Integrand size = 22, antiderivative size = 143 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=-\frac {2 a^3 c}{5 e (e x)^{5/2}}-\frac {2 a^3 d}{3 e^2 (e x)^{3/2}}-\frac {6 a^2 b c}{e^3 \sqrt {e x}}+\frac {6 a^2 b d \sqrt {e x}}{e^4}+\frac {2 a b^2 c (e x)^{3/2}}{e^5}+\frac {6 a b^2 d (e x)^{5/2}}{5 e^6}+\frac {2 b^3 c (e x)^{7/2}}{7 e^7}+\frac {2 b^3 d (e x)^{9/2}}{9 e^8} \] Output:
-2/5*a^3*c/e/(e*x)^(5/2)-2/3*a^3*d/e^2/(e*x)^(3/2)-6*a^2*b*c/e^3/(e*x)^(1/ 2)+6*a^2*b*d*(e*x)^(1/2)/e^4+2*a*b^2*c*(e*x)^(3/2)/e^5+6/5*a*b^2*d*(e*x)^( 5/2)/e^6+2/7*b^3*c*(e*x)^(7/2)/e^7+2/9*b^3*d*(e*x)^(9/2)/e^8
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=\frac {2 x \left (945 a^2 b x^2 (-c+d x)+63 a b^2 x^4 (5 c+3 d x)-21 a^3 (3 c+5 d x)+5 b^3 x^6 (9 c+7 d x)\right )}{315 (e x)^{7/2}} \] Input:
Integrate[((c + d*x)*(a + b*x^2)^3)/(e*x)^(7/2),x]
Output:
(2*x*(945*a^2*b*x^2*(-c + d*x) + 63*a*b^2*x^4*(5*c + 3*d*x) - 21*a^3*(3*c + 5*d*x) + 5*b^3*x^6*(9*c + 7*d*x)))/(315*(e*x)^(7/2))
Time = 0.43 (sec) , antiderivative size = 143, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^3 (c+d x)}{(e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {a^3 c}{(e x)^{7/2}}+\frac {a^3 d}{e (e x)^{5/2}}+\frac {3 a^2 b c}{e^2 (e x)^{3/2}}+\frac {3 a^2 b d}{e^3 \sqrt {e x}}+\frac {3 a b^2 c \sqrt {e x}}{e^4}+\frac {3 a b^2 d (e x)^{3/2}}{e^5}+\frac {b^3 c (e x)^{5/2}}{e^6}+\frac {b^3 d (e x)^{7/2}}{e^7}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^3 c}{5 e (e x)^{5/2}}-\frac {2 a^3 d}{3 e^2 (e x)^{3/2}}-\frac {6 a^2 b c}{e^3 \sqrt {e x}}+\frac {6 a^2 b d \sqrt {e x}}{e^4}+\frac {2 a b^2 c (e x)^{3/2}}{e^5}+\frac {6 a b^2 d (e x)^{5/2}}{5 e^6}+\frac {2 b^3 c (e x)^{7/2}}{7 e^7}+\frac {2 b^3 d (e x)^{9/2}}{9 e^8}\) |
Input:
Int[((c + d*x)*(a + b*x^2)^3)/(e*x)^(7/2),x]
Output:
(-2*a^3*c)/(5*e*(e*x)^(5/2)) - (2*a^3*d)/(3*e^2*(e*x)^(3/2)) - (6*a^2*b*c) /(e^3*Sqrt[e*x]) + (6*a^2*b*d*Sqrt[e*x])/e^4 + (2*a*b^2*c*(e*x)^(3/2))/e^5 + (6*a*b^2*d*(e*x)^(5/2))/(5*e^6) + (2*b^3*c*(e*x)^(7/2))/(7*e^7) + (2*b^ 3*d*(e*x)^(9/2))/(9*e^8)
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 = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {2 x \left (-35 b^{3} d \,x^{7}-45 b^{3} c \,x^{6}-189 a \,b^{2} d \,x^{5}-315 a \,b^{2} c \,x^{4}-945 a^{2} b d \,x^{3}+945 a^{2} b c \,x^{2}+105 a^{3} d x +63 c \,a^{3}\right )}{315 \left (e x \right )^{\frac {7}{2}}}\) | \(81\) |
orering | \(-\frac {2 x \left (-35 b^{3} d \,x^{7}-45 b^{3} c \,x^{6}-189 a \,b^{2} d \,x^{5}-315 a \,b^{2} c \,x^{4}-945 a^{2} b d \,x^{3}+945 a^{2} b c \,x^{2}+105 a^{3} d x +63 c \,a^{3}\right )}{315 \left (e x \right )^{\frac {7}{2}}}\) | \(81\) |
pseudoelliptic | \(-\frac {2 \left (-\frac {5}{9} b^{3} d \,x^{7}-\frac {5}{7} b^{3} c \,x^{6}-3 a \,b^{2} d \,x^{5}-5 a \,b^{2} c \,x^{4}-15 a^{2} b d \,x^{3}+15 a^{2} b c \,x^{2}+\frac {5}{3} a^{3} d x +c \,a^{3}\right )}{5 \sqrt {e x}\, e^{3} x^{2}}\) | \(85\) |
trager | \(-\frac {2 \left (-35 b^{3} d \,x^{7}-45 b^{3} c \,x^{6}-189 a \,b^{2} d \,x^{5}-315 a \,b^{2} c \,x^{4}-945 a^{2} b d \,x^{3}+945 a^{2} b c \,x^{2}+105 a^{3} d x +63 c \,a^{3}\right ) \sqrt {e x}}{315 e^{4} x^{3}}\) | \(86\) |
risch | \(-\frac {2 \left (-35 b^{3} d \,x^{7}-45 b^{3} c \,x^{6}-189 a \,b^{2} d \,x^{5}-315 a \,b^{2} c \,x^{4}-945 a^{2} b d \,x^{3}+945 a^{2} b c \,x^{2}+105 a^{3} d x +63 c \,a^{3}\right )}{315 e^{3} x^{2} \sqrt {e x}}\) | \(86\) |
derivativedivides | \(\frac {\frac {2 b^{3} d \left (e x \right )^{\frac {9}{2}}}{9}+\frac {2 b^{3} c e \left (e x \right )^{\frac {7}{2}}}{7}+\frac {6 a \,b^{2} d \,e^{2} \left (e x \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} c \,e^{3} \left (e x \right )^{\frac {3}{2}}+6 a^{2} b d \,e^{4} \sqrt {e x}-\frac {2 a^{3} c \,e^{7}}{5 \left (e x \right )^{\frac {5}{2}}}-\frac {6 a^{2} b c \,e^{5}}{\sqrt {e x}}-\frac {2 a^{3} d \,e^{6}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{8}}\) | \(117\) |
default | \(\frac {\frac {2 b^{3} d \left (e x \right )^{\frac {9}{2}}}{9}+\frac {2 b^{3} c e \left (e x \right )^{\frac {7}{2}}}{7}+\frac {6 a \,b^{2} d \,e^{2} \left (e x \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} c \,e^{3} \left (e x \right )^{\frac {3}{2}}+6 a^{2} b d \,e^{4} \sqrt {e x}-\frac {2 a^{3} c \,e^{7}}{5 \left (e x \right )^{\frac {5}{2}}}-\frac {6 a^{2} b c \,e^{5}}{\sqrt {e x}}-\frac {2 a^{3} d \,e^{6}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{8}}\) | \(117\) |
Input:
int((d*x+c)*(b*x^2+a)^3/(e*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/315*x*(-35*b^3*d*x^7-45*b^3*c*x^6-189*a*b^2*d*x^5-315*a*b^2*c*x^4-945*a ^2*b*d*x^3+945*a^2*b*c*x^2+105*a^3*d*x+63*a^3*c)/(e*x)^(7/2)
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.59 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=\frac {2 \, {\left (35 \, b^{3} d x^{7} + 45 \, b^{3} c x^{6} + 189 \, a b^{2} d x^{5} + 315 \, a b^{2} c x^{4} + 945 \, a^{2} b d x^{3} - 945 \, a^{2} b c x^{2} - 105 \, a^{3} d x - 63 \, a^{3} c\right )} \sqrt {e x}}{315 \, e^{4} x^{3}} \] Input:
integrate((d*x+c)*(b*x^2+a)^3/(e*x)^(7/2),x, algorithm="fricas")
Output:
2/315*(35*b^3*d*x^7 + 45*b^3*c*x^6 + 189*a*b^2*d*x^5 + 315*a*b^2*c*x^4 + 9 45*a^2*b*d*x^3 - 945*a^2*b*c*x^2 - 105*a^3*d*x - 63*a^3*c)*sqrt(e*x)/(e^4* x^3)
Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=- \frac {2 a^{3} c x}{5 \left (e x\right )^{\frac {7}{2}}} - \frac {2 a^{3} d x^{2}}{3 \left (e x\right )^{\frac {7}{2}}} - \frac {6 a^{2} b c x^{3}}{\left (e x\right )^{\frac {7}{2}}} + \frac {6 a^{2} b d x^{4}}{\left (e x\right )^{\frac {7}{2}}} + \frac {2 a b^{2} c x^{5}}{\left (e x\right )^{\frac {7}{2}}} + \frac {6 a b^{2} d x^{6}}{5 \left (e x\right )^{\frac {7}{2}}} + \frac {2 b^{3} c x^{7}}{7 \left (e x\right )^{\frac {7}{2}}} + \frac {2 b^{3} d x^{8}}{9 \left (e x\right )^{\frac {7}{2}}} \] Input:
integrate((d*x+c)*(b*x**2+a)**3/(e*x)**(7/2),x)
Output:
-2*a**3*c*x/(5*(e*x)**(7/2)) - 2*a**3*d*x**2/(3*(e*x)**(7/2)) - 6*a**2*b*c *x**3/(e*x)**(7/2) + 6*a**2*b*d*x**4/(e*x)**(7/2) + 2*a*b**2*c*x**5/(e*x)* *(7/2) + 6*a*b**2*d*x**6/(5*(e*x)**(7/2)) + 2*b**3*c*x**7/(7*(e*x)**(7/2)) + 2*b**3*d*x**8/(9*(e*x)**(7/2))
Time = 0.03 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {21 \, {\left (45 \, a^{2} b c e^{2} x^{2} + 5 \, a^{3} d e^{2} x + 3 \, a^{3} c e^{2}\right )}}{\left (e x\right )^{\frac {5}{2}} e^{2}} - \frac {35 \, \left (e x\right )^{\frac {9}{2}} b^{3} d + 45 \, \left (e x\right )^{\frac {7}{2}} b^{3} c e + 189 \, \left (e x\right )^{\frac {5}{2}} a b^{2} d e^{2} + 315 \, \left (e x\right )^{\frac {3}{2}} a b^{2} c e^{3} + 945 \, \sqrt {e x} a^{2} b d e^{4}}{e^{7}}\right )}}{315 \, e} \] Input:
integrate((d*x+c)*(b*x^2+a)^3/(e*x)^(7/2),x, algorithm="maxima")
Output:
-2/315*(21*(45*a^2*b*c*e^2*x^2 + 5*a^3*d*e^2*x + 3*a^3*c*e^2)/((e*x)^(5/2) *e^2) - (35*(e*x)^(9/2)*b^3*d + 45*(e*x)^(7/2)*b^3*c*e + 189*(e*x)^(5/2)*a *b^2*d*e^2 + 315*(e*x)^(3/2)*a*b^2*c*e^3 + 945*sqrt(e*x)*a^2*b*d*e^4)/e^7) /e
Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {21 \, {\left (45 \, a^{2} b c e^{3} x^{2} + 5 \, a^{3} d e^{3} x + 3 \, a^{3} c e^{3}\right )}}{\sqrt {e x} e^{2} x^{2}} - \frac {35 \, \sqrt {e x} b^{3} d e^{36} x^{4} + 45 \, \sqrt {e x} b^{3} c e^{36} x^{3} + 189 \, \sqrt {e x} a b^{2} d e^{36} x^{2} + 315 \, \sqrt {e x} a b^{2} c e^{36} x + 945 \, \sqrt {e x} a^{2} b d e^{36}}{e^{36}}\right )}}{315 \, e^{4}} \] Input:
integrate((d*x+c)*(b*x^2+a)^3/(e*x)^(7/2),x, algorithm="giac")
Output:
-2/315*(21*(45*a^2*b*c*e^3*x^2 + 5*a^3*d*e^3*x + 3*a^3*c*e^3)/(sqrt(e*x)*e ^2*x^2) - (35*sqrt(e*x)*b^3*d*e^36*x^4 + 45*sqrt(e*x)*b^3*c*e^36*x^3 + 189 *sqrt(e*x)*a*b^2*d*e^36*x^2 + 315*sqrt(e*x)*a*b^2*c*e^36*x + 945*sqrt(e*x) *a^2*b*d*e^36)/e^36)/e^4
Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=\frac {2\,b^3\,c\,{\left (e\,x\right )}^{7/2}}{7\,e^7}-\frac {\frac {2\,d\,a^3\,e^2\,x}{3}+\frac {2\,c\,a^3\,e^2}{5}+6\,b\,c\,a^2\,e^2\,x^2}{e^3\,{\left (e\,x\right )}^{5/2}}+\frac {2\,b^3\,d\,{\left (e\,x\right )}^{9/2}}{9\,e^8}+\frac {2\,a\,b^2\,c\,{\left (e\,x\right )}^{3/2}}{e^5}+\frac {6\,a^2\,b\,d\,\sqrt {e\,x}}{e^4}+\frac {6\,a\,b^2\,d\,{\left (e\,x\right )}^{5/2}}{5\,e^6} \] Input:
int(((a + b*x^2)^3*(c + d*x))/(e*x)^(7/2),x)
Output:
(2*b^3*c*(e*x)^(7/2))/(7*e^7) - ((2*a^3*c*e^2)/5 + (2*a^3*d*e^2*x)/3 + 6*a ^2*b*c*e^2*x^2)/(e^3*(e*x)^(5/2)) + (2*b^3*d*(e*x)^(9/2))/(9*e^8) + (2*a*b ^2*c*(e*x)^(3/2))/e^5 + (6*a^2*b*d*(e*x)^(1/2))/e^4 + (6*a*b^2*d*(e*x)^(5/ 2))/(5*e^6)
Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.60 \[ \int \frac {(c+d x) \left (a+b x^2\right )^3}{(e x)^{7/2}} \, dx=\frac {2 \sqrt {e}\, \left (35 b^{3} d \,x^{7}+45 b^{3} c \,x^{6}+189 a \,b^{2} d \,x^{5}+315 a \,b^{2} c \,x^{4}+945 a^{2} b d \,x^{3}-945 a^{2} b c \,x^{2}-105 a^{3} d x -63 a^{3} c \right )}{315 \sqrt {x}\, e^{4} x^{2}} \] Input:
int((d*x+c)*(b*x^2+a)^3/(e*x)^(7/2),x)
Output:
(2*sqrt(e)*( - 63*a**3*c - 105*a**3*d*x - 945*a**2*b*c*x**2 + 945*a**2*b*d *x**3 + 315*a*b**2*c*x**4 + 189*a*b**2*d*x**5 + 45*b**3*c*x**6 + 35*b**3*d *x**7))/(315*sqrt(x)*e**4*x**2)