Integrand size = 20, antiderivative size = 150 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 c^3 \left (b c^2+a d^2\right )}{3 d^6 (c+d x)^{3/2}}-\frac {2 c^2 \left (5 b c^2+3 a d^2\right )}{d^6 \sqrt {c+d x}}-\frac {2 c \left (10 b c^2+3 a d^2\right ) \sqrt {c+d x}}{d^6}+\frac {2 \left (10 b c^2+a d^2\right ) (c+d x)^{3/2}}{3 d^6}-\frac {2 b c (c+d x)^{5/2}}{d^6}+\frac {2 b (c+d x)^{7/2}}{7 d^6} \] Output:
2/3*c^3*(a*d^2+b*c^2)/d^6/(d*x+c)^(3/2)-2*c^2*(3*a*d^2+5*b*c^2)/d^6/(d*x+c )^(1/2)-2*c*(3*a*d^2+10*b*c^2)*(d*x+c)^(1/2)/d^6+2/3*(a*d^2+10*b*c^2)*(d*x +c)^(3/2)/d^6-2*b*c*(d*x+c)^(5/2)/d^6+2/7*b*(d*x+c)^(7/2)/d^6
Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (7 a d^2 \left (16 c^3+24 c^2 d x+6 c d^2 x^2-d^3 x^3\right )+b \left (256 c^5+384 c^4 d x+96 c^3 d^2 x^2-16 c^2 d^3 x^3+6 c d^4 x^4-3 d^5 x^5\right )\right )}{21 d^6 (c+d x)^{3/2}} \] Input:
Integrate[(x^3*(a + b*x^2))/(c + d*x)^(5/2),x]
Output:
(-2*(7*a*d^2*(16*c^3 + 24*c^2*d*x + 6*c*d^2*x^2 - d^3*x^3) + b*(256*c^5 + 384*c^4*d*x + 96*c^3*d^2*x^2 - 16*c^2*d^3*x^3 + 6*c*d^4*x^4 - 3*d^5*x^5))) /(21*d^6*(c + d*x)^(3/2))
Time = 0.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (\frac {-3 a c d^2-10 b c^3}{d^5 \sqrt {c+d x}}+\frac {\sqrt {c+d x} \left (a d^2+10 b c^2\right )}{d^5}+\frac {3 a c^2 d^2+5 b c^4}{d^5 (c+d x)^{3/2}}+\frac {c^3 \left (-a d^2-b c^2\right )}{d^5 (c+d x)^{5/2}}-\frac {5 b c (c+d x)^{3/2}}{d^5}+\frac {b (c+d x)^{5/2}}{d^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{3/2} \left (a d^2+10 b c^2\right )}{3 d^6}-\frac {2 c \sqrt {c+d x} \left (3 a d^2+10 b c^2\right )}{d^6}-\frac {2 c^2 \left (3 a d^2+5 b c^2\right )}{d^6 \sqrt {c+d x}}+\frac {2 c^3 \left (a d^2+b c^2\right )}{3 d^6 (c+d x)^{3/2}}+\frac {2 b (c+d x)^{7/2}}{7 d^6}-\frac {2 b c (c+d x)^{5/2}}{d^6}\) |
Input:
Int[(x^3*(a + b*x^2))/(c + d*x)^(5/2),x]
Output:
(2*c^3*(b*c^2 + a*d^2))/(3*d^6*(c + d*x)^(3/2)) - (2*c^2*(5*b*c^2 + 3*a*d^ 2))/(d^6*Sqrt[c + d*x]) - (2*c*(10*b*c^2 + 3*a*d^2)*Sqrt[c + d*x])/d^6 + ( 2*(10*b*c^2 + a*d^2)*(c + d*x)^(3/2))/(3*d^6) - (2*b*c*(c + d*x)^(5/2))/d^ 6 + (2*b*(c + d*x)^(7/2))/(7*d^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 = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(-\frac {32 \left (-\frac {x^{3} \left (\frac {3 b \,x^{2}}{7}+a \right ) d^{5}}{16}+\frac {3 \left (\frac {b \,x^{2}}{7}+a \right ) x^{2} c \,d^{4}}{8}+\frac {3 x \left (-\frac {2 b \,x^{2}}{21}+a \right ) c^{2} d^{3}}{2}+c^{3} \left (\frac {6 b \,x^{2}}{7}+a \right ) d^{2}+\frac {24 b \,c^{4} d x}{7}+\frac {16 b \,c^{5}}{7}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{6}}\) | \(93\) |
risch | \(-\frac {2 \left (-3 b \,d^{3} x^{3}+12 b c \,d^{2} x^{2}-7 a x \,d^{3}-37 b \,c^{2} d x +56 a \,d^{2} c +158 b \,c^{3}\right ) \sqrt {d x +c}}{21 d^{6}}-\frac {2 c^{2} \left (9 a x \,d^{3}+15 b \,c^{2} d x +8 a \,d^{2} c +14 b \,c^{3}\right )}{3 d^{6} \left (d x +c \right )^{\frac {3}{2}}}\) | \(106\) |
gosper | \(-\frac {2 \left (-3 b \,d^{5} x^{5}+6 b c \,d^{4} x^{4}-7 x^{3} a \,d^{5}-16 b \,c^{2} d^{3} x^{3}+42 x^{2} a c \,d^{4}+96 b \,c^{3} d^{2} x^{2}+168 a \,c^{2} d^{3} x +384 b \,c^{4} d x +112 a \,c^{3} d^{2}+256 b \,c^{5}\right )}{21 \left (d x +c \right )^{\frac {3}{2}} d^{6}}\) | \(109\) |
trager | \(-\frac {2 \left (-3 b \,d^{5} x^{5}+6 b c \,d^{4} x^{4}-7 x^{3} a \,d^{5}-16 b \,c^{2} d^{3} x^{3}+42 x^{2} a c \,d^{4}+96 b \,c^{3} d^{2} x^{2}+168 a \,c^{2} d^{3} x +384 b \,c^{4} d x +112 a \,c^{3} d^{2}+256 b \,c^{5}\right )}{21 \left (d x +c \right )^{\frac {3}{2}} d^{6}}\) | \(109\) |
orering | \(-\frac {2 \left (-3 b \,d^{5} x^{5}+6 b c \,d^{4} x^{4}-7 x^{3} a \,d^{5}-16 b \,c^{2} d^{3} x^{3}+42 x^{2} a c \,d^{4}+96 b \,c^{3} d^{2} x^{2}+168 a \,c^{2} d^{3} x +384 b \,c^{4} d x +112 a \,c^{3} d^{2}+256 b \,c^{5}\right )}{21 \left (d x +c \right )^{\frac {3}{2}} d^{6}}\) | \(109\) |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}-2 b c \left (d x +c \right )^{\frac {5}{2}}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {20 b \,c^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}-6 a c \,d^{2} \sqrt {d x +c}-20 b \,c^{3} \sqrt {d x +c}-\frac {2 c^{2} \left (3 a \,d^{2}+5 b \,c^{2}\right )}{\sqrt {d x +c}}+\frac {2 c^{3} \left (a \,d^{2}+b \,c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{6}}\) | \(129\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}-2 b c \left (d x +c \right )^{\frac {5}{2}}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {20 b \,c^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}-6 a c \,d^{2} \sqrt {d x +c}-20 b \,c^{3} \sqrt {d x +c}-\frac {2 c^{2} \left (3 a \,d^{2}+5 b \,c^{2}\right )}{\sqrt {d x +c}}+\frac {2 c^{3} \left (a \,d^{2}+b \,c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{6}}\) | \(129\) |
Input:
int(x^3*(b*x^2+a)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-32/3/(d*x+c)^(3/2)*(-1/16*x^3*(3/7*b*x^2+a)*d^5+3/8*(1/7*b*x^2+a)*x^2*c*d ^4+3/2*x*(-2/21*b*x^2+a)*c^2*d^3+c^3*(6/7*b*x^2+a)*d^2+24/7*b*c^4*d*x+16/7 *b*c^5)/d^6
Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b d^{5} x^{5} - 6 \, b c d^{4} x^{4} - 256 \, b c^{5} - 112 \, a c^{3} d^{2} + {\left (16 \, b c^{2} d^{3} + 7 \, a d^{5}\right )} x^{3} - 6 \, {\left (16 \, b c^{3} d^{2} + 7 \, a c d^{4}\right )} x^{2} - 24 \, {\left (16 \, b c^{4} d + 7 \, a c^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{21 \, {\left (d^{8} x^{2} + 2 \, c d^{7} x + c^{2} d^{6}\right )}} \] Input:
integrate(x^3*(b*x^2+a)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
2/21*(3*b*d^5*x^5 - 6*b*c*d^4*x^4 - 256*b*c^5 - 112*a*c^3*d^2 + (16*b*c^2* d^3 + 7*a*d^5)*x^3 - 6*(16*b*c^3*d^2 + 7*a*c*d^4)*x^2 - 24*(16*b*c^4*d + 7 *a*c^2*d^3)*x)*sqrt(d*x + c)/(d^8*x^2 + 2*c*d^7*x + c^2*d^6)
Time = 1.68 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {b c \left (c + d x\right )^{\frac {5}{2}}}{d^{2}} + \frac {b \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {c^{3} \left (a d^{2} + b c^{2}\right )}{3 d^{2} \left (c + d x\right )^{\frac {3}{2}}} - \frac {c^{2} \cdot \left (3 a d^{2} + 5 b c^{2}\right )}{d^{2} \sqrt {c + d x}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a d^{2} + 10 b c^{2}\right )}{3 d^{2}} + \frac {\sqrt {c + d x} \left (- 3 a c d^{2} - 10 b c^{3}\right )}{d^{2}}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {\frac {a x^{4}}{4} + \frac {b x^{6}}{6}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(b*x**2+a)/(d*x+c)**(5/2),x)
Output:
Piecewise((2*(-b*c*(c + d*x)**(5/2)/d**2 + b*(c + d*x)**(7/2)/(7*d**2) + c **3*(a*d**2 + b*c**2)/(3*d**2*(c + d*x)**(3/2)) - c**2*(3*a*d**2 + 5*b*c** 2)/(d**2*sqrt(c + d*x)) + (c + d*x)**(3/2)*(a*d**2 + 10*b*c**2)/(3*d**2) + sqrt(c + d*x)*(-3*a*c*d**2 - 10*b*c**3)/d**2)/d**4, Ne(d, 0)), ((a*x**4/4 + b*x**6/6)/c**(5/2), True))
Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (d x + c\right )}^{\frac {7}{2}} b - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} b c + 7 \, {\left (10 \, b c^{2} + a d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 21 \, {\left (10 \, b c^{3} + 3 \, a c d^{2}\right )} \sqrt {d x + c}}{d^{2}} + \frac {7 \, {\left (b c^{5} + a c^{3} d^{2} - 3 \, {\left (5 \, b c^{4} + 3 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{2}}\right )}}{21 \, d^{4}} \] Input:
integrate(x^3*(b*x^2+a)/(d*x+c)^(5/2),x, algorithm="maxima")
Output:
2/21*((3*(d*x + c)^(7/2)*b - 21*(d*x + c)^(5/2)*b*c + 7*(10*b*c^2 + a*d^2) *(d*x + c)^(3/2) - 21*(10*b*c^3 + 3*a*c*d^2)*sqrt(d*x + c))/d^2 + 7*(b*c^5 + a*c^3*d^2 - 3*(5*b*c^4 + 3*a*c^2*d^2)*(d*x + c))/((d*x + c)^(3/2)*d^2)) /d^4
Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (d x + c\right )} b c^{4} - b c^{5} + 9 \, {\left (d x + c\right )} a c^{2} d^{2} - a c^{3} d^{2}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{6}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {7}{2}} b d^{36} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} b c d^{36} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d^{36} - 210 \, \sqrt {d x + c} b c^{3} d^{36} + 7 \, {\left (d x + c\right )}^{\frac {3}{2}} a d^{38} - 63 \, \sqrt {d x + c} a c d^{38}\right )}}{21 \, d^{42}} \] Input:
integrate(x^3*(b*x^2+a)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
-2/3*(15*(d*x + c)*b*c^4 - b*c^5 + 9*(d*x + c)*a*c^2*d^2 - a*c^3*d^2)/((d* x + c)^(3/2)*d^6) + 2/21*(3*(d*x + c)^(7/2)*b*d^36 - 21*(d*x + c)^(5/2)*b* c*d^36 + 70*(d*x + c)^(3/2)*b*c^2*d^36 - 210*sqrt(d*x + c)*b*c^3*d^36 + 7* (d*x + c)^(3/2)*a*d^38 - 63*sqrt(d*x + c)*a*c*d^38)/d^42
Time = 8.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {\frac {2\,b\,c^5}{3}-\left (10\,b\,c^4+6\,a\,c^2\,d^2\right )\,\left (c+d\,x\right )+\frac {2\,a\,c^3\,d^2}{3}}{d^6\,{\left (c+d\,x\right )}^{3/2}}-\frac {\left (20\,b\,c^3+6\,a\,c\,d^2\right )\,\sqrt {c+d\,x}}{d^6}+\frac {2\,b\,{\left (c+d\,x\right )}^{7/2}}{7\,d^6}+\frac {\left (20\,b\,c^2+2\,a\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^6}-\frac {2\,b\,c\,{\left (c+d\,x\right )}^{5/2}}{d^6} \] Input:
int((x^3*(a + b*x^2))/(c + d*x)^(5/2),x)
Output:
((2*b*c^5)/3 - (10*b*c^4 + 6*a*c^2*d^2)*(c + d*x) + (2*a*c^3*d^2)/3)/(d^6* (c + d*x)^(3/2)) - ((20*b*c^3 + 6*a*c*d^2)*(c + d*x)^(1/2))/d^6 + (2*b*(c + d*x)^(7/2))/(7*d^6) + ((2*a*d^2 + 20*b*c^2)*(c + d*x)^(3/2))/(3*d^6) - ( 2*b*c*(c + d*x)^(5/2))/d^6
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.77 \[ \int \frac {x^3 \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {\frac {2}{7} b \,d^{5} x^{5}-\frac {4}{7} b c \,d^{4} x^{4}+\frac {2}{3} a \,d^{5} x^{3}+\frac {32}{21} b \,c^{2} d^{3} x^{3}-4 a c \,d^{4} x^{2}-\frac {64}{7} b \,c^{3} d^{2} x^{2}-16 a \,c^{2} d^{3} x -\frac {256}{7} b \,c^{4} d x -\frac {32}{3} a \,c^{3} d^{2}-\frac {512}{21} b \,c^{5}}{\sqrt {d x +c}\, d^{6} \left (d x +c \right )} \] Input:
int(x^3*(b*x^2+a)/(d*x+c)^(5/2),x)
Output:
(2*( - 112*a*c**3*d**2 - 168*a*c**2*d**3*x - 42*a*c*d**4*x**2 + 7*a*d**5*x **3 - 256*b*c**5 - 384*b*c**4*d*x - 96*b*c**3*d**2*x**2 + 16*b*c**2*d**3*x **3 - 6*b*c*d**4*x**4 + 3*b*d**5*x**5))/(21*sqrt(c + d*x)*d**6*(c + d*x))