Integrand size = 22, antiderivative size = 88 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=-\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a (2 A b+a B) x^{-\frac {5}{2}+n}}{5-2 n}-\frac {2 b (A b+2 a B) x^{-\frac {5}{2}+2 n}}{5-4 n}-\frac {2 b^2 B x^{-\frac {5}{2}+3 n}}{5-6 n} \] 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 = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=2 \left (-\frac {a^2 A}{5 x^{5/2}}-\frac {a (2 A b+a B) x^{-\frac {5}{2}+n}}{5-2 n}-\frac {b (A b+2 a B) x^{-\frac {5}{2}+2 n}}{5-4 n}-\frac {b^2 B x^{-\frac {5}{2}+3 n}}{5-6 n}\right ) \] Input:
Integrate[((a + b*x^n)^2*(A + B*x^n))/x^(7/2),x]
Output:
2*(-1/5*(a^2*A)/x^(5/2) - (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.
\(\displaystyle \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (\frac {a^2 A}{x^{7/2}}+a x^{n-\frac {7}{2}} (a B+2 A b)+b x^{2 n-\frac {7}{2}} (2 a B+A b)+b^2 B x^{3 n-\frac {7}{2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a x^{n-\frac {5}{2}} (a B+2 A b)}{5-2 n}-\frac {2 b x^{2 n-\frac {5}{2}} (2 a B+A b)}{5-4 n}-\frac {2 b^2 B x^{3 n-\frac {5}{2}}}{5-6 n}\) |
Input:
Int[((a + b*x^n)^2*(A + B*x^n))/x^(7/2),x]
Output:
(-2*a^2*A)/(5*x^(5/2)) - (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 {24 \left (4 n^{3}-44 n^{2}+109 n -74\right ) \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{5 x^{\frac {5}{2}} \left (48 n^{3}-220 n^{2}+300 n -125\right )}+\frac {8 x^{2} \left (22 n^{2}-126 n +149\right ) \left (\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x^{\frac {9}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x^{\frac {9}{2}}}-\frac {7 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{2 x^{\frac {9}{2}}}\right )}{5 \left (48 n^{3}-220 n^{2}+300 n -125\right )}-\frac {32 x^{3} \left (-8+3 n \right ) \left (\frac {2 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x^{\frac {11}{2}}}+\frac {4 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x^{\frac {11}{2}}}-\frac {16 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x^{\frac {11}{2}}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x^{\frac {11}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{x^{\frac {11}{2}}}-\frac {8 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x^{\frac {11}{2}}}+\frac {63 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{4 x^{\frac {11}{2}}}\right )}{5 \left (48 n^{3}-220 n^{2}+300 n -125\right )}+\frac {16 x^{4} \left (\frac {6 b^{2} x^{2 n} n^{3} \left (A +B \,x^{n}\right )}{x^{\frac {13}{2}}}-\frac {27 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x^{\frac {13}{2}}}+\frac {6 b^{2} x^{3 n} n^{3} B}{x^{\frac {13}{2}}}+\frac {12 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{3} b}{x^{\frac {13}{2}}}-\frac {54 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x^{\frac {13}{2}}}+\frac {239 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{2 x^{\frac {13}{2}}}-\frac {27 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x^{\frac {13}{2}}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{3}}{x^{\frac {13}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{3}}{x^{\frac {13}{2}}}-\frac {27 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{2 x^{\frac {13}{2}}}+\frac {239 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{4 x^{\frac {13}{2}}}-\frac {693 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{8 x^{\frac {13}{2}}}\right )}{5 \left (48 n^{3}-220 n^{2}+300 n -125\right )}\) | \(621\) |
Input:
int((a+b*x^n)^2*(A+B*x^n)/x^(7/2),x,method=_RETURNVERBOSE)
Output:
-24/5/x^(5/2)*(4*n^3-44*n^2+109*n-74)/(48*n^3-220*n^2+300*n-125)*(a+b*x^n) ^2*(A+B*x^n)+8/5*x^2*(22*n^2-126*n+149)/(48*n^3-220*n^2+300*n-125)*(2*(a+b *x^n)*(A+B*x^n)/x^(9/2)*b*x^n*n+(a+b*x^n)^2*B*x^n*n/x^(9/2)-7/2*(a+b*x^n)^ 2*(A+B*x^n)/x^(9/2))-32/5*x^3*(-8+3*n)/(48*n^3-220*n^2+300*n-125)*(2*b^2*( x^n)^2*n^2/x^(11/2)*(A+B*x^n)+4*(a+b*x^n)*B*(x^n)^2*n^2/x^(11/2)*b-16*(a+b *x^n)*(A+B*x^n)/x^(11/2)*b*x^n*n+2*(a+b*x^n)*(A+B*x^n)/x^(11/2)*b*x^n*n^2+ (a+b*x^n)^2*B*x^n*n^2/x^(11/2)-8*(a+b*x^n)^2*B*x^n*n/x^(11/2)+63/4*(a+b*x^ n)^2*(A+B*x^n)/x^(11/2))+16/5/(48*n^3-220*n^2+300*n-125)*x^4*(6*b^2*(x^n)^ 2*n^3/x^(13/2)*(A+B*x^n)-27*b^2*(x^n)^2*n^2/x^(13/2)*(A+B*x^n)+6*b^2*(x^n) ^3*n^3/x^(13/2)*B+12*(a+b*x^n)*B*(x^n)^2*n^3/x^(13/2)*b-54*(a+b*x^n)*B*(x^ n)^2*n^2/x^(13/2)*b+239/2*(a+b*x^n)*(A+B*x^n)/x^(13/2)*b*x^n*n-27*(a+b*x^n )*(A+B*x^n)/x^(13/2)*b*x^n*n^2+2*(a+b*x^n)*(A+B*x^n)/x^(13/2)*b*x^n*n^3+(a +b*x^n)^2*B*x^n*n^3/x^(13/2)-27/2*(a+b*x^n)^2*B*x^n*n^2/x^(13/2)+239/4*(a+ b*x^n)^2*B*x^n*n/x^(13/2)-693/8*(a+b*x^n)^2*(A+B*x^n)/x^(13/2))
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (78) = 156\).
Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.22 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\frac {2 \, {\left (5 \, {\left (8 \, B b^{2} n^{2} - 30 \, B b^{2} n + 25 \, B b^{2}\right )} \sqrt {x} x^{3 \, n} + 5 \, {\left (50 \, B a b + 25 \, A b^{2} + 12 \, {\left (2 \, B a b + A b^{2}\right )} n^{2} - 40 \, {\left (2 \, B a b + A b^{2}\right )} n\right )} \sqrt {x} x^{2 \, n} + 5 \, {\left (25 \, B a^{2} + 50 \, A a b + 24 \, {\left (B a^{2} + 2 \, A a b\right )} n^{2} - 50 \, {\left (B a^{2} + 2 \, A a b\right )} n\right )} \sqrt {x} x^{n} - {\left (48 \, A a^{2} n^{3} - 220 \, A a^{2} n^{2} + 300 \, A a^{2} n - 125 \, A a^{2}\right )} \sqrt {x}\right )}}{5 \, {\left (48 \, n^{3} - 220 \, n^{2} + 300 \, n - 125\right )} x^{3}} \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(7/2),x, algorithm="fricas")
Output:
2/5*(5*(8*B*b^2*n^2 - 30*B*b^2*n + 25*B*b^2)*sqrt(x)*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)*sqrt(x)*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)*sqrt(x)*x^n - (48*A*a^2*n^3 - 220*A*a^2*n^2 + 300*A*a^2*n - 125 *A*a^2)*sqrt(x))/((48*n^3 - 220*n^2 + 300*n - 125)*x^3)
Timed out. \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((a+b*x**n)**2*(A+B*x**n)/x**(7/2),x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(7/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(n-7/2>0)', see `assume?` for mor e details)
\[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2}}{x^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(7/2),x, algorithm="giac")
Output:
integrate((B*x^n + A)*(b*x^n + a)^2/x^(7/2), x)
Time = 4.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\frac {B\,a^2\,x^{n-\frac {5}{2}}}{n-\frac {5}{2}}-\frac {2\,A\,a^2}{5\,x^{5/2}}+\frac {2\,A\,b^2\,x^{2\,n-\frac {5}{2}}}{4\,n-5}+\frac {2\,B\,b^2\,x^{3\,n-\frac {5}{2}}}{6\,n-5}+\frac {2\,A\,a\,b\,x^{n-\frac {5}{2}}}{n-\frac {5}{2}}+\frac {4\,B\,a\,b\,x^{2\,n-\frac {5}{2}}}{4\,n-5} \] Input:
int(((A + B*x^n)*(a + b*x^n)^2)/x^(7/2),x)
Output:
(B*a^2*x^(n - 5/2))/(n - 5/2) - (2*A*a^2)/(5*x^(5/2)) + (2*A*b^2*x^(2*n - 5/2))/(4*n - 5) + (2*B*b^2*x^(3*n - 5/2))/(6*n - 5) + (2*A*a*b*x^(n - 5/2) )/(n - 5/2) + (4*B*a*b*x^(2*n - 5/2))/(4*n - 5)
Time = 0.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.77 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{7/2}} \, dx=\frac {16 x^{3 n} b^{3} n^{2}-60 x^{3 n} b^{3} n +50 x^{3 n} b^{3}+72 x^{2 n} a \,b^{2} n^{2}-240 x^{2 n} a \,b^{2} n +150 x^{2 n} a \,b^{2}+144 x^{n} a^{2} b \,n^{2}-300 x^{n} a^{2} b n +150 x^{n} a^{2} b -\frac {96 a^{3} n^{3}}{5}+88 a^{3} n^{2}-120 a^{3} n +50 a^{3}}{\sqrt {x}\, x^{2} \left (48 n^{3}-220 n^{2}+300 n -125\right )} \] Input:
int((a+b*x^n)^2*(A+B*x^n)/x^(7/2),x)
Output:
(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*sqrt(x)*x**2*(48*n**3 - 220*n**2 + 300*n - 125))