Integrand size = 22, antiderivative size = 92 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 a c^2}{3 e (e x)^{3/2}}-\frac {4 a c d}{e^2 \sqrt {e x}}+\frac {2 \left (b c^2+a d^2\right ) \sqrt {e x}}{e^3}+\frac {4 b c d (e x)^{3/2}}{3 e^4}+\frac {2 b d^2 (e x)^{5/2}}{5 e^5} \] Output:
-2/3*a*c^2/e/(e*x)^(3/2)-4*a*c*d/e^2/(e*x)^(1/2)+2*(a*d^2+b*c^2)*(e*x)^(1/ 2)/e^3+4/3*b*c*d*(e*x)^(3/2)/e^4+2/5*b*d^2*(e*x)^(5/2)/e^5
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.62 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2 x \left (-5 a \left (c^2+6 c d x-3 d^2 x^2\right )+b x^2 \left (15 c^2+10 c d x+3 d^2 x^2\right )\right )}{15 (e x)^{5/2}} \] Input:
Integrate[((c + d*x)^2*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(2*x*(-5*a*(c^2 + 6*c*d*x - 3*d^2*x^2) + b*x^2*(15*c^2 + 10*c*d*x + 3*d^2* x^2)))/(15*(e*x)^(5/2))
Time = 0.37 (sec) , antiderivative size = 92, 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 = {522, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^2}{(e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {a d^2+b c^2}{e^2 \sqrt {e x}}+\frac {a c^2}{(e x)^{5/2}}+\frac {2 a c d}{e (e x)^{3/2}}+\frac {2 b c d \sqrt {e x}}{e^3}+\frac {b d^2 (e x)^{3/2}}{e^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a d^2+b c^2\right )}{e^3}-\frac {2 a c^2}{3 e (e x)^{3/2}}-\frac {4 a c d}{e^2 \sqrt {e x}}+\frac {4 b c d (e x)^{3/2}}{3 e^4}+\frac {2 b d^2 (e x)^{5/2}}{5 e^5}\) |
Input:
Int[((c + d*x)^2*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(-2*a*c^2)/(3*e*(e*x)^(3/2)) - (4*a*c*d)/(e^2*Sqrt[e*x]) + (2*(b*c^2 + a*d ^2)*Sqrt[e*x])/e^3 + (4*b*c*d*(e*x)^(3/2))/(3*e^4) + (2*b*d^2*(e*x)^(5/2)) /(5*e^5)
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {2 x \left (-3 b \,d^{2} x^{4}-10 b c d \,x^{3}-15 a \,d^{2} x^{2}-15 b \,c^{2} x^{2}+30 a d x c +5 a \,c^{2}\right )}{15 \left (e x \right )^{\frac {5}{2}}}\) | \(57\) |
orering | \(-\frac {2 x \left (-3 b \,d^{2} x^{4}-10 b c d \,x^{3}-15 a \,d^{2} x^{2}-15 b \,c^{2} x^{2}+30 a d x c +5 a \,c^{2}\right )}{15 \left (e x \right )^{\frac {5}{2}}}\) | \(57\) |
trager | \(-\frac {2 \left (-3 b \,d^{2} x^{4}-10 b c d \,x^{3}-15 a \,d^{2} x^{2}-15 b \,c^{2} x^{2}+30 a d x c +5 a \,c^{2}\right ) \sqrt {e x}}{15 e^{3} x^{2}}\) | \(62\) |
risch | \(-\frac {2 \left (-3 b \,d^{2} x^{4}-10 b c d \,x^{3}-15 a \,d^{2} x^{2}-15 b \,c^{2} x^{2}+30 a d x c +5 a \,c^{2}\right )}{15 e^{2} x \sqrt {e x}}\) | \(62\) |
pseudoelliptic | \(-\frac {2 \left (-3 b \,d^{2} x^{4}-10 b c d \,x^{3}-15 a \,d^{2} x^{2}-15 b \,c^{2} x^{2}+30 a d x c +5 a \,c^{2}\right )}{15 e^{2} x \sqrt {e x}}\) | \(62\) |
derivativedivides | \(\frac {\frac {2 b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}+\frac {4 b c d e \left (e x \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e^{2} \sqrt {e x}+2 b \,c^{2} e^{2} \sqrt {e x}-\frac {4 a c d \,e^{3}}{\sqrt {e x}}-\frac {2 a \,c^{2} e^{4}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{5}}\) | \(82\) |
default | \(\frac {\frac {2 b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}+\frac {4 b c d e \left (e x \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e^{2} \sqrt {e x}+2 b \,c^{2} e^{2} \sqrt {e x}-\frac {4 a c d \,e^{3}}{\sqrt {e x}}-\frac {2 a \,c^{2} e^{4}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{5}}\) | \(82\) |
Input:
int((d*x+c)^2*(b*x^2+a)/(e*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/15*x*(-3*b*d^2*x^4-10*b*c*d*x^3-15*a*d^2*x^2-15*b*c^2*x^2+30*a*c*d*x+5* a*c^2)/(e*x)^(5/2)
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b d^{2} x^{4} + 10 \, b c d x^{3} - 30 \, a c d x - 5 \, a c^{2} + 15 \, {\left (b c^{2} + a d^{2}\right )} x^{2}\right )} \sqrt {e x}}{15 \, e^{3} x^{2}} \] Input:
integrate((d*x+c)^2*(b*x^2+a)/(e*x)^(5/2),x, algorithm="fricas")
Output:
2/15*(3*b*d^2*x^4 + 10*b*c*d*x^3 - 30*a*c*d*x - 5*a*c^2 + 15*(b*c^2 + a*d^ 2)*x^2)*sqrt(e*x)/(e^3*x^2)
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=- \frac {2 a c^{2} x}{3 \left (e x\right )^{\frac {5}{2}}} - \frac {4 a c d x^{2}}{\left (e x\right )^{\frac {5}{2}}} + \frac {2 a d^{2} x^{3}}{\left (e x\right )^{\frac {5}{2}}} + \frac {2 b c^{2} x^{3}}{\left (e x\right )^{\frac {5}{2}}} + \frac {4 b c d x^{4}}{3 \left (e x\right )^{\frac {5}{2}}} + \frac {2 b d^{2} x^{5}}{5 \left (e x\right )^{\frac {5}{2}}} \] Input:
integrate((d*x+c)**2*(b*x**2+a)/(e*x)**(5/2),x)
Output:
-2*a*c**2*x/(3*(e*x)**(5/2)) - 4*a*c*d*x**2/(e*x)**(5/2) + 2*a*d**2*x**3/( e*x)**(5/2) + 2*b*c**2*x**3/(e*x)**(5/2) + 4*b*c*d*x**4/(3*(e*x)**(5/2)) + 2*b*d**2*x**5/(5*(e*x)**(5/2))
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {5 \, {\left (6 \, a c d e x + a c^{2} e\right )}}{\left (e x\right )^{\frac {3}{2}} e} - \frac {3 \, \left (e x\right )^{\frac {5}{2}} b d^{2} + 10 \, \left (e x\right )^{\frac {3}{2}} b c d e + 15 \, {\left (b c^{2} + a d^{2}\right )} \sqrt {e x} e^{2}}{e^{4}}\right )}}{15 \, e} \] Input:
integrate((d*x+c)^2*(b*x^2+a)/(e*x)^(5/2),x, algorithm="maxima")
Output:
-2/15*(5*(6*a*c*d*e*x + a*c^2*e)/((e*x)^(3/2)*e) - (3*(e*x)^(5/2)*b*d^2 + 10*(e*x)^(3/2)*b*c*d*e + 15*(b*c^2 + a*d^2)*sqrt(e*x)*e^2)/e^4)/e
Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {5 \, {\left (6 \, a c d e x + a c^{2} e\right )}}{\sqrt {e x} e^{2} x} - \frac {3 \, \sqrt {e x} b d^{2} e^{18} x^{2} + 10 \, \sqrt {e x} b c d e^{18} x + 15 \, \sqrt {e x} b c^{2} e^{18} + 15 \, \sqrt {e x} a d^{2} e^{18}}{e^{20}}\right )}}{15 \, e} \] Input:
integrate((d*x+c)^2*(b*x^2+a)/(e*x)^(5/2),x, algorithm="giac")
Output:
-2/15*(5*(6*a*c*d*e*x + a*c^2*e)/(sqrt(e*x)*e^2*x) - (3*sqrt(e*x)*b*d^2*e^ 18*x^2 + 10*sqrt(e*x)*b*c*d*e^18*x + 15*sqrt(e*x)*b*c^2*e^18 + 15*sqrt(e*x )*a*d^2*e^18)/e^20)/e
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2\,\sqrt {e\,x}\,\left (b\,c^2+a\,d^2\right )}{e^3}-\frac {\frac {2\,a\,e\,c^2}{3}+4\,a\,d\,e\,x\,c}{e^2\,{\left (e\,x\right )}^{3/2}}+\frac {2\,b\,d^2\,{\left (e\,x\right )}^{5/2}}{5\,e^5}+\frac {4\,b\,c\,d\,{\left (e\,x\right )}^{3/2}}{3\,e^4} \] Input:
int(((a + b*x^2)*(c + d*x)^2)/(e*x)^(5/2),x)
Output:
(2*(e*x)^(1/2)*(a*d^2 + b*c^2))/e^3 - ((2*a*c^2*e)/3 + 4*a*c*d*e*x)/(e^2*( e*x)^(3/2)) + (2*b*d^2*(e*x)^(5/2))/(5*e^5) + (4*b*c*d*(e*x)^(3/2))/(3*e^4 )
Time = 0.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2 \sqrt {e}\, \left (3 b \,d^{2} x^{4}+10 b c d \,x^{3}+15 a \,d^{2} x^{2}+15 b \,c^{2} x^{2}-30 a c d x -5 a \,c^{2}\right )}{15 \sqrt {x}\, e^{3} x} \] Input:
int((d*x+c)^2*(b*x^2+a)/(e*x)^(5/2),x)
Output:
(2*sqrt(e)*( - 5*a*c**2 - 30*a*c*d*x + 15*a*d**2*x**2 + 15*b*c**2*x**2 + 1 0*b*c*d*x**3 + 3*b*d**2*x**4))/(15*sqrt(x)*e**3*x)