Integrand size = 22, antiderivative size = 64 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=-\frac {2 a}{3 d (d x)^{3/2}}-\frac {2 b x^n}{d (3-2 n) (d x)^{3/2}}-\frac {2 c x^{2 n}}{d (3-4 n) (d x)^{3/2}} \] Output:
-2/3*a/d/(d*x)^(3/2)-2*b*x^n/d/(3-2*n)/(d*x)^(3/2)-2*c*x^(2*n)/d/(3-4*n)/( d*x)^(3/2)
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\frac {2 x \left (-a+3 x^n \left (\frac {b}{-3+2 n}+\frac {c x^n}{-3+4 n}\right )\right )}{3 (d x)^{5/2}} \] Input:
Integrate[(a + b*x^n + c*x^(2*n))/(d*x)^(5/2),x]
Output:
(2*x*(-a + 3*x^n*(b/(-3 + 2*n) + (c*x^n)/(-3 + 4*n))))/(3*(d*x)^(5/2))
Time = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1691, 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.
\(\displaystyle \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1691 |
\(\displaystyle \int \left (\frac {a}{(d x)^{5/2}}+\frac {b x^n}{(d x)^{5/2}}+\frac {c x^{2 n}}{(d x)^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a}{3 d (d x)^{3/2}}-\frac {2 b x^{n+1}}{(3-2 n) (d x)^{5/2}}-\frac {2 c x^{2 n+1}}{(3-4 n) (d x)^{5/2}}\) |
Input:
Int[(a + b*x^n + c*x^(2*n))/(d*x)^(5/2),x]
Output:
(-2*b*x^(1 + n))/((3 - 2*n)*(d*x)^(5/2)) - (2*c*x^(1 + 2*n))/((3 - 4*n)*(d *x)^(5/2)) - (2*a)/(3*d*(d*x)^(3/2))
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^m*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && IGtQ[p, 0] && !IntegerQ [Simplify[(m + 1)/n]]
Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(56)=112\).
Time = 0.09 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.89
method | result | size |
orering | \(-\frac {2 x \left (8 n^{2}-48 n +49\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )}{3 \left (-3+4 n \right ) \left (-3+2 n \right ) \left (d x \right )^{\frac {5}{2}}}+\frac {4 x^{2} \left (-5+2 n \right ) \left (\frac {\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}}{\left (d x \right )^{\frac {5}{2}}}-\frac {5 \left (a +b \,x^{n}+c \,x^{2 n}\right ) d}{2 \left (d x \right )^{\frac {7}{2}}}\right )}{\left (-3+4 n \right ) \left (-3+2 n \right )}-\frac {8 x^{3} \left (\frac {\frac {b \,x^{n} n^{2}}{x^{2}}-\frac {b \,x^{n} n}{x^{2}}+\frac {4 c \,x^{2 n} n^{2}}{x^{2}}-\frac {2 c \,x^{2 n} n}{x^{2}}}{\left (d x \right )^{\frac {5}{2}}}-\frac {5 \left (\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}\right ) d}{\left (d x \right )^{\frac {7}{2}}}+\frac {35 \left (a +b \,x^{n}+c \,x^{2 n}\right ) d^{2}}{4 \left (d x \right )^{\frac {9}{2}}}\right )}{3 \left (8 n^{2}-18 n +9\right )}\) | \(249\) |
Input:
int((a+b*x^n+c*x^(2*n))/(d*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*x*(8*n^2-48*n+49)/(-3+4*n)/(-3+2*n)*(a+b*x^n+c*x^(2*n))/(d*x)^(5/2)+4 *x^2*(-5+2*n)/(-3+4*n)/(-3+2*n)*((b*x^n*n/x+2*c*x^(2*n)*n/x)/(d*x)^(5/2)-5 /2*(a+b*x^n+c*x^(2*n))/(d*x)^(7/2)*d)-8/3/(8*n^2-18*n+9)*x^3*((b*x^n*n^2/x ^2-b*x^n*n/x^2+4*c*x^(2*n)*n^2/x^2-2*c*x^(2*n)*n/x^2)/(d*x)^(5/2)-5*(b*x^n *n/x+2*c*x^(2*n)*n/x)/(d*x)^(7/2)*d+35/4*(a+b*x^n+c*x^(2*n))/(d*x)^(9/2)*d ^2)
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.36 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, {\left (2 \, c n - 3 \, c\right )} \sqrt {d x} x^{2 \, n} + 3 \, {\left (4 \, b n - 3 \, b\right )} \sqrt {d x} x^{n} - {\left (8 \, a n^{2} - 18 \, a n + 9 \, a\right )} \sqrt {d x}\right )}}{3 \, {\left (8 \, d^{3} n^{2} - 18 \, d^{3} n + 9 \, d^{3}\right )} x^{2}} \] Input:
integrate((a+b*x^n+c*x^(2*n))/(d*x)^(5/2),x, algorithm="fricas")
Output:
2/3*(3*(2*c*n - 3*c)*sqrt(d*x)*x^(2*n) + 3*(4*b*n - 3*b)*sqrt(d*x)*x^n - ( 8*a*n^2 - 18*a*n + 9*a)*sqrt(d*x))/((8*d^3*n^2 - 18*d^3*n + 9*d^3)*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (56) = 112\).
Time = 161.92 (sec) , antiderivative size = 384, normalized size of antiderivative = 6.00 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\begin {cases} a \left (\begin {cases} - \frac {2}{3 d \left (d x\right )^{\frac {3}{2}}} & \text {for}\: d \neq 0 \\\tilde {\infty } x & \text {otherwise} \end {cases}\right ) - \frac {4 b x^{\frac {7}{4}}}{3 \left (d x\right )^{\frac {5}{2}}} + \frac {c x^{\frac {5}{2}} \log {\left (x \right )}}{\left (d x\right )^{\frac {5}{2}}} & \text {for}\: n = \frac {3}{4} \\a \left (\begin {cases} - \frac {2}{3 d \left (d x\right )^{\frac {3}{2}}} & \text {for}\: d \neq 0 \\\tilde {\infty } x & \text {otherwise} \end {cases}\right ) + \frac {b x^{\frac {5}{2}} \log {\left (x \right )}}{\left (d x\right )^{\frac {5}{2}}} + c \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {3}{2}}}{3 d^{4}} & \text {for}\: d \neq 0 \\\tilde {\infty } x^{4} & \text {otherwise} \end {cases}\right ) & \text {for}\: n = \frac {3}{2} \\- \frac {16 a n^{2} x}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} + \frac {36 a n x}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} - \frac {18 a x}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} + \frac {24 b n x x^{n}}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} - \frac {18 b x x^{n}}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} + \frac {12 c n x x^{2 n}}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} - \frac {18 c x x^{2 n}}{24 n^{2} \left (d x\right )^{\frac {5}{2}} - 54 n \left (d x\right )^{\frac {5}{2}} + 27 \left (d x\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*x**n+c*x**(2*n))/(d*x)**(5/2),x)
Output:
Piecewise((a*Piecewise((-2/(3*d*(d*x)**(3/2)), Ne(d, 0)), (zoo*x, True)) - 4*b*x**(7/4)/(3*(d*x)**(5/2)) + c*x**(5/2)*log(x)/(d*x)**(5/2), Eq(n, 3/4 )), (a*Piecewise((-2/(3*d*(d*x)**(3/2)), Ne(d, 0)), (zoo*x, True)) + b*x** (5/2)*log(x)/(d*x)**(5/2) + c*Piecewise((2*(d*x)**(3/2)/(3*d**4), Ne(d, 0) ), (zoo*x**4, True)), Eq(n, 3/2)), (-16*a*n**2*x/(24*n**2*(d*x)**(5/2) - 5 4*n*(d*x)**(5/2) + 27*(d*x)**(5/2)) + 36*a*n*x/(24*n**2*(d*x)**(5/2) - 54* n*(d*x)**(5/2) + 27*(d*x)**(5/2)) - 18*a*x/(24*n**2*(d*x)**(5/2) - 54*n*(d *x)**(5/2) + 27*(d*x)**(5/2)) + 24*b*n*x*x**n/(24*n**2*(d*x)**(5/2) - 54*n *(d*x)**(5/2) + 27*(d*x)**(5/2)) - 18*b*x*x**n/(24*n**2*(d*x)**(5/2) - 54* n*(d*x)**(5/2) + 27*(d*x)**(5/2)) + 12*c*n*x*x**(2*n)/(24*n**2*(d*x)**(5/2 ) - 54*n*(d*x)**(5/2) + 27*(d*x)**(5/2)) - 18*c*x*x**(2*n)/(24*n**2*(d*x)* *(5/2) - 54*n*(d*x)**(5/2) + 27*(d*x)**(5/2)), True))
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=-\frac {2 \, a}{3 \, \left (d x\right )^{\frac {3}{2}} d} + \frac {2 \, c x^{2 \, n}}{d^{\frac {5}{2}} {\left (4 \, n - 3\right )} x^{\frac {3}{2}}} + \frac {2 \, b x^{n}}{d^{\frac {5}{2}} {\left (2 \, n - 3\right )} x^{\frac {3}{2}}} \] Input:
integrate((a+b*x^n+c*x^(2*n))/(d*x)^(5/2),x, algorithm="maxima")
Output:
-2/3*a/((d*x)^(3/2)*d) + 2*c*x^(2*n)/(d^(5/2)*(4*n - 3)*x^(3/2)) + 2*b*x^n /(d^(5/2)*(2*n - 3)*x^(3/2))
\[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\int { \frac {c x^{2 \, n} + b x^{n} + a}{\left (d x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*x^n+c*x^(2*n))/(d*x)^(5/2),x, algorithm="giac")
Output:
integrate((c*x^(2*n) + b*x^n + a)/(d*x)^(5/2), x)
Time = 10.84 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\frac {2\,c\,x^{2\,n}}{d^2\,x\,\left (4\,n-3\right )\,\sqrt {d\,x}}-\frac {2\,a\,\sqrt {d\,x}}{3\,d^3\,x^2}+\frac {2\,b\,x^n}{d^2\,x\,\left (2\,n-3\right )\,\sqrt {d\,x}} \] Input:
int((a + b*x^n + c*x^(2*n))/(d*x)^(5/2),x)
Output:
(2*c*x^(2*n))/(d^2*x*(4*n - 3)*(d*x)^(1/2)) - (2*a*(d*x)^(1/2))/(3*d^3*x^2 ) + (2*b*x^n)/(d^2*x*(2*n - 3)*(d*x)^(1/2))
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {a+b x^n+c x^{2 n}}{(d x)^{5/2}} \, dx=\frac {2 \sqrt {d}\, \left (6 x^{2 n} c n -9 x^{2 n} c +12 x^{n} b n -9 x^{n} b -8 a \,n^{2}+18 a n -9 a \right )}{3 \sqrt {x}\, d^{3} x \left (8 n^{2}-18 n +9\right )} \] Input:
int((a+b*x^n+c*x^(2*n))/(d*x)^(5/2),x)
Output:
(2*sqrt(d)*(6*x**(2*n)*c*n - 9*x**(2*n)*c + 12*x**n*b*n - 9*x**n*b - 8*a*n **2 + 18*a*n - 9*a))/(3*sqrt(x)*d**3*x*(8*n**2 - 18*n + 9))