Integrand size = 17, antiderivative size = 71 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2 (b c-a d)^2 (c+d x)^{7/2}}{7 d^3}-\frac {4 b (b c-a d) (c+d x)^{9/2}}{9 d^3}+\frac {2 b^2 (c+d x)^{11/2}}{11 d^3} \] Output:
2/7*(-a*d+b*c)^2*(d*x+c)^(7/2)/d^3-4/9*b*(-a*d+b*c)*(d*x+c)^(9/2)/d^3+2/11 *b^2*(d*x+c)^(11/2)/d^3
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2 (c+d x)^{7/2} \left (99 a^2 d^2+22 a b d (-2 c+7 d x)+b^2 \left (8 c^2-28 c d x+63 d^2 x^2\right )\right )}{693 d^3} \] Input:
Integrate[(a + b*x)^2*(c + d*x)^(5/2),x]
Output:
(2*(c + d*x)^(7/2)*(99*a^2*d^2 + 22*a*b*d*(-2*c + 7*d*x) + b^2*(8*c^2 - 28 *c*d*x + 63*d^2*x^2)))/(693*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.
\(\displaystyle \int (a+b x)^2 (c+d x)^{5/2} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {2 b (c+d x)^{7/2} (b c-a d)}{d^2}+\frac {(c+d x)^{5/2} (a d-b c)^2}{d^2}+\frac {b^2 (c+d x)^{9/2}}{d^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 b (c+d x)^{9/2} (b c-a d)}{9 d^3}+\frac {2 (c+d x)^{7/2} (b c-a d)^2}{7 d^3}+\frac {2 b^2 (c+d x)^{11/2}}{11 d^3}\) |
Input:
Int[(a + b*x)^2*(c + d*x)^(5/2),x]
Output:
(2*(b*c - a*d)^2*(c + d*x)^(7/2))/(7*d^3) - (4*b*(b*c - a*d)*(c + d*x)^(9/ 2))/(9*d^3) + (2*b^2*(c + d*x)^(11/2))/(11*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 = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {2 \left (x d +c \right )^{\frac {7}{2}} \left (\left (\frac {7}{11} b^{2} x^{2}+\frac {14}{9} a b x +a^{2}\right ) d^{2}-\frac {4 c \left (\frac {7 b x}{11}+a \right ) b d}{9}+\frac {8 b^{2} c^{2}}{99}\right )}{7 d^{3}}\) | \(54\) |
derivativedivides | \(\frac {\frac {2 b^{2} \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {4 b \left (a d -b c \right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {7}{2}}}{7}}{d^{3}}\) | \(56\) |
default | \(\frac {\frac {2 b^{2} \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {4 b \left (a d -b c \right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {7}{2}}}{7}}{d^{3}}\) | \(56\) |
gosper | \(\frac {2 \left (x d +c \right )^{\frac {7}{2}} \left (63 d^{2} x^{2} b^{2}+154 x a b \,d^{2}-28 x \,b^{2} c d +99 a^{2} d^{2}-44 a b c d +8 b^{2} c^{2}\right )}{693 d^{3}}\) | \(63\) |
orering | \(\frac {2 \left (x d +c \right )^{\frac {7}{2}} \left (63 d^{2} x^{2} b^{2}+154 x a b \,d^{2}-28 x \,b^{2} c d +99 a^{2} d^{2}-44 a b c d +8 b^{2} c^{2}\right )}{693 d^{3}}\) | \(63\) |
trager | \(\frac {2 \left (63 b^{2} d^{5} x^{5}+154 a b \,d^{5} x^{4}+161 b^{2} c \,d^{4} x^{4}+99 a^{2} d^{5} x^{3}+418 a b c \,d^{4} x^{3}+113 b^{2} c^{2} d^{3} x^{3}+297 a^{2} c \,d^{4} x^{2}+330 a b \,c^{2} d^{3} x^{2}+3 b^{2} c^{3} d^{2} x^{2}+297 a^{2} c^{2} d^{3} x +22 a b \,c^{3} d^{2} x -4 b^{2} c^{4} d x +99 a^{2} c^{3} d^{2}-44 a b \,c^{4} d +8 b^{2} c^{5}\right ) \sqrt {x d +c}}{693 d^{3}}\) | \(182\) |
risch | \(\frac {2 \left (63 b^{2} d^{5} x^{5}+154 a b \,d^{5} x^{4}+161 b^{2} c \,d^{4} x^{4}+99 a^{2} d^{5} x^{3}+418 a b c \,d^{4} x^{3}+113 b^{2} c^{2} d^{3} x^{3}+297 a^{2} c \,d^{4} x^{2}+330 a b \,c^{2} d^{3} x^{2}+3 b^{2} c^{3} d^{2} x^{2}+297 a^{2} c^{2} d^{3} x +22 a b \,c^{3} d^{2} x -4 b^{2} c^{4} d x +99 a^{2} c^{3} d^{2}-44 a b \,c^{4} d +8 b^{2} c^{5}\right ) \sqrt {x d +c}}{693 d^{3}}\) | \(182\) |
Input:
int((b*x+a)^2*(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/7*(d*x+c)^(7/2)*((7/11*b^2*x^2+14/9*a*b*x+a^2)*d^2-4/9*c*(7/11*b*x+a)*b* d+8/99*b^2*c^2)/d^3
Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (59) = 118\).
Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.45 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2 \, {\left (63 \, b^{2} d^{5} x^{5} + 8 \, b^{2} c^{5} - 44 \, a b c^{4} d + 99 \, a^{2} c^{3} d^{2} + 7 \, {\left (23 \, b^{2} c d^{4} + 22 \, a b d^{5}\right )} x^{4} + {\left (113 \, b^{2} c^{2} d^{3} + 418 \, a b c d^{4} + 99 \, a^{2} d^{5}\right )} x^{3} + 3 \, {\left (b^{2} c^{3} d^{2} + 110 \, a b c^{2} d^{3} + 99 \, a^{2} c d^{4}\right )} x^{2} - {\left (4 \, b^{2} c^{4} d - 22 \, a b c^{3} d^{2} - 297 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{693 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(5/2),x, algorithm="fricas")
Output:
2/693*(63*b^2*d^5*x^5 + 8*b^2*c^5 - 44*a*b*c^4*d + 99*a^2*c^3*d^2 + 7*(23* b^2*c*d^4 + 22*a*b*d^5)*x^4 + (113*b^2*c^2*d^3 + 418*a*b*c*d^4 + 99*a^2*d^ 5)*x^3 + 3*(b^2*c^3*d^2 + 110*a*b*c^2*d^3 + 99*a^2*c*d^4)*x^2 - (4*b^2*c^4 *d - 22*a*b*c^3*d^2 - 297*a^2*c^2*d^3)*x)*sqrt(d*x + c)/d^3
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (65) = 130\).
Time = 0.33 (sec) , antiderivative size = 355, normalized size of antiderivative = 5.00 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\begin {cases} \frac {2 a^{2} c^{3} \sqrt {c + d x}}{7 d} + \frac {6 a^{2} c^{2} x \sqrt {c + d x}}{7} + \frac {6 a^{2} c d x^{2} \sqrt {c + d x}}{7} + \frac {2 a^{2} d^{2} x^{3} \sqrt {c + d x}}{7} - \frac {8 a b c^{4} \sqrt {c + d x}}{63 d^{2}} + \frac {4 a b c^{3} x \sqrt {c + d x}}{63 d} + \frac {20 a b c^{2} x^{2} \sqrt {c + d x}}{21} + \frac {76 a b c d x^{3} \sqrt {c + d x}}{63} + \frac {4 a b d^{2} x^{4} \sqrt {c + d x}}{9} + \frac {16 b^{2} c^{5} \sqrt {c + d x}}{693 d^{3}} - \frac {8 b^{2} c^{4} x \sqrt {c + d x}}{693 d^{2}} + \frac {2 b^{2} c^{3} x^{2} \sqrt {c + d x}}{231 d} + \frac {226 b^{2} c^{2} x^{3} \sqrt {c + d x}}{693} + \frac {46 b^{2} c d x^{4} \sqrt {c + d x}}{99} + \frac {2 b^{2} d^{2} x^{5} \sqrt {c + d x}}{11} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*(d*x+c)**(5/2),x)
Output:
Piecewise((2*a**2*c**3*sqrt(c + d*x)/(7*d) + 6*a**2*c**2*x*sqrt(c + d*x)/7 + 6*a**2*c*d*x**2*sqrt(c + d*x)/7 + 2*a**2*d**2*x**3*sqrt(c + d*x)/7 - 8* a*b*c**4*sqrt(c + d*x)/(63*d**2) + 4*a*b*c**3*x*sqrt(c + d*x)/(63*d) + 20* a*b*c**2*x**2*sqrt(c + d*x)/21 + 76*a*b*c*d*x**3*sqrt(c + d*x)/63 + 4*a*b* d**2*x**4*sqrt(c + d*x)/9 + 16*b**2*c**5*sqrt(c + d*x)/(693*d**3) - 8*b**2 *c**4*x*sqrt(c + d*x)/(693*d**2) + 2*b**2*c**3*x**2*sqrt(c + d*x)/(231*d) + 226*b**2*c**2*x**3*sqrt(c + d*x)/693 + 46*b**2*c*d*x**4*sqrt(c + d*x)/99 + 2*b**2*d**2*x**5*sqrt(c + d*x)/11, Ne(d, 0)), (c**(5/2)*(a**2*x + a*b*x **2 + b**2*x**3/3), True))
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2 \, {\left (63 \, {\left (d x + c\right )}^{\frac {11}{2}} b^{2} - 154 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {9}{2}} + 99 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}}\right )}}{693 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(5/2),x, algorithm="maxima")
Output:
2/693*(63*(d*x + c)^(11/2)*b^2 - 154*(b^2*c - a*b*d)*(d*x + c)^(9/2) + 99* (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x + c)^(7/2))/d^3
Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (59) = 118\).
Time = 0.13 (sec) , antiderivative size = 558, normalized size of antiderivative = 7.86 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)^2*(d*x+c)^(5/2),x, algorithm="giac")
Output:
2/3465*(3465*sqrt(d*x + c)*a^2*c^3 + 3465*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^2*c^2 + 2310*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b*c^3/d + 693 *(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2*c + 231*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b^2 *c^3/d^2 + 1386*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a*b*c^2/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)*a^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)*b^2*c^2 /d^2 + 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*c/d + 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)*b^2*c/d^2 + 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 + 315*sqrt(d*x + c)*c^4 )*a*b/d + 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^2/d^2)/d
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2\,{\left (c+d\,x\right )}^{7/2}\,\left (63\,b^2\,{\left (c+d\,x\right )}^2+99\,a^2\,d^2+99\,b^2\,c^2-154\,b^2\,c\,\left (c+d\,x\right )+154\,a\,b\,d\,\left (c+d\,x\right )-198\,a\,b\,c\,d\right )}{693\,d^3} \] Input:
int((a + b*x)^2*(c + d*x)^(5/2),x)
Output:
(2*(c + d*x)^(7/2)*(63*b^2*(c + d*x)^2 + 99*a^2*d^2 + 99*b^2*c^2 - 154*b^2 *c*(c + d*x) + 154*a*b*d*(c + d*x) - 198*a*b*c*d))/(693*d^3)
Time = 0.17 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.54 \[ \int (a+b x)^2 (c+d x)^{5/2} \, dx=\frac {2 \sqrt {d x +c}\, \left (63 b^{2} d^{5} x^{5}+154 a b \,d^{5} x^{4}+161 b^{2} c \,d^{4} x^{4}+99 a^{2} d^{5} x^{3}+418 a b c \,d^{4} x^{3}+113 b^{2} c^{2} d^{3} x^{3}+297 a^{2} c \,d^{4} x^{2}+330 a b \,c^{2} d^{3} x^{2}+3 b^{2} c^{3} d^{2} x^{2}+297 a^{2} c^{2} d^{3} x +22 a b \,c^{3} d^{2} x -4 b^{2} c^{4} d x +99 a^{2} c^{3} d^{2}-44 a b \,c^{4} d +8 b^{2} c^{5}\right )}{693 d^{3}} \] Input:
int((b*x+a)^2*(d*x+c)^(5/2),x)
Output:
(2*sqrt(c + d*x)*(99*a**2*c**3*d**2 + 297*a**2*c**2*d**3*x + 297*a**2*c*d* *4*x**2 + 99*a**2*d**5*x**3 - 44*a*b*c**4*d + 22*a*b*c**3*d**2*x + 330*a*b *c**2*d**3*x**2 + 418*a*b*c*d**4*x**3 + 154*a*b*d**5*x**4 + 8*b**2*c**5 - 4*b**2*c**4*d*x + 3*b**2*c**3*d**2*x**2 + 113*b**2*c**2*d**3*x**3 + 161*b* *2*c*d**4*x**4 + 63*b**2*d**5*x**5))/(693*d**3)