Integrand size = 22, antiderivative size = 125 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 a c^3}{3 e (e x)^{3/2}}-\frac {6 a c^2 d}{e^2 \sqrt {e x}}+\frac {2 c \left (b c^2+3 a d^2\right ) \sqrt {e x}}{e^3}+\frac {2 d \left (3 b c^2+a d^2\right ) (e x)^{3/2}}{3 e^4}+\frac {6 b c d^2 (e x)^{5/2}}{5 e^5}+\frac {2 b d^3 (e x)^{7/2}}{7 e^6} \] Output:
-2/3*a*c^3/e/(e*x)^(3/2)-6*a*c^2*d/e^2/(e*x)^(1/2)+2*c*(3*a*d^2+b*c^2)*(e* x)^(1/2)/e^3+2/3*d*(a*d^2+3*b*c^2)*(e*x)^(3/2)/e^4+6/5*b*c*d^2*(e*x)^(5/2) /e^5+2/7*b*d^3*(e*x)^(7/2)/e^6
Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {x \left (-70 a \left (c^3+9 c^2 d x-9 c d^2 x^2-d^3 x^3\right )+6 b x^2 \left (35 c^3+35 c^2 d x+21 c d^2 x^2+5 d^3 x^3\right )\right )}{105 (e x)^{5/2}} \] Input:
Integrate[((c + d*x)^3*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(x*(-70*a*(c^3 + 9*c^2*d*x - 9*c*d^2*x^2 - d^3*x^3) + 6*b*x^2*(35*c^3 + 35 *c^2*d*x + 21*c*d^2*x^2 + 5*d^3*x^3)))/(105*(e*x)^(5/2))
Time = 0.43 (sec) , antiderivative size = 125, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^3}{(e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {d \sqrt {e x} \left (a d^2+3 b c^2\right )}{e^3}+\frac {c \left (3 a d^2+b c^2\right )}{e^2 \sqrt {e x}}+\frac {a c^3}{(e x)^{5/2}}+\frac {3 a c^2 d}{e (e x)^{3/2}}+\frac {3 b c d^2 (e x)^{3/2}}{e^4}+\frac {b d^3 (e x)^{5/2}}{e^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 d (e x)^{3/2} \left (a d^2+3 b c^2\right )}{3 e^4}+\frac {2 c \sqrt {e x} \left (3 a d^2+b c^2\right )}{e^3}-\frac {2 a c^3}{3 e (e x)^{3/2}}-\frac {6 a c^2 d}{e^2 \sqrt {e x}}+\frac {6 b c d^2 (e x)^{5/2}}{5 e^5}+\frac {2 b d^3 (e x)^{7/2}}{7 e^6}\) |
Input:
Int[((c + d*x)^3*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(-2*a*c^3)/(3*e*(e*x)^(3/2)) - (6*a*c^2*d)/(e^2*Sqrt[e*x]) + (2*c*(b*c^2 + 3*a*d^2)*Sqrt[e*x])/e^3 + (2*d*(3*b*c^2 + a*d^2)*(e*x)^(3/2))/(3*e^4) + ( 6*b*c*d^2*(e*x)^(5/2))/(5*e^5) + (2*b*d^3*(e*x)^(7/2))/(7*e^6)
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 = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {2 x \left (-15 b \,d^{3} x^{5}-63 b c \,d^{2} x^{4}-35 a \,x^{3} d^{3}-105 b \,c^{2} d \,x^{3}-315 a \,d^{2} x^{2} c -105 b \,c^{3} x^{2}+315 a d x \,c^{2}+35 c^{3} a \right )}{105 \left (e x \right )^{\frac {5}{2}}}\) | \(81\) |
orering | \(-\frac {2 x \left (-15 b \,d^{3} x^{5}-63 b c \,d^{2} x^{4}-35 a \,x^{3} d^{3}-105 b \,c^{2} d \,x^{3}-315 a \,d^{2} x^{2} c -105 b \,c^{3} x^{2}+315 a d x \,c^{2}+35 c^{3} a \right )}{105 \left (e x \right )^{\frac {5}{2}}}\) | \(81\) |
pseudoelliptic | \(-\frac {2 \left (-\frac {3 b \,d^{3} x^{5}}{7}-\frac {9 b c \,d^{2} x^{4}}{5}+\left (-a \,d^{3}-3 b \,c^{2} d \right ) x^{3}+\left (-9 a \,d^{2} c -3 b \,c^{3}\right ) x^{2}+9 a d x \,c^{2}+c^{3} a \right )}{3 \sqrt {e x}\, e^{2} x}\) | \(83\) |
trager | \(-\frac {2 \left (-15 b \,d^{3} x^{5}-63 b c \,d^{2} x^{4}-35 a \,x^{3} d^{3}-105 b \,c^{2} d \,x^{3}-315 a \,d^{2} x^{2} c -105 b \,c^{3} x^{2}+315 a d x \,c^{2}+35 c^{3} a \right ) \sqrt {e x}}{105 e^{3} x^{2}}\) | \(86\) |
risch | \(-\frac {2 \left (-15 b \,d^{3} x^{5}-63 b c \,d^{2} x^{4}-35 a \,x^{3} d^{3}-105 b \,c^{2} d \,x^{3}-315 a \,d^{2} x^{2} c -105 b \,c^{3} x^{2}+315 a d x \,c^{2}+35 c^{3} a \right )}{105 e^{2} x \sqrt {e x}}\) | \(86\) |
derivativedivides | \(\frac {\frac {2 b \,d^{3} \left (e x \right )^{\frac {7}{2}}}{7}+\frac {6 b c \,d^{2} e \left (e x \right )^{\frac {5}{2}}}{5}+\frac {2 a \,d^{3} e^{2} \left (e x \right )^{\frac {3}{2}}}{3}+2 b \,c^{2} d \,e^{2} \left (e x \right )^{\frac {3}{2}}+6 a c \,d^{2} e^{3} \sqrt {e x}+2 b \,c^{3} e^{3} \sqrt {e x}-\frac {6 a \,c^{2} d \,e^{4}}{\sqrt {e x}}-\frac {2 a \,c^{3} e^{5}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{6}}\) | \(116\) |
default | \(\frac {\frac {2 b \,d^{3} \left (e x \right )^{\frac {7}{2}}}{7}+\frac {6 b c \,d^{2} e \left (e x \right )^{\frac {5}{2}}}{5}+\frac {2 a \,d^{3} e^{2} \left (e x \right )^{\frac {3}{2}}}{3}+2 b \,c^{2} d \,e^{2} \left (e x \right )^{\frac {3}{2}}+6 a c \,d^{2} e^{3} \sqrt {e x}+2 b \,c^{3} e^{3} \sqrt {e x}-\frac {6 a \,c^{2} d \,e^{4}}{\sqrt {e x}}-\frac {2 a \,c^{3} e^{5}}{3 \left (e x \right )^{\frac {3}{2}}}}{e^{6}}\) | \(116\) |
Input:
int((d*x+c)^3*(b*x^2+a)/(e*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/105*x*(-15*b*d^3*x^5-63*b*c*d^2*x^4-35*a*d^3*x^3-105*b*c^2*d*x^3-315*a* c*d^2*x^2-105*b*c^3*x^2+315*a*c^2*d*x+35*a*c^3)/(e*x)^(5/2)
Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2 \, {\left (15 \, b d^{3} x^{5} + 63 \, b c d^{2} x^{4} - 315 \, a c^{2} d x - 35 \, a c^{3} + 35 \, {\left (3 \, b c^{2} d + a d^{3}\right )} x^{3} + 105 \, {\left (b c^{3} + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {e x}}{105 \, e^{3} x^{2}} \] Input:
integrate((d*x+c)^3*(b*x^2+a)/(e*x)^(5/2),x, algorithm="fricas")
Output:
2/105*(15*b*d^3*x^5 + 63*b*c*d^2*x^4 - 315*a*c^2*d*x - 35*a*c^3 + 35*(3*b* c^2*d + a*d^3)*x^3 + 105*(b*c^3 + 3*a*c*d^2)*x^2)*sqrt(e*x)/(e^3*x^2)
Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.17 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=- \frac {2 a c^{3} x}{3 \left (e x\right )^{\frac {5}{2}}} - \frac {6 a c^{2} d x^{2}}{\left (e x\right )^{\frac {5}{2}}} + \frac {6 a c d^{2} x^{3}}{\left (e x\right )^{\frac {5}{2}}} + \frac {2 a d^{3} x^{4}}{3 \left (e x\right )^{\frac {5}{2}}} + \frac {2 b c^{3} x^{3}}{\left (e x\right )^{\frac {5}{2}}} + \frac {2 b c^{2} d x^{4}}{\left (e x\right )^{\frac {5}{2}}} + \frac {6 b c d^{2} x^{5}}{5 \left (e x\right )^{\frac {5}{2}}} + \frac {2 b d^{3} x^{6}}{7 \left (e x\right )^{\frac {5}{2}}} \] Input:
integrate((d*x+c)**3*(b*x**2+a)/(e*x)**(5/2),x)
Output:
-2*a*c**3*x/(3*(e*x)**(5/2)) - 6*a*c**2*d*x**2/(e*x)**(5/2) + 6*a*c*d**2*x **3/(e*x)**(5/2) + 2*a*d**3*x**4/(3*(e*x)**(5/2)) + 2*b*c**3*x**3/(e*x)**( 5/2) + 2*b*c**2*d*x**4/(e*x)**(5/2) + 6*b*c*d**2*x**5/(5*(e*x)**(5/2)) + 2 *b*d**3*x**6/(7*(e*x)**(5/2))
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {35 \, {\left (9 \, a c^{2} d e x + a c^{3} e\right )}}{\left (e x\right )^{\frac {3}{2}} e} - \frac {15 \, \left (e x\right )^{\frac {7}{2}} b d^{3} + 63 \, \left (e x\right )^{\frac {5}{2}} b c d^{2} e + 35 \, {\left (3 \, b c^{2} d + a d^{3}\right )} \left (e x\right )^{\frac {3}{2}} e^{2} + 105 \, {\left (b c^{3} + 3 \, a c d^{2}\right )} \sqrt {e x} e^{3}}{e^{5}}\right )}}{105 \, e} \] Input:
integrate((d*x+c)^3*(b*x^2+a)/(e*x)^(5/2),x, algorithm="maxima")
Output:
-2/105*(35*(9*a*c^2*d*e*x + a*c^3*e)/((e*x)^(3/2)*e) - (15*(e*x)^(7/2)*b*d ^3 + 63*(e*x)^(5/2)*b*c*d^2*e + 35*(3*b*c^2*d + a*d^3)*(e*x)^(3/2)*e^2 + 1 05*(b*c^3 + 3*a*c*d^2)*sqrt(e*x)*e^3)/e^5)/e
Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.09 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {35 \, {\left (9 \, a c^{2} d e x + a c^{3} e\right )}}{\sqrt {e x} e^{2} x} - \frac {15 \, \sqrt {e x} b d^{3} e^{33} x^{3} + 63 \, \sqrt {e x} b c d^{2} e^{33} x^{2} + 105 \, \sqrt {e x} b c^{2} d e^{33} x + 35 \, \sqrt {e x} a d^{3} e^{33} x + 105 \, \sqrt {e x} b c^{3} e^{33} + 315 \, \sqrt {e x} a c d^{2} e^{33}}{e^{35}}\right )}}{105 \, e} \] Input:
integrate((d*x+c)^3*(b*x^2+a)/(e*x)^(5/2),x, algorithm="giac")
Output:
-2/105*(35*(9*a*c^2*d*e*x + a*c^3*e)/(sqrt(e*x)*e^2*x) - (15*sqrt(e*x)*b*d ^3*e^33*x^3 + 63*sqrt(e*x)*b*c*d^2*e^33*x^2 + 105*sqrt(e*x)*b*c^2*d*e^33*x + 35*sqrt(e*x)*a*d^3*e^33*x + 105*sqrt(e*x)*b*c^3*e^33 + 315*sqrt(e*x)*a* c*d^2*e^33)/e^35)/e
Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2\,c\,\sqrt {e\,x}\,\left (b\,c^2+3\,a\,d^2\right )}{e^3}-\frac {\frac {2\,a\,e\,c^3}{3}+6\,a\,d\,e\,x\,c^2}{e^2\,{\left (e\,x\right )}^{3/2}}+\frac {2\,d\,{\left (e\,x\right )}^{3/2}\,\left (3\,b\,c^2+a\,d^2\right )}{3\,e^4}+\frac {2\,b\,d^3\,{\left (e\,x\right )}^{7/2}}{7\,e^6}+\frac {6\,b\,c\,d^2\,{\left (e\,x\right )}^{5/2}}{5\,e^5} \] Input:
int(((a + b*x^2)*(c + d*x)^3)/(e*x)^(5/2),x)
Output:
(2*c*(e*x)^(1/2)*(3*a*d^2 + b*c^2))/e^3 - ((2*a*c^3*e)/3 + 6*a*c^2*d*e*x)/ (e^2*(e*x)^(3/2)) + (2*d*(e*x)^(3/2)*(a*d^2 + 3*b*c^2))/(3*e^4) + (2*b*d^3 *(e*x)^(7/2))/(7*e^6) + (6*b*c*d^2*(e*x)^(5/2))/(5*e^5)
Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.69 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {2 \sqrt {e}\, \left (15 b \,d^{3} x^{5}+63 b c \,d^{2} x^{4}+35 a \,d^{3} x^{3}+105 b \,c^{2} d \,x^{3}+315 a c \,d^{2} x^{2}+105 b \,c^{3} x^{2}-315 a \,c^{2} d x -35 a \,c^{3}\right )}{105 \sqrt {x}\, e^{3} x} \] Input:
int((d*x+c)^3*(b*x^2+a)/(e*x)^(5/2),x)
Output:
(2*sqrt(e)*( - 35*a*c**3 - 315*a*c**2*d*x + 315*a*c*d**2*x**2 + 35*a*d**3* x**3 + 105*b*c**3*x**2 + 105*b*c**2*d*x**3 + 63*b*c*d**2*x**4 + 15*b*d**3* x**5))/(105*sqrt(x)*e**3*x)