Integrand size = 22, antiderivative size = 88 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2}{5} a^2 A x^{5/2}+\frac {2 a (2 A b+a B) x^{\frac {5}{2}+n}}{5+2 n}+\frac {2 b (A b+2 a B) x^{\frac {5}{2}+2 n}}{5+4 n}+\frac {2 b^2 B x^{\frac {5}{2}+3 n}}{5+6 n} \] Output:
2/5*a^2*A*x^(5/2)+2*a*(2*A*b+B*a)*x^(5/2+n)/(5+2*n)+2*b*(A*b+2*B*a)*x^(5/2 +2*n)/(5+4*n)+2*b^2*B*x^(5/2+3*n)/(5+6*n)
Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=2 \left (\frac {1}{5} a^2 A x^{5/2}+\frac {a (2 A b+a B) x^{\frac {5}{2}+n}}{5+2 n}+\frac {b (A b+2 a B) x^{\frac {5}{2}+2 n}}{5+4 n}+\frac {b^2 B x^{\frac {5}{2}+3 n}}{5+6 n}\right ) \] Input:
Integrate[x^(3/2)*(a + b*x^n)^2*(A + B*x^n),x]
Output:
2*((a^2*A*x^(5/2))/5 + (a*(2*A*b + a*B)*x^(5/2 + n))/(5 + 2*n) + (b*(A*b + 2*a*B)*x^(5/2 + 2*n))/(5 + 4*n) + (b^2*B*x^(5/2 + 3*n))/(5 + 6*n))
Time = 0.38 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {950, 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.
\(\displaystyle \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (a^2 A x^{3/2}+a x^{n+\frac {3}{2}} (a B+2 A b)+b x^{2 n+\frac {3}{2}} (2 a B+A b)+b^2 B x^{3 n+\frac {3}{2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} a^2 A x^{5/2}+\frac {2 a x^{n+\frac {5}{2}} (a B+2 A b)}{2 n+5}+\frac {2 b x^{2 n+\frac {5}{2}} (2 a B+A b)}{4 n+5}+\frac {2 b^2 B x^{3 n+\frac {5}{2}}}{6 n+5}\) |
Input:
Int[x^(3/2)*(a + b*x^n)^2*(A + B*x^n),x]
Output:
(2*a^2*A*x^(5/2))/5 + (2*a*(2*A*b + a*B)*x^(5/2 + n))/(5 + 2*n) + (2*b*(A* b + 2*a*B)*x^(5/2 + 2*n))/(5 + 4*n) + (2*b^2*B*x^(5/2 + 3*n))/(5 + 6*n)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^ n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGt Q[p, 0] && IGtQ[q, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(620\) vs. \(2(78)=156\).
Time = 0.28 (sec) , antiderivative size = 621, normalized size of antiderivative = 7.06
method | result | size |
orering | \(\frac {8 \left (12 n^{3}+88 n^{2}+147 n +68\right ) x^{\frac {5}{2}} \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{5 \left (48 n^{3}+220 n^{2}+300 n +125\right )}-\frac {8 x^{2} \left (22 n^{2}+54 n +29\right ) \left (\frac {3 \sqrt {x}\, \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{2}+2 \sqrt {x}\, \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n +\sqrt {x}\, \left (a +b \,x^{n}\right )^{2} B \,x^{n} n \right )}{5 \left (48 n^{3}+220 n^{2}+300 n +125\right )}+\frac {32 x^{3} \left (2+3 n \right ) \left (\frac {3 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{4 \sqrt {x}}+\frac {4 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{\sqrt {x}}+\frac {2 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{\sqrt {x}}+\frac {2 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{\sqrt {x}}+\frac {4 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{\sqrt {x}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{\sqrt {x}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{\sqrt {x}}\right )}{5 \left (48 n^{3}+220 n^{2}+300 n +125\right )}-\frac {16 x^{4} \left (-\frac {\left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{2 x^{\frac {3}{2}}}-\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{4 x^{\frac {3}{2}}}-\frac {3 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{8 x^{\frac {3}{2}}}+\frac {3 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x^{\frac {3}{2}}}+\frac {6 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x^{\frac {3}{2}}}+\frac {3 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x^{\frac {3}{2}}}+\frac {3 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{2 x^{\frac {3}{2}}}+\frac {6 b^{2} x^{2 n} n^{3} \left (A +B \,x^{n}\right )}{x^{\frac {3}{2}}}+\frac {6 b^{2} x^{3 n} n^{3} B}{x^{\frac {3}{2}}}+\frac {12 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{3} b}{x^{\frac {3}{2}}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{3}}{x^{\frac {3}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{3}}{x^{\frac {3}{2}}}\right )}{5 \left (48 n^{3}+220 n^{2}+300 n +125\right )}\) | \(621\) |
Input:
int(x^(3/2)*(a+b*x^n)^2*(A+B*x^n),x,method=_RETURNVERBOSE)
Output:
8/5*(12*n^3+88*n^2+147*n+68)*x^(5/2)/(48*n^3+220*n^2+300*n+125)*(a+b*x^n)^ 2*(A+B*x^n)-8/5*x^2*(22*n^2+54*n+29)/(48*n^3+220*n^2+300*n+125)*(3/2*x^(1/ 2)*(a+b*x^n)^2*(A+B*x^n)+2*x^(1/2)*(a+b*x^n)*(A+B*x^n)*b*x^n*n+x^(1/2)*(a+ b*x^n)^2*B*x^n*n)+32/5*x^3*(2+3*n)/(48*n^3+220*n^2+300*n+125)*(3/4*(a+b*x^ n)^2*(A+B*x^n)/x^(1/2)+4/x^(1/2)*(a+b*x^n)*(A+B*x^n)*b*x^n*n+2/x^(1/2)*(a+ b*x^n)^2*B*x^n*n+2/x^(1/2)*b^2*(x^n)^2*n^2*(A+B*x^n)+4/x^(1/2)*(a+b*x^n)*B *(x^n)^2*n^2*b+2/x^(1/2)*(a+b*x^n)*(A+B*x^n)*b*x^n*n^2+1/x^(1/2)*(a+b*x^n) ^2*B*x^n*n^2)-16/5/(48*n^3+220*n^2+300*n+125)*x^4*(-1/2*(a+b*x^n)*(A+B*x^n )/x^(3/2)*b*x^n*n-1/4*(a+b*x^n)^2*B*x^n*n/x^(3/2)-3/8*(a+b*x^n)^2*(A+B*x^n )/x^(3/2)+3/x^(3/2)*b^2*(x^n)^2*n^2*(A+B*x^n)+6/x^(3/2)*(a+b*x^n)*B*(x^n)^ 2*n^2*b+3/x^(3/2)*(a+b*x^n)*(A+B*x^n)*b*x^n*n^2+3/2/x^(3/2)*(a+b*x^n)^2*B* x^n*n^2+6/x^(3/2)*b^2*(x^n)^2*n^3*(A+B*x^n)+6/x^(3/2)*b^2*(x^n)^3*n^3*B+12 /x^(3/2)*(a+b*x^n)*B*(x^n)^2*n^3*b+2/x^(3/2)*(a+b*x^n)*(A+B*x^n)*b*x^n*n^3 +1/x^(3/2)*(a+b*x^n)^2*B*x^n*n^3)
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (78) = 156\).
Time = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.17 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2 \, {\left (5 \, {\left (8 \, B b^{2} n^{2} + 30 \, B b^{2} n + 25 \, B b^{2}\right )} x^{\frac {5}{2}} x^{3 \, n} + 5 \, {\left (50 \, B a b + 25 \, A b^{2} + 12 \, {\left (2 \, B a b + A b^{2}\right )} n^{2} + 40 \, {\left (2 \, B a b + A b^{2}\right )} n\right )} x^{\frac {5}{2}} x^{2 \, n} + 5 \, {\left (25 \, B a^{2} + 50 \, A a b + 24 \, {\left (B a^{2} + 2 \, A a b\right )} n^{2} + 50 \, {\left (B a^{2} + 2 \, A a b\right )} n\right )} x^{\frac {5}{2}} x^{n} + {\left (48 \, A a^{2} n^{3} + 220 \, A a^{2} n^{2} + 300 \, A a^{2} n + 125 \, A a^{2}\right )} x^{\frac {5}{2}}\right )}}{5 \, {\left (48 \, n^{3} + 220 \, n^{2} + 300 \, n + 125\right )}} \] Input:
integrate(x^(3/2)*(a+b*x^n)^2*(A+B*x^n),x, algorithm="fricas")
Output:
2/5*(5*(8*B*b^2*n^2 + 30*B*b^2*n + 25*B*b^2)*x^(5/2)*x^(3*n) + 5*(50*B*a*b + 25*A*b^2 + 12*(2*B*a*b + A*b^2)*n^2 + 40*(2*B*a*b + A*b^2)*n)*x^(5/2)*x ^(2*n) + 5*(25*B*a^2 + 50*A*a*b + 24*(B*a^2 + 2*A*a*b)*n^2 + 50*(B*a^2 + 2 *A*a*b)*n)*x^(5/2)*x^n + (48*A*a^2*n^3 + 220*A*a^2*n^2 + 300*A*a^2*n + 125 *A*a^2)*x^(5/2))/(48*n^3 + 220*n^2 + 300*n + 125)
Leaf count of result is larger than twice the leaf count of optimal. 842 vs. \(2 (82) = 164\).
Time = 177.03 (sec) , antiderivative size = 842, normalized size of antiderivative = 9.57 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx =\text {Too large to display} \] Input:
integrate(x**(3/2)*(a+b*x**n)**2*(A+B*x**n),x)
Output:
Piecewise((2*A*a**2*x**(5/2)/5 + 2*A*a*b*log(x) - 2*A*b**2/(5*x**(5/2)) + B*a**2*log(x) - 4*B*a*b/(5*x**(5/2)) - B*b**2/(5*x**5), Eq(n, -5/2)), (2*A *a**2*x**(5/2)/5 + 8*A*a*b*x**(5/4)/5 + A*b**2*log(x) + 4*B*a**2*x**(5/4)/ 5 + 2*B*a*b*log(x) - 4*B*b**2/(5*x**(5/4)), Eq(n, -5/4)), (2*A*a**2*x**(5/ 2)/5 + 6*A*a*b*x**(5/3)/5 + 6*A*b**2*x**(5/6)/5 + 3*B*a**2*x**(5/3)/5 + 12 *B*a*b*x**(5/6)/5 + B*b**2*log(x), Eq(n, -5/6)), (96*A*a**2*n**3*x**(5/2)/ (240*n**3 + 1100*n**2 + 1500*n + 625) + 440*A*a**2*n**2*x**(5/2)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 600*A*a**2*n*x**(5/2)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 250*A*a**2*x**(5/2)/(240*n**3 + 1100*n**2 + 1500*n + 62 5) + 480*A*a*b*n**2*x**(5/2)*x**n/(240*n**3 + 1100*n**2 + 1500*n + 625) + 1000*A*a*b*n*x**(5/2)*x**n/(240*n**3 + 1100*n**2 + 1500*n + 625) + 500*A*a *b*x**(5/2)*x**n/(240*n**3 + 1100*n**2 + 1500*n + 625) + 120*A*b**2*n**2*x **(5/2)*x**(2*n)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 400*A*b**2*n*x**( 5/2)*x**(2*n)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 250*A*b**2*x**(5/2)* x**(2*n)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 240*B*a**2*n**2*x**(5/2)* x**n/(240*n**3 + 1100*n**2 + 1500*n + 625) + 500*B*a**2*n*x**(5/2)*x**n/(2 40*n**3 + 1100*n**2 + 1500*n + 625) + 250*B*a**2*x**(5/2)*x**n/(240*n**3 + 1100*n**2 + 1500*n + 625) + 240*B*a*b*n**2*x**(5/2)*x**(2*n)/(240*n**3 + 1100*n**2 + 1500*n + 625) + 800*B*a*b*n*x**(5/2)*x**(2*n)/(240*n**3 + 1100 *n**2 + 1500*n + 625) + 500*B*a*b*x**(5/2)*x**(2*n)/(240*n**3 + 1100*n*...
Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2}{5} \, A a^{2} x^{\frac {5}{2}} + \frac {2 \, B b^{2} x^{3 \, n + \frac {5}{2}}}{6 \, n + 5} + \frac {4 \, B a b x^{2 \, n + \frac {5}{2}}}{4 \, n + 5} + \frac {2 \, A b^{2} x^{2 \, n + \frac {5}{2}}}{4 \, n + 5} + \frac {2 \, B a^{2} x^{n + \frac {5}{2}}}{2 \, n + 5} + \frac {4 \, A a b x^{n + \frac {5}{2}}}{2 \, n + 5} \] Input:
integrate(x^(3/2)*(a+b*x^n)^2*(A+B*x^n),x, algorithm="maxima")
Output:
2/5*A*a^2*x^(5/2) + 2*B*b^2*x^(3*n + 5/2)/(6*n + 5) + 4*B*a*b*x^(2*n + 5/2 )/(4*n + 5) + 2*A*b^2*x^(2*n + 5/2)/(4*n + 5) + 2*B*a^2*x^(n + 5/2)/(2*n + 5) + 4*A*a*b*x^(n + 5/2)/(2*n + 5)
Time = 0.14 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2}{5} \, A a^{2} x^{\frac {5}{2}} + \frac {2 \, B b^{2} x^{\frac {5}{2}} \sqrt {x}^{6 \, n}}{6 \, n + 5} + \frac {4 \, B a b x^{\frac {5}{2}} \sqrt {x}^{4 \, n}}{4 \, n + 5} + \frac {2 \, A b^{2} x^{\frac {5}{2}} \sqrt {x}^{4 \, n}}{4 \, n + 5} + \frac {2 \, B a^{2} x^{\frac {5}{2}} \sqrt {x}^{2 \, n}}{2 \, n + 5} + \frac {4 \, A a b x^{\frac {5}{2}} \sqrt {x}^{2 \, n}}{2 \, n + 5} \] Input:
integrate(x^(3/2)*(a+b*x^n)^2*(A+B*x^n),x, algorithm="giac")
Output:
2/5*A*a^2*x^(5/2) + 2*B*b^2*x^(5/2)*sqrt(x)^(6*n)/(6*n + 5) + 4*B*a*b*x^(5 /2)*sqrt(x)^(4*n)/(4*n + 5) + 2*A*b^2*x^(5/2)*sqrt(x)^(4*n)/(4*n + 5) + 2* B*a^2*x^(5/2)*sqrt(x)^(2*n)/(2*n + 5) + 4*A*a*b*x^(5/2)*sqrt(x)^(2*n)/(2*n + 5)
Time = 4.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2\,A\,a^2\,x^{5/2}}{5}+\frac {x^n\,x^{5/2}\,\left (2\,B\,a^2+4\,A\,b\,a\right )}{2\,n+5}+\frac {x^{2\,n}\,x^{5/2}\,\left (2\,A\,b^2+4\,B\,a\,b\right )}{4\,n+5}+\frac {2\,B\,b^2\,x^{3\,n}\,x^{5/2}}{6\,n+5} \] Input:
int(x^(3/2)*(A + B*x^n)*(a + b*x^n)^2,x)
Output:
(2*A*a^2*x^(5/2))/5 + (x^n*x^(5/2)*(2*B*a^2 + 4*A*a*b))/(2*n + 5) + (x^(2* n)*x^(5/2)*(2*A*b^2 + 4*B*a*b))/(4*n + 5) + (2*B*b^2*x^(3*n)*x^(5/2))/(6*n + 5)
Time = 0.22 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int x^{3/2} \left (a+b x^n\right )^2 \left (A+B x^n\right ) \, dx=\frac {2 \sqrt {x}\, x^{2} \left (40 x^{3 n} b^{3} n^{2}+150 x^{3 n} b^{3} n +125 x^{3 n} b^{3}+180 x^{2 n} a \,b^{2} n^{2}+600 x^{2 n} a \,b^{2} n +375 x^{2 n} a \,b^{2}+360 x^{n} a^{2} b \,n^{2}+750 x^{n} a^{2} b n +375 x^{n} a^{2} b +48 a^{3} n^{3}+220 a^{3} n^{2}+300 a^{3} n +125 a^{3}\right )}{240 n^{3}+1100 n^{2}+1500 n +625} \] Input:
int(x^(3/2)*(a+b*x^n)^2*(A+B*x^n),x)
Output:
(2*sqrt(x)*x**2*(40*x**(3*n)*b**3*n**2 + 150*x**(3*n)*b**3*n + 125*x**(3*n )*b**3 + 180*x**(2*n)*a*b**2*n**2 + 600*x**(2*n)*a*b**2*n + 375*x**(2*n)*a *b**2 + 360*x**n*a**2*b*n**2 + 750*x**n*a**2*b*n + 375*x**n*a**2*b + 48*a* *3*n**3 + 220*a**3*n**2 + 300*a**3*n + 125*a**3))/(5*(48*n**3 + 220*n**2 + 300*n + 125))