Integrand size = 22, antiderivative size = 112 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{e^4 \sqrt {d+e x}}+\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right ) \sqrt {d+e x}}{e^4}-\frac {2 c (3 B d-A e) (d+e x)^{3/2}}{3 e^4}+\frac {2 B c (d+e x)^{5/2}}{5 e^4} \]
-2/3*c*(-A*e+3*B*d)*(e*x+d)^(3/2)/e^4+2/5*B*c*(e*x+d)^(5/2)/e^4+2*(-A*e+B* d)*(a*e^2+c*d^2)/e^4/(e*x+d)^(1/2)+2*(-2*A*c*d*e+B*a*e^2+3*B*c*d^2)*(e*x+d )^(1/2)/e^4
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {-10 A e \left (3 a e^2+c \left (8 d^2+4 d e x-e^2 x^2\right )\right )+6 B \left (5 a e^2 (2 d+e x)+c \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{15 e^4 \sqrt {d+e x}} \]
(-10*A*e*(3*a*e^2 + c*(8*d^2 + 4*d*e*x - e^2*x^2)) + 6*B*(5*a*e^2*(2*d + e *x) + c*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)))/(15*e^4*Sqrt[d + e* x])
Time = 0.24 (sec) , antiderivative size = 112, 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 = {652, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 \sqrt {d+e x}}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^{3/2}}+\frac {c \sqrt {d+e x} (A e-3 B d)}{e^3}+\frac {B c (d+e x)^{3/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}+\frac {2 \left (a e^2+c d^2\right ) (B d-A e)}{e^4 \sqrt {d+e x}}-\frac {2 c (d+e x)^{3/2} (3 B d-A e)}{3 e^4}+\frac {2 B c (d+e x)^{5/2}}{5 e^4}\) |
(2*(B*d - A*e)*(c*d^2 + a*e^2))/(e^4*Sqrt[d + e*x]) + (2*(3*B*c*d^2 - 2*A* c*d*e + a*B*e^2)*Sqrt[d + e*x])/e^4 - (2*c*(3*B*d - A*e)*(d + e*x)^(3/2))/ (3*e^4) + (2*B*c*(d + e*x)^(5/2))/(5*e^4)
3.15.32.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.74
method | result | size |
pseudoelliptic | \(\frac {\left (\left (6 B \,x^{3}+10 A \,x^{2}\right ) c -30 a \left (-B x +A \right )\right ) e^{3}-40 d \left (x \left (\frac {3 B x}{10}+A \right ) c -\frac {3 B a}{2}\right ) e^{2}-80 c \,d^{2} \left (-\frac {3 B x}{5}+A \right ) e +96 B c \,d^{3}}{15 \sqrt {e x +d}\, e^{4}}\) | \(83\) |
gosper | \(-\frac {2 \left (-3 B c \,x^{3} e^{3}-5 A c \,e^{3} x^{2}+6 B \,x^{2} c d \,e^{2}+20 A c d \,e^{2} x -15 B x a \,e^{3}-24 B c \,d^{2} e x +15 A a \,e^{3}+40 A c \,d^{2} e -30 B a d \,e^{2}-48 B c \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\) | \(101\) |
trager | \(-\frac {2 \left (-3 B c \,x^{3} e^{3}-5 A c \,e^{3} x^{2}+6 B \,x^{2} c d \,e^{2}+20 A c d \,e^{2} x -15 B x a \,e^{3}-24 B c \,d^{2} e x +15 A a \,e^{3}+40 A c \,d^{2} e -30 B a d \,e^{2}-48 B c \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\) | \(101\) |
risch | \(-\frac {2 \left (-3 B c \,x^{2} e^{2}-5 A c x \,e^{2}+9 B c d e x +25 A c d e -15 B a \,e^{2}-33 B c \,d^{2}\right ) \sqrt {e x +d}}{15 e^{4}}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{e^{4} \sqrt {e x +d}}\) | \(101\) |
derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A c e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B c d \left (e x +d \right )^{\frac {3}{2}}-4 A c d e \sqrt {e x +d}+2 a \,e^{2} B \sqrt {e x +d}+6 B c \,d^{2} \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(120\) |
default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A c e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B c d \left (e x +d \right )^{\frac {3}{2}}-4 A c d e \sqrt {e x +d}+2 a \,e^{2} B \sqrt {e x +d}+6 B c \,d^{2} \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(120\) |
1/15*(((6*B*x^3+10*A*x^2)*c-30*a*(-B*x+A))*e^3-40*d*(x*(3/10*B*x+A)*c-3/2* B*a)*e^2-80*c*d^2*(-3/5*B*x+A)*e+96*B*c*d^3)/(e*x+d)^(1/2)/e^4
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, B c e^{3} x^{3} + 48 \, B c d^{3} - 40 \, A c d^{2} e + 30 \, B a d e^{2} - 15 \, A a e^{3} - {\left (6 \, B c d e^{2} - 5 \, A c e^{3}\right )} x^{2} + {\left (24 \, B c d^{2} e - 20 \, A c d e^{2} + 15 \, B a e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{5} x + d e^{4}\right )}} \]
2/15*(3*B*c*e^3*x^3 + 48*B*c*d^3 - 40*A*c*d^2*e + 30*B*a*d*e^2 - 15*A*a*e^ 3 - (6*B*c*d*e^2 - 5*A*c*e^3)*x^2 + (24*B*c*d^2*e - 20*A*c*d*e^2 + 15*B*a* e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)
Time = 1.93 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A c e - 3 B c d\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right )}{e^{3}} + \frac {\left (- A e + B d\right ) \left (a e^{2} + c d^{2}\right )}{e^{3} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(B*c*(d + e*x)**(5/2)/(5*e**3) + (d + e*x)**(3/2)*(A*c*e - 3* B*c*d)/(3*e**3) + sqrt(d + e*x)*(-2*A*c*d*e + B*a*e**2 + 3*B*c*d**2)/e**3 + (-A*e + B*d)*(a*e**2 + c*d**2)/(e**3*sqrt(d + e*x)))/e, Ne(e, 0)), ((A*a *x + A*c*x**3/3 + B*a*x**2/2 + B*c*x**4/4)/d**(3/2), True))
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} B c - 5 \, {\left (3 \, B c d - A c e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} \sqrt {e x + d}}{e^{3}} + \frac {15 \, {\left (B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}\right )}}{\sqrt {e x + d} e^{3}}\right )}}{15 \, e} \]
2/15*((3*(e*x + d)^(5/2)*B*c - 5*(3*B*c*d - A*c*e)*(e*x + d)^(3/2) + 15*(3 *B*c*d^2 - 2*A*c*d*e + B*a*e^2)*sqrt(e*x + d))/e^3 + 15*(B*c*d^3 - A*c*d^2 *e + B*a*d*e^2 - A*a*e^3)/(sqrt(e*x + d)*e^3))/e
Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B c e^{16} - 15 \, {\left (e x + d\right )}^{\frac {3}{2}} B c d e^{16} + 45 \, \sqrt {e x + d} B c d^{2} e^{16} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A c e^{17} - 30 \, \sqrt {e x + d} A c d e^{17} + 15 \, \sqrt {e x + d} B a e^{18}\right )}}{15 \, e^{20}} \]
2*(B*c*d^3 - A*c*d^2*e + B*a*d*e^2 - A*a*e^3)/(sqrt(e*x + d)*e^4) + 2/15*( 3*(e*x + d)^(5/2)*B*c*e^16 - 15*(e*x + d)^(3/2)*B*c*d*e^16 + 45*sqrt(e*x + d)*B*c*d^2*e^16 + 5*(e*x + d)^(3/2)*A*c*e^17 - 30*sqrt(e*x + d)*A*c*d*e^1 7 + 15*sqrt(e*x + d)*B*a*e^18)/e^20
Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {d+e\,x}\,\left (6\,B\,c\,d^2-4\,A\,c\,d\,e+2\,B\,a\,e^2\right )}{e^4}-\frac {-2\,B\,c\,d^3+2\,A\,c\,d^2\,e-2\,B\,a\,d\,e^2+2\,A\,a\,e^3}{e^4\,\sqrt {d+e\,x}}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,c\,\left (A\,e-3\,B\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4} \]