Integrand size = 20, antiderivative size = 83 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{9/2}}{9 e^3}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{11/2}}{11 e^3}+\frac {2 b B (d+e x)^{13/2}}{13 e^3} \]
2/9*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(9/2)/e^3-2/11*(-A*b*e-B*a*e+2*B*b*d)*(e *x+d)^(11/2)/e^3+2/13*b*B*(e*x+d)^(13/2)/e^3
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\frac {2 (d+e x)^{9/2} \left (13 A b e (-2 d+9 e x)+13 a e (-2 B d+11 A e+9 B e x)+b B \left (8 d^2-36 d e x+99 e^2 x^2\right )\right )}{1287 e^3} \]
(2*(d + e*x)^(9/2)*(13*A*b*e*(-2*d + 9*e*x) + 13*a*e*(-2*B*d + 11*A*e + 9* B*e*x) + b*B*(8*d^2 - 36*d*e*x + 99*e^2*x^2)))/(1287*e^3)
Time = 0.22 (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.
\(\displaystyle \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(d+e x)^{9/2} (a B e+A b e-2 b B d)}{e^2}+\frac {(d+e x)^{7/2} (a e-b d) (A e-B d)}{e^2}+\frac {b B (d+e x)^{11/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (d+e x)^{11/2} (-a B e-A b e+2 b B d)}{11 e^3}+\frac {2 (d+e x)^{9/2} (b d-a e) (B d-A e)}{9 e^3}+\frac {2 b B (d+e x)^{13/2}}{13 e^3}\) |
(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(9/2))/(9*e^3) - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(11/2))/(11*e^3) + (2*b*B*(d + e*x)^(13/2))/(13*e^3)
3.18.18.3.1 Defintions of rubi rules used
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 = 0.96 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\left (\frac {9}{13} b \,x^{2}+\frac {9}{11} a x \right ) B +A \left (\frac {9 b x}{11}+a \right )\right ) e^{2}-\frac {2 d \left (\left (\frac {18 b x}{13}+a \right ) B +A b \right ) e}{11}+\frac {8 b B \,d^{2}}{143}\right ) \left (e x +d \right )^{\frac {9}{2}}}{9 e^{3}}\) | \(63\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {9}{2}} \left (99 b B \,x^{2} e^{2}+117 A b \,e^{2} x +117 B a \,e^{2} x -36 B b d e x +143 A a \,e^{2}-26 A b d e -26 B a d e +8 b B \,d^{2}\right )}{1287 e^{3}}\) | \(73\) |
derivativedivides | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {13}{2}}}{13}+\frac {2 \left (\left (a e -b d \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a e -b d \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}}{e^{3}}\) | \(73\) |
default | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {13}{2}}}{13}+\frac {2 \left (\left (a e -b d \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (a e -b d \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}}{e^{3}}\) | \(73\) |
trager | \(\frac {2 \left (99 B b \,e^{6} x^{6}+117 A b \,e^{6} x^{5}+117 B a \,e^{6} x^{5}+360 B b d \,e^{5} x^{5}+143 A a \,e^{6} x^{4}+442 A b d \,e^{5} x^{4}+442 B a d \,e^{5} x^{4}+458 B b \,d^{2} e^{4} x^{4}+572 A a d \,e^{5} x^{3}+598 A b \,d^{2} e^{4} x^{3}+598 B a \,d^{2} e^{4} x^{3}+212 B b \,d^{3} e^{3} x^{3}+858 A a \,d^{2} e^{4} x^{2}+312 A b \,d^{3} e^{3} x^{2}+312 B a \,d^{3} e^{3} x^{2}+3 B b \,d^{4} e^{2} x^{2}+572 A a \,d^{3} e^{3} x +13 A b \,d^{4} e^{2} x +13 B a \,d^{4} e^{2} x -4 B b \,d^{5} e x +143 A a \,d^{4} e^{2}-26 A b \,d^{5} e -26 B a \,d^{5} e +8 B b \,d^{6}\right ) \sqrt {e x +d}}{1287 e^{3}}\) | \(277\) |
risch | \(\frac {2 \left (99 B b \,e^{6} x^{6}+117 A b \,e^{6} x^{5}+117 B a \,e^{6} x^{5}+360 B b d \,e^{5} x^{5}+143 A a \,e^{6} x^{4}+442 A b d \,e^{5} x^{4}+442 B a d \,e^{5} x^{4}+458 B b \,d^{2} e^{4} x^{4}+572 A a d \,e^{5} x^{3}+598 A b \,d^{2} e^{4} x^{3}+598 B a \,d^{2} e^{4} x^{3}+212 B b \,d^{3} e^{3} x^{3}+858 A a \,d^{2} e^{4} x^{2}+312 A b \,d^{3} e^{3} x^{2}+312 B a \,d^{3} e^{3} x^{2}+3 B b \,d^{4} e^{2} x^{2}+572 A a \,d^{3} e^{3} x +13 A b \,d^{4} e^{2} x +13 B a \,d^{4} e^{2} x -4 B b \,d^{5} e x +143 A a \,d^{4} e^{2}-26 A b \,d^{5} e -26 B a \,d^{5} e +8 B b \,d^{6}\right ) \sqrt {e x +d}}{1287 e^{3}}\) | \(277\) |
2/9*(((9/13*b*x^2+9/11*a*x)*B+A*(9/11*b*x+a))*e^2-2/11*d*((18/13*b*x+a)*B+ A*b)*e+8/143*b*B*d^2)*(e*x+d)^(9/2)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (71) = 142\).
Time = 0.23 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.77 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\frac {2 \, {\left (99 \, B b e^{6} x^{6} + 8 \, B b d^{6} + 143 \, A a d^{4} e^{2} - 26 \, {\left (B a + A b\right )} d^{5} e + 9 \, {\left (40 \, B b d e^{5} + 13 \, {\left (B a + A b\right )} e^{6}\right )} x^{5} + {\left (458 \, B b d^{2} e^{4} + 143 \, A a e^{6} + 442 \, {\left (B a + A b\right )} d e^{5}\right )} x^{4} + 2 \, {\left (106 \, B b d^{3} e^{3} + 286 \, A a d e^{5} + 299 \, {\left (B a + A b\right )} d^{2} e^{4}\right )} x^{3} + 3 \, {\left (B b d^{4} e^{2} + 286 \, A a d^{2} e^{4} + 104 \, {\left (B a + A b\right )} d^{3} e^{3}\right )} x^{2} - {\left (4 \, B b d^{5} e - 572 \, A a d^{3} e^{3} - 13 \, {\left (B a + A b\right )} d^{4} e^{2}\right )} x\right )} \sqrt {e x + d}}{1287 \, e^{3}} \]
2/1287*(99*B*b*e^6*x^6 + 8*B*b*d^6 + 143*A*a*d^4*e^2 - 26*(B*a + A*b)*d^5* e + 9*(40*B*b*d*e^5 + 13*(B*a + A*b)*e^6)*x^5 + (458*B*b*d^2*e^4 + 143*A*a *e^6 + 442*(B*a + A*b)*d*e^5)*x^4 + 2*(106*B*b*d^3*e^3 + 286*A*a*d*e^5 + 2 99*(B*a + A*b)*d^2*e^4)*x^3 + 3*(B*b*d^4*e^2 + 286*A*a*d^2*e^4 + 104*(B*a + A*b)*d^3*e^3)*x^2 - (4*B*b*d^5*e - 572*A*a*d^3*e^3 - 13*(B*a + A*b)*d^4* e^2)*x)*sqrt(e*x + d)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 578 vs. \(2 (87) = 174\).
Time = 0.58 (sec) , antiderivative size = 578, normalized size of antiderivative = 6.96 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\begin {cases} \frac {2 A a d^{4} \sqrt {d + e x}}{9 e} + \frac {8 A a d^{3} x \sqrt {d + e x}}{9} + \frac {4 A a d^{2} e x^{2} \sqrt {d + e x}}{3} + \frac {8 A a d e^{2} x^{3} \sqrt {d + e x}}{9} + \frac {2 A a e^{3} x^{4} \sqrt {d + e x}}{9} - \frac {4 A b d^{5} \sqrt {d + e x}}{99 e^{2}} + \frac {2 A b d^{4} x \sqrt {d + e x}}{99 e} + \frac {16 A b d^{3} x^{2} \sqrt {d + e x}}{33} + \frac {92 A b d^{2} e x^{3} \sqrt {d + e x}}{99} + \frac {68 A b d e^{2} x^{4} \sqrt {d + e x}}{99} + \frac {2 A b e^{3} x^{5} \sqrt {d + e x}}{11} - \frac {4 B a d^{5} \sqrt {d + e x}}{99 e^{2}} + \frac {2 B a d^{4} x \sqrt {d + e x}}{99 e} + \frac {16 B a d^{3} x^{2} \sqrt {d + e x}}{33} + \frac {92 B a d^{2} e x^{3} \sqrt {d + e x}}{99} + \frac {68 B a d e^{2} x^{4} \sqrt {d + e x}}{99} + \frac {2 B a e^{3} x^{5} \sqrt {d + e x}}{11} + \frac {16 B b d^{6} \sqrt {d + e x}}{1287 e^{3}} - \frac {8 B b d^{5} x \sqrt {d + e x}}{1287 e^{2}} + \frac {2 B b d^{4} x^{2} \sqrt {d + e x}}{429 e} + \frac {424 B b d^{3} x^{3} \sqrt {d + e x}}{1287} + \frac {916 B b d^{2} e x^{4} \sqrt {d + e x}}{1287} + \frac {80 B b d e^{2} x^{5} \sqrt {d + e x}}{143} + \frac {2 B b e^{3} x^{6} \sqrt {d + e x}}{13} & \text {for}\: e \neq 0 \\d^{\frac {7}{2}} \left (A a x + \frac {A b x^{2}}{2} + \frac {B a x^{2}}{2} + \frac {B b x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*A*a*d**4*sqrt(d + e*x)/(9*e) + 8*A*a*d**3*x*sqrt(d + e*x)/9 + 4*A*a*d**2*e*x**2*sqrt(d + e*x)/3 + 8*A*a*d*e**2*x**3*sqrt(d + e*x)/9 + 2 *A*a*e**3*x**4*sqrt(d + e*x)/9 - 4*A*b*d**5*sqrt(d + e*x)/(99*e**2) + 2*A* b*d**4*x*sqrt(d + e*x)/(99*e) + 16*A*b*d**3*x**2*sqrt(d + e*x)/33 + 92*A*b *d**2*e*x**3*sqrt(d + e*x)/99 + 68*A*b*d*e**2*x**4*sqrt(d + e*x)/99 + 2*A* b*e**3*x**5*sqrt(d + e*x)/11 - 4*B*a*d**5*sqrt(d + e*x)/(99*e**2) + 2*B*a* d**4*x*sqrt(d + e*x)/(99*e) + 16*B*a*d**3*x**2*sqrt(d + e*x)/33 + 92*B*a*d **2*e*x**3*sqrt(d + e*x)/99 + 68*B*a*d*e**2*x**4*sqrt(d + e*x)/99 + 2*B*a* e**3*x**5*sqrt(d + e*x)/11 + 16*B*b*d**6*sqrt(d + e*x)/(1287*e**3) - 8*B*b *d**5*x*sqrt(d + e*x)/(1287*e**2) + 2*B*b*d**4*x**2*sqrt(d + e*x)/(429*e) + 424*B*b*d**3*x**3*sqrt(d + e*x)/1287 + 916*B*b*d**2*e*x**4*sqrt(d + e*x) /1287 + 80*B*b*d*e**2*x**5*sqrt(d + e*x)/143 + 2*B*b*e**3*x**6*sqrt(d + e* x)/13, Ne(e, 0)), (d**(7/2)*(A*a*x + A*b*x**2/2 + B*a*x**2/2 + B*b*x**3/3) , True))
Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\frac {2 \, {\left (99 \, {\left (e x + d\right )}^{\frac {13}{2}} B b - 117 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} {\left (e x + d\right )}^{\frac {11}{2}} + 143 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} {\left (e x + d\right )}^{\frac {9}{2}}\right )}}{1287 \, e^{3}} \]
2/1287*(99*(e*x + d)^(13/2)*B*b - 117*(2*B*b*d - (B*a + A*b)*e)*(e*x + d)^ (11/2) + 143*(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e)*(e*x + d)^(9/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 1044 vs. \(2 (71) = 142\).
Time = 0.29 (sec) , antiderivative size = 1044, normalized size of antiderivative = 12.58 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\text {Too large to display} \]
2/45045*(45045*sqrt(e*x + d)*A*a*d^4 + 60060*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*d^3 + 15015*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d^4/e + 15015*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*d^4/e + 18018*(3*(e*x + d )^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*a*d^2 + 3003*(3*( e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d^4/e^2 + 12012*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)* B*a*d^3/e + 12012*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b*d^3/e + 5148*(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*d + 5148*(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^3/e^2 + 7722*(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*a*d^2/e + 7722*(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^2/e + 143*(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*a + 858*(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^2/e^2 + 572*(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*a*d/e + 572*(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 + ...
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int (a+b x) (A+B x) (d+e x)^{7/2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{9/2}\,\left (99\,B\,b\,{\left (d+e\,x\right )}^2+143\,A\,a\,e^2+143\,B\,b\,d^2+117\,A\,b\,e\,\left (d+e\,x\right )+117\,B\,a\,e\,\left (d+e\,x\right )-234\,B\,b\,d\,\left (d+e\,x\right )-143\,A\,b\,d\,e-143\,B\,a\,d\,e\right )}{1287\,e^3} \]