a^2*d^2 + 35*b^2*c^2 - 42*b^2* c*(c + d*x) + 42*a*b*d*(c + d*x) - 70*a*b*c*d))/(105*d^3)
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.38 \[ \int (a+b x)^2 \sqrt {c+d x} \, dx=\frac {2 \sqrt {d x +c}\, \left (15 b^{2} d^{3} x^{3}+42 a b \,d^{3} x^{2}+3 b^{2} c \,d^{2} x^{2}+35 a^{2} d^{3} x +14 a b c \,d^{2} x -4 b^{2} c^{2} d x +35 a^{2} c \,d^{2}-28 a b \,c^{2} d +8 b^{2} c^{3}\right )}{105 d^{3}} \] Input:
int((b*x+a)^2*(d*x+c)^(1/2),x)
Output:
(2*sqrt(c + d*x)*(35*a**2*c*d**2 + 35*a**2*d**3*x - 28*a*b*c**2*d + 14*a*b *c*d**2*x + 42*a*b*d**3*x**2 + 8*b**2*c**3 - 4*b**2*c**2*d*x + 3*b**2*c*d* *2*x**2 + 15*b**2*d**3*x**3))/(105*d**3)