Integrand size = 20, antiderivative size = 83 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{7/2}}{7 e^3}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{9/2}}{9 e^3}+\frac {2 b B (d+e x)^{11/2}}{11 e^3} \] Output:
2/7*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(7/2)/e^3-2/9*(-A*b*e-B*a*e+2*B*b*d)*(e* x+d)^(9/2)/e^3+2/11*b*B*(e*x+d)^(11/2)/e^3
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2 (d+e x)^{7/2} \left (11 A b e (-2 d+7 e x)+11 a e (-2 B d+9 A e+7 B e x)+b B \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3} \] Input:
Integrate[(a + b*x)*(A + B*x)*(d + e*x)^(5/2),x]
Output:
(2*(d + e*x)^(7/2)*(11*A*b*e*(-2*d + 7*e*x) + 11*a*e*(-2*B*d + 9*A*e + 7*B *e*x) + b*B*(8*d^2 - 28*d*e*x + 63*e^2*x^2)))/(693*e^3)
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {86, 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.
\(\displaystyle \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(d+e x)^{7/2} (a B e+A b e-2 b B d)}{e^2}+\frac {(d+e x)^{5/2} (a e-b d) (A e-B d)}{e^2}+\frac {b B (d+e x)^{9/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (d+e x)^{9/2} (-a B e-A b e+2 b B d)}{9 e^3}+\frac {2 (d+e x)^{7/2} (b d-a e) (B d-A e)}{7 e^3}+\frac {2 b B (d+e x)^{11/2}}{11 e^3}\) |
Input:
Int[(a + b*x)*(A + B*x)*(d + e*x)^(5/2),x]
Output:
(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(7/2))/(7*e^3) - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(9/2))/(9*e^3) + (2*b*B*(d + e*x)^(11/2))/(11*e^3)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.67 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (\left (\frac {7 x \left (\frac {9 b x}{11}+a \right ) B}{9}+A \left (\frac {7 b x}{9}+a \right )\right ) e^{2}-\frac {2 \left (\left (\frac {14 b x}{11}+a \right ) B +A b \right ) d e}{9}+\frac {8 b B \,d^{2}}{99}\right )}{7 e^{3}}\) | \(60\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 b B \,x^{2} e^{2}+77 A x b \,e^{2}+77 B x a \,e^{2}-28 B x b d e +99 A a \,e^{2}-22 A b d e -22 B a d e +8 b B \,d^{2}\right )}{693 e^{3}}\) | \(73\) |
derivativedivides | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a e -d b \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\) | \(73\) |
default | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a e -d b \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\) | \(73\) |
orering | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 b B \,x^{2} e^{2}+77 A x b \,e^{2}+77 B x a \,e^{2}-28 B x b d e +99 A a \,e^{2}-22 A b d e -22 B a d e +8 b B \,d^{2}\right )}{693 e^{3}}\) | \(73\) |
trager | \(\frac {2 \left (63 b B \,e^{5} x^{5}+77 A b \,e^{5} x^{4}+77 B a \,e^{5} x^{4}+161 B b d \,e^{4} x^{4}+99 A a \,e^{5} x^{3}+209 A b d \,e^{4} x^{3}+209 B a d \,e^{4} x^{3}+113 B b \,d^{2} e^{3} x^{3}+297 A a d \,e^{4} x^{2}+165 A b \,d^{2} e^{3} x^{2}+165 B a \,d^{2} e^{3} x^{2}+3 B b \,d^{3} e^{2} x^{2}+297 A a \,d^{2} e^{3} x +11 A b \,d^{3} e^{2} x +11 B a \,d^{3} e^{2} x -4 B b \,d^{4} e x +99 A a \,d^{3} e^{2}-22 A b \,d^{4} e -22 B a \,d^{4} e +8 B b \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\) | \(225\) |
risch | \(\frac {2 \left (63 b B \,e^{5} x^{5}+77 A b \,e^{5} x^{4}+77 B a \,e^{5} x^{4}+161 B b d \,e^{4} x^{4}+99 A a \,e^{5} x^{3}+209 A b d \,e^{4} x^{3}+209 B a d \,e^{4} x^{3}+113 B b \,d^{2} e^{3} x^{3}+297 A a d \,e^{4} x^{2}+165 A b \,d^{2} e^{3} x^{2}+165 B a \,d^{2} e^{3} x^{2}+3 B b \,d^{3} e^{2} x^{2}+297 A a \,d^{2} e^{3} x +11 A b \,d^{3} e^{2} x +11 B a \,d^{3} e^{2} x -4 B b \,d^{4} e x +99 A a \,d^{3} e^{2}-22 A b \,d^{4} e -22 B a \,d^{4} e +8 B b \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\) | \(225\) |
Input:
int((b*x+a)*(B*x+A)*(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/7*(e*x+d)^(7/2)*((7/9*x*(9/11*b*x+a)*B+A*(7/9*b*x+a))*e^2-2/9*((14/11*b* x+a)*B+A*b)*d*e+8/99*b*B*d^2)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (71) = 142\).
Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.28 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2 \, {\left (63 \, B b e^{5} x^{5} + 8 \, B b d^{5} + 99 \, A a d^{3} e^{2} - 22 \, {\left (B a + A b\right )} d^{4} e + 7 \, {\left (23 \, B b d e^{4} + 11 \, {\left (B a + A b\right )} e^{5}\right )} x^{4} + {\left (113 \, B b d^{2} e^{3} + 99 \, A a e^{5} + 209 \, {\left (B a + A b\right )} d e^{4}\right )} x^{3} + 3 \, {\left (B b d^{3} e^{2} + 99 \, A a d e^{4} + 55 \, {\left (B a + A b\right )} d^{2} e^{3}\right )} x^{2} - {\left (4 \, B b d^{4} e - 297 \, A a d^{2} e^{3} - 11 \, {\left (B a + A b\right )} d^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/693*(63*B*b*e^5*x^5 + 8*B*b*d^5 + 99*A*a*d^3*e^2 - 22*(B*a + A*b)*d^4*e + 7*(23*B*b*d*e^4 + 11*(B*a + A*b)*e^5)*x^4 + (113*B*b*d^2*e^3 + 99*A*a*e^ 5 + 209*(B*a + A*b)*d*e^4)*x^3 + 3*(B*b*d^3*e^2 + 99*A*a*d*e^4 + 55*(B*a + A*b)*d^2*e^3)*x^2 - (4*B*b*d^4*e - 297*A*a*d^2*e^3 - 11*(B*a + A*b)*d^3*e ^2)*x)*sqrt(e*x + d)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (87) = 174\).
Time = 0.48 (sec) , antiderivative size = 476, normalized size of antiderivative = 5.73 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\begin {cases} \frac {2 A a d^{3} \sqrt {d + e x}}{7 e} + \frac {6 A a d^{2} x \sqrt {d + e x}}{7} + \frac {6 A a d e x^{2} \sqrt {d + e x}}{7} + \frac {2 A a e^{2} x^{3} \sqrt {d + e x}}{7} - \frac {4 A b d^{4} \sqrt {d + e x}}{63 e^{2}} + \frac {2 A b d^{3} x \sqrt {d + e x}}{63 e} + \frac {10 A b d^{2} x^{2} \sqrt {d + e x}}{21} + \frac {38 A b d e x^{3} \sqrt {d + e x}}{63} + \frac {2 A b e^{2} x^{4} \sqrt {d + e x}}{9} - \frac {4 B a d^{4} \sqrt {d + e x}}{63 e^{2}} + \frac {2 B a d^{3} x \sqrt {d + e x}}{63 e} + \frac {10 B a d^{2} x^{2} \sqrt {d + e x}}{21} + \frac {38 B a d e x^{3} \sqrt {d + e x}}{63} + \frac {2 B a e^{2} x^{4} \sqrt {d + e x}}{9} + \frac {16 B b d^{5} \sqrt {d + e x}}{693 e^{3}} - \frac {8 B b d^{4} x \sqrt {d + e x}}{693 e^{2}} + \frac {2 B b d^{3} x^{2} \sqrt {d + e x}}{231 e} + \frac {226 B b d^{2} x^{3} \sqrt {d + e x}}{693} + \frac {46 B b d e x^{4} \sqrt {d + e x}}{99} + \frac {2 B b e^{2} x^{5} \sqrt {d + e x}}{11} & \text {for}\: e \neq 0 \\d^{\frac {5}{2}} \left (A a x + \frac {A b x^{2}}{2} + \frac {B a x^{2}}{2} + \frac {B b x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)**(5/2),x)
Output:
Piecewise((2*A*a*d**3*sqrt(d + e*x)/(7*e) + 6*A*a*d**2*x*sqrt(d + e*x)/7 + 6*A*a*d*e*x**2*sqrt(d + e*x)/7 + 2*A*a*e**2*x**3*sqrt(d + e*x)/7 - 4*A*b* d**4*sqrt(d + e*x)/(63*e**2) + 2*A*b*d**3*x*sqrt(d + e*x)/(63*e) + 10*A*b* d**2*x**2*sqrt(d + e*x)/21 + 38*A*b*d*e*x**3*sqrt(d + e*x)/63 + 2*A*b*e**2 *x**4*sqrt(d + e*x)/9 - 4*B*a*d**4*sqrt(d + e*x)/(63*e**2) + 2*B*a*d**3*x* sqrt(d + e*x)/(63*e) + 10*B*a*d**2*x**2*sqrt(d + e*x)/21 + 38*B*a*d*e*x**3 *sqrt(d + e*x)/63 + 2*B*a*e**2*x**4*sqrt(d + e*x)/9 + 16*B*b*d**5*sqrt(d + e*x)/(693*e**3) - 8*B*b*d**4*x*sqrt(d + e*x)/(693*e**2) + 2*B*b*d**3*x**2 *sqrt(d + e*x)/(231*e) + 226*B*b*d**2*x**3*sqrt(d + e*x)/693 + 46*B*b*d*e* x**4*sqrt(d + e*x)/99 + 2*B*b*e**2*x**5*sqrt(d + e*x)/11, Ne(e, 0)), (d**( 5/2)*(A*a*x + A*b*x**2/2 + B*a*x**2/2 + B*b*x**3/3), True))
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} B b - 77 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} {\left (e x + d\right )}^{\frac {9}{2}} + 99 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} {\left (e x + d\right )}^{\frac {7}{2}}\right )}}{693 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(5/2),x, algorithm="maxima")
Output:
2/693*(63*(e*x + d)^(11/2)*B*b - 77*(2*B*b*d - (B*a + A*b)*e)*(e*x + d)^(9 /2) + 99*(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e)*(e*x + d)^(7/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (71) = 142\).
Time = 0.13 (sec) , antiderivative size = 735, normalized size of antiderivative = 8.86 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(5/2),x, algorithm="giac")
Output:
2/3465*(3465*sqrt(e*x + d)*A*a*d^3 + 3465*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*d^2 + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d^3/e + 115 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*d^3/e + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*a*d + 231*(3*(e*x + d)^( 5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d^3/e^2 + 693*(3*( e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*d^2/e + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b* d^2/e + 99*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)* d^2 - 35*sqrt(e*x + d)*d^3)*A*a + 297*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5 /2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*B*b*d^2/e^2 + 297*( 5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqr t(e*x + d)*d^3)*B*a*d/e + 297*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*A*b*d/e + 33*(35*(e*x + d)^ (9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3 /2)*d^3 + 315*sqrt(e*x + d)*d^4)*B*b*d/e^2 + 11*(35*(e*x + d)^(9/2) - 180* (e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 31 5*sqrt(e*x + d)*d^4)*B*a/e + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)* d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)* d^4)*A*b/e + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d )^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2)*d^4 - 693...
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{7/2}\,\left (63\,B\,b\,{\left (d+e\,x\right )}^2+99\,A\,a\,e^2+99\,B\,b\,d^2+77\,A\,b\,e\,\left (d+e\,x\right )+77\,B\,a\,e\,\left (d+e\,x\right )-154\,B\,b\,d\,\left (d+e\,x\right )-99\,A\,b\,d\,e-99\,B\,a\,d\,e\right )}{693\,e^3} \] Input:
int((A + B*x)*(a + b*x)*(d + e*x)^(5/2),x)
Output:
(2*(d + e*x)^(7/2)*(63*B*b*(d + e*x)^2 + 99*A*a*e^2 + 99*B*b*d^2 + 77*A*b* e*(d + e*x) + 77*B*a*e*(d + e*x) - 154*B*b*d*(d + e*x) - 99*A*b*d*e - 99*B *a*d*e))/(693*e^3)
Time = 0.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.17 \[ \int (a+b x) (A+B x) (d+e x)^{5/2} \, dx=\frac {2 \sqrt {e x +d}\, \left (63 b^{2} e^{5} x^{5}+154 a b \,e^{5} x^{4}+161 b^{2} d \,e^{4} x^{4}+99 a^{2} e^{5} x^{3}+418 a b d \,e^{4} x^{3}+113 b^{2} d^{2} e^{3} x^{3}+297 a^{2} d \,e^{4} x^{2}+330 a b \,d^{2} e^{3} x^{2}+3 b^{2} d^{3} e^{2} x^{2}+297 a^{2} d^{2} e^{3} x +22 a b \,d^{3} e^{2} x -4 b^{2} d^{4} e x +99 a^{2} d^{3} e^{2}-44 a b \,d^{4} e +8 b^{2} d^{5}\right )}{693 e^{3}} \] Input:
int((b*x+a)*(B*x+A)*(e*x+d)^(5/2),x)
Output:
(2*sqrt(d + e*x)*(99*a**2*d**3*e**2 + 297*a**2*d**2*e**3*x + 297*a**2*d*e* *4*x**2 + 99*a**2*e**5*x**3 - 44*a*b*d**4*e + 22*a*b*d**3*e**2*x + 330*a*b *d**2*e**3*x**2 + 418*a*b*d*e**4*x**3 + 154*a*b*e**5*x**4 + 8*b**2*d**5 - 4*b**2*d**4*e*x + 3*b**2*d**3*e**2*x**2 + 113*b**2*d**2*e**3*x**3 + 161*b* *2*d*e**4*x**4 + 63*b**2*e**5*x**5))/(693*e**3)