Integrand size = 17, antiderivative size = 100 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=-\frac {2 (b c-a d)^3 (c+d x)^{5/2}}{5 d^4}+\frac {6 b (b c-a d)^2 (c+d x)^{7/2}}{7 d^4}-\frac {2 b^2 (b c-a d) (c+d x)^{9/2}}{3 d^4}+\frac {2 b^3 (c+d x)^{11/2}}{11 d^4} \]
-2/5*(-a*d+b*c)^3*(d*x+c)^(5/2)/d^4+6/7*b*(-a*d+b*c)^2*(d*x+c)^(7/2)/d^4-2 /3*b^2*(-a*d+b*c)*(d*x+c)^(9/2)/d^4+2/11*b^3*(d*x+c)^(11/2)/d^4
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\frac {2 (c+d x)^{5/2} \left (231 a^3 d^3+99 a^2 b d^2 (-2 c+5 d x)+11 a b^2 d \left (8 c^2-20 c d x+35 d^2 x^2\right )+b^3 \left (-16 c^3+40 c^2 d x-70 c d^2 x^2+105 d^3 x^3\right )\right )}{1155 d^4} \]
(2*(c + d*x)^(5/2)*(231*a^3*d^3 + 99*a^2*b*d^2*(-2*c + 5*d*x) + 11*a*b^2*d *(8*c^2 - 20*c*d*x + 35*d^2*x^2) + b^3*(-16*c^3 + 40*c^2*d*x - 70*c*d^2*x^ 2 + 105*d^3*x^3)))/(1155*d^4)
Time = 0.22 (sec) , antiderivative size = 100, 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.
\(\displaystyle \int (a+b x)^3 (c+d x)^{3/2} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (c+d x)^{7/2} (b c-a d)}{d^3}+\frac {3 b (c+d x)^{5/2} (b c-a d)^2}{d^3}+\frac {(c+d x)^{3/2} (a d-b c)^3}{d^3}+\frac {b^3 (c+d x)^{9/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^2 (c+d x)^{9/2} (b c-a d)}{3 d^4}+\frac {6 b (c+d x)^{7/2} (b c-a d)^2}{7 d^4}-\frac {2 (c+d x)^{5/2} (b c-a d)^3}{5 d^4}+\frac {2 b^3 (c+d x)^{11/2}}{11 d^4}\) |
(-2*(b*c - a*d)^3*(c + d*x)^(5/2))/(5*d^4) + (6*b*(b*c - a*d)^2*(c + d*x)^ (7/2))/(7*d^4) - (2*b^2*(b*c - a*d)*(c + d*x)^(9/2))/(3*d^4) + (2*b^3*(c + d*x)^(11/2))/(11*d^4)
3.14.89.3.1 Defintions of rubi rules used
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 = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a d -b c \right ) b^{2} \left (d x +c \right )^{\frac {9}{2}}}{3}+\frac {6 \left (a d -b c \right )^{2} b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{4}}\) | \(78\) |
default | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a d -b c \right ) b^{2} \left (d x +c \right )^{\frac {9}{2}}}{3}+\frac {6 \left (a d -b c \right )^{2} b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{4}}\) | \(78\) |
pseudoelliptic | \(\frac {2 \left (\left (\frac {5}{11} b^{3} x^{3}+\frac {5}{3} a \,b^{2} x^{2}+\frac {15}{7} a^{2} b x +a^{3}\right ) d^{3}-\frac {6 \left (\frac {35}{99} b^{2} x^{2}+\frac {10}{9} a b x +a^{2}\right ) b c \,d^{2}}{7}+\frac {8 b^{2} c^{2} \left (\frac {5 b x}{11}+a \right ) d}{21}-\frac {16 b^{3} c^{3}}{231}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5 d^{4}}\) | \(93\) |
gosper | \(\frac {2 \left (d x +c \right )^{\frac {5}{2}} \left (105 d^{3} x^{3} b^{3}+385 x^{2} a \,b^{2} d^{3}-70 x^{2} b^{3} c \,d^{2}+495 x \,a^{2} b \,d^{3}-220 x a \,b^{2} c \,d^{2}+40 x \,b^{3} c^{2} d +231 a^{3} d^{3}-198 a^{2} b c \,d^{2}+88 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{1155 d^{4}}\) | \(116\) |
trager | \(\frac {2 \left (105 b^{3} d^{5} x^{5}+385 a \,b^{2} d^{5} x^{4}+140 b^{3} c \,d^{4} x^{4}+495 a^{2} b \,d^{5} x^{3}+550 a \,b^{2} c \,d^{4} x^{3}+5 b^{3} c^{2} d^{3} x^{3}+231 a^{3} d^{5} x^{2}+792 a^{2} b c \,d^{4} x^{2}+33 a \,b^{2} c^{2} d^{3} x^{2}-6 b^{3} c^{3} d^{2} x^{2}+462 a^{3} c \,d^{4} x +99 a^{2} b \,c^{2} d^{3} x -44 a \,b^{2} c^{3} d^{2} x +8 b^{3} c^{4} d x +231 a^{3} c^{2} d^{3}-198 a^{2} b \,c^{3} d^{2}+88 a \,b^{2} c^{4} d -16 b^{3} c^{5}\right ) \sqrt {d x +c}}{1155 d^{4}}\) | \(228\) |
risch | \(\frac {2 \left (105 b^{3} d^{5} x^{5}+385 a \,b^{2} d^{5} x^{4}+140 b^{3} c \,d^{4} x^{4}+495 a^{2} b \,d^{5} x^{3}+550 a \,b^{2} c \,d^{4} x^{3}+5 b^{3} c^{2} d^{3} x^{3}+231 a^{3} d^{5} x^{2}+792 a^{2} b c \,d^{4} x^{2}+33 a \,b^{2} c^{2} d^{3} x^{2}-6 b^{3} c^{3} d^{2} x^{2}+462 a^{3} c \,d^{4} x +99 a^{2} b \,c^{2} d^{3} x -44 a \,b^{2} c^{3} d^{2} x +8 b^{3} c^{4} d x +231 a^{3} c^{2} d^{3}-198 a^{2} b \,c^{3} d^{2}+88 a \,b^{2} c^{4} d -16 b^{3} c^{5}\right ) \sqrt {d x +c}}{1155 d^{4}}\) | \(228\) |
2/d^4*(1/11*b^3*(d*x+c)^(11/2)+1/3*(a*d-b*c)*b^2*(d*x+c)^(9/2)+3/7*(a*d-b* c)^2*b*(d*x+c)^(7/2)+1/5*(a*d-b*c)^3*(d*x+c)^(5/2))
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (84) = 168\).
Time = 0.23 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.16 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{3} d^{5} x^{5} - 16 \, b^{3} c^{5} + 88 \, a b^{2} c^{4} d - 198 \, a^{2} b c^{3} d^{2} + 231 \, a^{3} c^{2} d^{3} + 35 \, {\left (4 \, b^{3} c d^{4} + 11 \, a b^{2} d^{5}\right )} x^{4} + 5 \, {\left (b^{3} c^{2} d^{3} + 110 \, a b^{2} c d^{4} + 99 \, a^{2} b d^{5}\right )} x^{3} - 3 \, {\left (2 \, b^{3} c^{3} d^{2} - 11 \, a b^{2} c^{2} d^{3} - 264 \, a^{2} b c d^{4} - 77 \, a^{3} d^{5}\right )} x^{2} + {\left (8 \, b^{3} c^{4} d - 44 \, a b^{2} c^{3} d^{2} + 99 \, a^{2} b c^{2} d^{3} + 462 \, a^{3} c d^{4}\right )} x\right )} \sqrt {d x + c}}{1155 \, d^{4}} \]
2/1155*(105*b^3*d^5*x^5 - 16*b^3*c^5 + 88*a*b^2*c^4*d - 198*a^2*b*c^3*d^2 + 231*a^3*c^2*d^3 + 35*(4*b^3*c*d^4 + 11*a*b^2*d^5)*x^4 + 5*(b^3*c^2*d^3 + 110*a*b^2*c*d^4 + 99*a^2*b*d^5)*x^3 - 3*(2*b^3*c^3*d^2 - 11*a*b^2*c^2*d^3 - 264*a^2*b*c*d^4 - 77*a^3*d^5)*x^2 + (8*b^3*c^4*d - 44*a*b^2*c^3*d^2 + 9 9*a^2*b*c^2*d^3 + 462*a^3*c*d^4)*x)*sqrt(d*x + c)/d^4
Time = 0.83 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (c + d x\right )^{\frac {11}{2}}}{11 d^{3}} + \frac {\left (c + d x\right )^{\frac {9}{2}} \cdot \left (3 a b^{2} d - 3 b^{3} c\right )}{9 d^{3}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \cdot \left (3 a^{2} b d^{2} - 6 a b^{2} c d + 3 b^{3} c^{2}\right )}{7 d^{3}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{5 d^{3}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(b**3*(c + d*x)**(11/2)/(11*d**3) + (c + d*x)**(9/2)*(3*a*b** 2*d - 3*b**3*c)/(9*d**3) + (c + d*x)**(7/2)*(3*a**2*b*d**2 - 6*a*b**2*c*d + 3*b**3*c**2)/(7*d**3) + (c + d*x)**(5/2)*(a**3*d**3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(5*d**3))/d, Ne(d, 0)), (c**(3/2)*Piecewise(( a**3*x, Eq(b, 0)), ((a + b*x)**4/(4*b), True)), True))
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\frac {2 \, {\left (105 \, {\left (d x + c\right )}^{\frac {11}{2}} b^{3} - 385 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}^{\frac {9}{2}} + 495 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - 231 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {5}{2}}\right )}}{1155 \, d^{4}} \]
2/1155*(105*(d*x + c)^(11/2)*b^3 - 385*(b^3*c - a*b^2*d)*(d*x + c)^(9/2) + 495*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*(d*x + c)^(7/2) - 231*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(5/2))/d^4
Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (84) = 168\).
Time = 0.31 (sec) , antiderivative size = 566, normalized size of antiderivative = 5.66 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx =\text {Too large to display} \]
2/3465*(3465*sqrt(d*x + c)*a^3*c^2 + 2310*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^3*c + 3465*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^2*b*c^2/d + 231 *(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^3 + 6 93*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a*b^2 *c^2/d^2 + 1386*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2*b*c/d + 99*(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^3*c^2/d^3 + 594*(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) *a*b^2*c/d^2 + 297*(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)*a^2*b/d + 22*(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 + 3 15*sqrt(d*x + c)*c^4)*b^3*c/d^3 + 33*(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)*a*b^2/d^2 + 5*(63*(d*x + c)^(11/2) - 385*(d*x + c)^(9/2)*c + 990 *(d*x + c)^(7/2)*c^2 - 1386*(d*x + c)^(5/2)*c^3 + 1155*(d*x + c)^(3/2)*c^4 - 693*sqrt(d*x + c)*c^5)*b^3/d^3)/d
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\frac {2\,b^3\,{\left (c+d\,x\right )}^{11/2}}{11\,d^4}-\frac {\left (6\,b^3\,c-6\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{9/2}}{9\,d^4}+\frac {2\,{\left (a\,d-b\,c\right )}^3\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}+\frac {6\,b\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4} \]
(2*b^3*(c + d*x)^(11/2))/(11*d^4) - ((6*b^3*c - 6*a*b^2*d)*(c + d*x)^(9/2) )/(9*d^4) + (2*(a*d - b*c)^3*(c + d*x)^(5/2))/(5*d^4) + (6*b*(a*d - b*c)^2 *(c + d*x)^(7/2))/(7*d^4)
Time = 0.00 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.26 \[ \int (a+b x)^3 (c+d x)^{3/2} \, dx=\frac {2 \sqrt {d x +c}\, \left (105 b^{3} d^{5} x^{5}+385 a \,b^{2} d^{5} x^{4}+140 b^{3} c \,d^{4} x^{4}+495 a^{2} b \,d^{5} x^{3}+550 a \,b^{2} c \,d^{4} x^{3}+5 b^{3} c^{2} d^{3} x^{3}+231 a^{3} d^{5} x^{2}+792 a^{2} b c \,d^{4} x^{2}+33 a \,b^{2} c^{2} d^{3} x^{2}-6 b^{3} c^{3} d^{2} x^{2}+462 a^{3} c \,d^{4} x +99 a^{2} b \,c^{2} d^{3} x -44 a \,b^{2} c^{3} d^{2} x +8 b^{3} c^{4} d x +231 a^{3} c^{2} d^{3}-198 a^{2} b \,c^{3} d^{2}+88 a \,b^{2} c^{4} d -16 b^{3} c^{5}\right )}{1155 d^{4}} \]
int(sqrt(c + d*x)*(a**3*c + a**3*d*x + 3*a**2*b*c*x + 3*a**2*b*d*x**2 + 3* a*b**2*c*x**2 + 3*a*b**2*d*x**3 + b**3*c*x**3 + b**3*d*x**4),x)
(2*sqrt(c + d*x)*(231*a**3*c**2*d**3 + 462*a**3*c*d**4*x + 231*a**3*d**5*x **2 - 198*a**2*b*c**3*d**2 + 99*a**2*b*c**2*d**3*x + 792*a**2*b*c*d**4*x** 2 + 495*a**2*b*d**5*x**3 + 88*a*b**2*c**4*d - 44*a*b**2*c**3*d**2*x + 33*a *b**2*c**2*d**3*x**2 + 550*a*b**2*c*d**4*x**3 + 385*a*b**2*d**5*x**4 - 16* b**3*c**5 + 8*b**3*c**4*d*x - 6*b**3*c**3*d**2*x**2 + 5*b**3*c**2*d**3*x** 3 + 140*b**3*c*d**4*x**4 + 105*b**3*d**5*x**5))/(1155*d**4)