Integrand size = 22, antiderivative size = 86 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=-\frac {2 a^2 A}{\sqrt {x}}-\frac {2 a (2 A b+a B) x^{-\frac {1}{2}+n}}{1-2 n}-\frac {2 b (A b+2 a B) x^{-\frac {1}{2}+2 n}}{1-4 n}-\frac {2 b^2 B x^{-\frac {1}{2}+3 n}}{1-6 n} \] Output:
-2*a^2*A/x^(1/2)-2*a*(2*A*b+B*a)*x^(-1/2+n)/(1-2*n)-2*b*(A*b+2*B*a)*x^(-1/ 2+2*n)/(1-4*n)-2*b^2*B*x^(-1/2+3*n)/(1-6*n)
Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=2 \left (-\frac {a^2 A}{\sqrt {x}}+\frac {a (2 A b+a B) x^{-\frac {1}{2}+n}}{-1+2 n}+\frac {b (A b+2 a B) x^{-\frac {1}{2}+2 n}}{-1+4 n}+\frac {b^2 B x^{-\frac {1}{2}+3 n}}{-1+6 n}\right ) \] Input:
Integrate[((a + b*x^n)^2*(A + B*x^n))/x^(3/2),x]
Output:
2*(-((a^2*A)/Sqrt[x]) + (a*(2*A*b + a*B)*x^(-1/2 + n))/(-1 + 2*n) + (b*(A* b + 2*a*B)*x^(-1/2 + 2*n))/(-1 + 4*n) + (b^2*B*x^(-1/2 + 3*n))/(-1 + 6*n))
Time = 0.36 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (\frac {a^2 A}{x^{3/2}}+a x^{n-\frac {3}{2}} (a B+2 A b)+b x^{2 n-\frac {3}{2}} (2 a B+A b)+b^2 B x^{3 n-\frac {3}{2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 A}{\sqrt {x}}-\frac {2 a x^{n-\frac {1}{2}} (a B+2 A b)}{1-2 n}-\frac {2 b x^{2 n-\frac {1}{2}} (2 a B+A b)}{1-4 n}-\frac {2 b^2 B x^{3 n-\frac {1}{2}}}{1-6 n}\) |
Input:
Int[((a + b*x^n)^2*(A + B*x^n))/x^(3/2),x]
Output:
(-2*a^2*A)/Sqrt[x] - (2*a*(2*A*b + a*B)*x^(-1/2 + n))/(1 - 2*n) - (2*b*(A* b + 2*a*B)*x^(-1/2 + 2*n))/(1 - 4*n) - (2*b^2*B*x^(-1/2 + 3*n))/(1 - 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. \(610\) vs. \(2(78)=156\).
Time = 0.28 (sec) , antiderivative size = 611, normalized size of antiderivative = 7.10
method | result | size |
orering | \(-\frac {8 \left (6 n^{2}-19 n +10\right ) \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{\sqrt {x}\, \left (24 n^{2}-10 n +1\right )}+\frac {8 x^{2} \left (22 n^{2}-54 n +29\right ) \left (\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x^{\frac {5}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x^{\frac {5}{2}}}-\frac {3 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{2 x^{\frac {5}{2}}}\right )}{48 n^{3}-44 n^{2}+12 n -1}-\frac {32 x^{3} \left (-4+3 n \right ) \left (\frac {2 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x^{\frac {7}{2}}}+\frac {4 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x^{\frac {7}{2}}}-\frac {8 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{x^{\frac {7}{2}}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x^{\frac {7}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{x^{\frac {7}{2}}}-\frac {4 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{x^{\frac {7}{2}}}+\frac {15 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{4 x^{\frac {7}{2}}}\right )}{48 n^{3}-44 n^{2}+12 n -1}+\frac {16 x^{4} \left (\frac {6 b^{2} x^{2 n} n^{3} \left (A +B \,x^{n}\right )}{x^{\frac {9}{2}}}-\frac {15 b^{2} x^{2 n} n^{2} \left (A +B \,x^{n}\right )}{x^{\frac {9}{2}}}+\frac {6 b^{2} x^{3 n} n^{3} B}{x^{\frac {9}{2}}}+\frac {12 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{3} b}{x^{\frac {9}{2}}}-\frac {30 \left (a +b \,x^{n}\right ) B \,x^{2 n} n^{2} b}{x^{\frac {9}{2}}}+\frac {71 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n}{2 x^{\frac {9}{2}}}-\frac {15 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{2}}{x^{\frac {9}{2}}}+\frac {2 \left (a +b \,x^{n}\right ) \left (A +B \,x^{n}\right ) b \,x^{n} n^{3}}{x^{\frac {9}{2}}}+\frac {\left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{3}}{x^{\frac {9}{2}}}-\frac {15 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n^{2}}{2 x^{\frac {9}{2}}}+\frac {71 \left (a +b \,x^{n}\right )^{2} B \,x^{n} n}{4 x^{\frac {9}{2}}}-\frac {105 \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{8 x^{\frac {9}{2}}}\right )}{48 n^{3}-44 n^{2}+12 n -1}\) | \(611\) |
Input:
int((a+b*x^n)^2*(A+B*x^n)/x^(3/2),x,method=_RETURNVERBOSE)
Output:
-8*(6*n^2-19*n+10)/x^(1/2)/(24*n^2-10*n+1)*(a+b*x^n)^2*(A+B*x^n)+8*x^2*(22 *n^2-54*n+29)/(48*n^3-44*n^2+12*n-1)*(2*(a+b*x^n)*(A+B*x^n)/x^(5/2)*b*x^n* n+(a+b*x^n)^2*B*x^n*n/x^(5/2)-3/2*(a+b*x^n)^2*(A+B*x^n)/x^(5/2))-32*x^3*(- 4+3*n)/(48*n^3-44*n^2+12*n-1)*(2*b^2*(x^n)^2*n^2/x^(7/2)*(A+B*x^n)+4*(a+b* x^n)*B*(x^n)^2*n^2/x^(7/2)*b-8*(a+b*x^n)*(A+B*x^n)/x^(7/2)*b*x^n*n+2*(a+b* x^n)*(A+B*x^n)/x^(7/2)*b*x^n*n^2+(a+b*x^n)^2*B*x^n*n^2/x^(7/2)-4*(a+b*x^n) ^2*B*x^n*n/x^(7/2)+15/4*(a+b*x^n)^2*(A+B*x^n)/x^(7/2))+16/(48*n^3-44*n^2+1 2*n-1)*x^4*(6*b^2*(x^n)^2*n^3/x^(9/2)*(A+B*x^n)-15*b^2*(x^n)^2*n^2/x^(9/2) *(A+B*x^n)+6*b^2*(x^n)^3*n^3/x^(9/2)*B+12*(a+b*x^n)*B*(x^n)^2*n^3/x^(9/2)* b-30*(a+b*x^n)*B*(x^n)^2*n^2/x^(9/2)*b+71/2*(a+b*x^n)*(A+B*x^n)/x^(9/2)*b* x^n*n-15*(a+b*x^n)*(A+B*x^n)/x^(9/2)*b*x^n*n^2+2*(a+b*x^n)*(A+B*x^n)/x^(9/ 2)*b*x^n*n^3+(a+b*x^n)^2*B*x^n*n^3/x^(9/2)-15/2*(a+b*x^n)^2*B*x^n*n^2/x^(9 /2)+71/4*(a+b*x^n)^2*B*x^n*n/x^(9/2)-105/8*(a+b*x^n)^2*(A+B*x^n)/x^(9/2))
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (78) = 156\).
Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.20 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=\frac {2 \, {\left ({\left (8 \, B b^{2} n^{2} - 6 \, B b^{2} n + B b^{2}\right )} \sqrt {x} x^{3 \, n} + {\left (2 \, B a b + A b^{2} + 12 \, {\left (2 \, B a b + A b^{2}\right )} n^{2} - 8 \, {\left (2 \, B a b + A b^{2}\right )} n\right )} \sqrt {x} x^{2 \, n} + {\left (B a^{2} + 2 \, A a b + 24 \, {\left (B a^{2} + 2 \, A a b\right )} n^{2} - 10 \, {\left (B a^{2} + 2 \, A a b\right )} n\right )} \sqrt {x} x^{n} - {\left (48 \, A a^{2} n^{3} - 44 \, A a^{2} n^{2} + 12 \, A a^{2} n - A a^{2}\right )} \sqrt {x}\right )}}{{\left (48 \, n^{3} - 44 \, n^{2} + 12 \, n - 1\right )} x} \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(3/2),x, algorithm="fricas")
Output:
2*((8*B*b^2*n^2 - 6*B*b^2*n + B*b^2)*sqrt(x)*x^(3*n) + (2*B*a*b + A*b^2 + 12*(2*B*a*b + A*b^2)*n^2 - 8*(2*B*a*b + A*b^2)*n)*sqrt(x)*x^(2*n) + (B*a^2 + 2*A*a*b + 24*(B*a^2 + 2*A*a*b)*n^2 - 10*(B*a^2 + 2*A*a*b)*n)*sqrt(x)*x^ n - (48*A*a^2*n^3 - 44*A*a^2*n^2 + 12*A*a^2*n - A*a^2)*sqrt(x))/((48*n^3 - 44*n^2 + 12*n - 1)*x)
Time = 1.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.21 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=- \frac {2 A a^{2}}{\sqrt {x}} + 2 A a b \left (\begin {cases} \frac {x^{n}}{\sqrt {x} \left (n - \frac {1}{2}\right )} & \text {for}\: n \neq \frac {1}{2} \\\frac {x^{n} \log {\left (x \right )}}{\sqrt {x}} & \text {otherwise} \end {cases}\right ) + A b^{2} \left (\begin {cases} \frac {x^{2 n}}{\sqrt {x} \left (2 n - \frac {1}{2}\right )} & \text {for}\: n \neq \frac {1}{4} \\\frac {x^{2 n} \log {\left (x \right )}}{\sqrt {x}} & \text {otherwise} \end {cases}\right ) + B a^{2} \left (\begin {cases} \frac {x^{n}}{\sqrt {x} \left (n - \frac {1}{2}\right )} & \text {for}\: n \neq \frac {1}{2} \\\frac {x^{n} \log {\left (x \right )}}{\sqrt {x}} & \text {otherwise} \end {cases}\right ) + 2 B a b \left (\begin {cases} \frac {x^{2 n}}{\sqrt {x} \left (2 n - \frac {1}{2}\right )} & \text {for}\: n \neq \frac {1}{4} \\\frac {x^{2 n} \log {\left (x \right )}}{\sqrt {x}} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} \frac {x^{3 n}}{\sqrt {x} \left (3 n - \frac {1}{2}\right )} & \text {for}\: n \neq \frac {1}{6} \\\frac {x^{3 n} \log {\left (x \right )}}{\sqrt {x}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((a+b*x**n)**2*(A+B*x**n)/x**(3/2),x)
Output:
-2*A*a**2/sqrt(x) + 2*A*a*b*Piecewise((x**n/(sqrt(x)*(n - 1/2)), Ne(n, 1/2 )), (x**n*log(x)/sqrt(x), True)) + A*b**2*Piecewise((x**(2*n)/(sqrt(x)*(2* n - 1/2)), Ne(n, 1/4)), (x**(2*n)*log(x)/sqrt(x), True)) + B*a**2*Piecewis e((x**n/(sqrt(x)*(n - 1/2)), Ne(n, 1/2)), (x**n*log(x)/sqrt(x), True)) + 2 *B*a*b*Piecewise((x**(2*n)/(sqrt(x)*(2*n - 1/2)), Ne(n, 1/4)), (x**(2*n)*l og(x)/sqrt(x), True)) + B*b**2*Piecewise((x**(3*n)/(sqrt(x)*(3*n - 1/2)), Ne(n, 1/6)), (x**(3*n)*log(x)/sqrt(x), True))
Exception generated. \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(3/2),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-3/2>0)', see `assume?` for mor e details)
\[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2}}{x^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+b*x^n)^2*(A+B*x^n)/x^(3/2),x, algorithm="giac")
Output:
integrate((B*x^n + A)*(b*x^n + a)^2/x^(3/2), x)
Time = 4.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=\frac {B\,a^2\,x^{n-\frac {1}{2}}}{n-\frac {1}{2}}-\frac {2\,A\,a^2}{\sqrt {x}}+\frac {2\,A\,b^2\,x^{2\,n-\frac {1}{2}}}{4\,n-1}+\frac {2\,B\,b^2\,x^{3\,n-\frac {1}{2}}}{6\,n-1}+\frac {2\,A\,a\,b\,x^{n-\frac {1}{2}}}{n-\frac {1}{2}}+\frac {4\,B\,a\,b\,x^{2\,n-\frac {1}{2}}}{4\,n-1} \] Input:
int(((A + B*x^n)*(a + b*x^n)^2)/x^(3/2),x)
Output:
(B*a^2*x^(n - 1/2))/(n - 1/2) - (2*A*a^2)/x^(1/2) + (2*A*b^2*x^(2*n - 1/2) )/(4*n - 1) + (2*B*b^2*x^(3*n - 1/2))/(6*n - 1) + (2*A*a*b*x^(n - 1/2))/(n - 1/2) + (4*B*a*b*x^(2*n - 1/2))/(4*n - 1)
Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x^n\right )^2 \left (A+B x^n\right )}{x^{3/2}} \, dx=\frac {16 x^{3 n} b^{3} n^{2}-12 x^{3 n} b^{3} n +2 x^{3 n} b^{3}+72 x^{2 n} a \,b^{2} n^{2}-48 x^{2 n} a \,b^{2} n +6 x^{2 n} a \,b^{2}+144 x^{n} a^{2} b \,n^{2}-60 x^{n} a^{2} b n +6 x^{n} a^{2} b -96 a^{3} n^{3}+88 a^{3} n^{2}-24 a^{3} n +2 a^{3}}{\sqrt {x}\, \left (48 n^{3}-44 n^{2}+12 n -1\right )} \] Input:
int((a+b*x^n)^2*(A+B*x^n)/x^(3/2),x)
Output:
(2*(8*x**(3*n)*b**3*n**2 - 6*x**(3*n)*b**3*n + x**(3*n)*b**3 + 36*x**(2*n) *a*b**2*n**2 - 24*x**(2*n)*a*b**2*n + 3*x**(2*n)*a*b**2 + 72*x**n*a**2*b*n **2 - 30*x**n*a**2*b*n + 3*x**n*a**2*b - 48*a**3*n**3 + 44*a**3*n**2 - 12* a**3*n + a**3))/(sqrt(x)*(48*n**3 - 44*n**2 + 12*n - 1))