Integrand size = 22, antiderivative size = 112 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{3 e^4 (d+e x)^{3/2}}-\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right )}{e^4 \sqrt {d+e x}}-\frac {2 c (3 B d-A e) \sqrt {d+e x}}{e^4}+\frac {2 B c (d+e x)^{3/2}}{3 e^4} \]
2/3*(-A*e+B*d)*(a*e^2+c*d^2)/e^4/(e*x+d)^(3/2)+2/3*B*c*(e*x+d)^(3/2)/e^4-2 *(-2*A*c*d*e+B*a*e^2+3*B*c*d^2)/e^4/(e*x+d)^(1/2)-2*c*(-A*e+3*B*d)*(e*x+d) ^(1/2)/e^4
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (a A e^3+a B e^2 (2 d+3 e x)-A c e \left (8 d^2+12 d e x+3 e^2 x^2\right )+B c \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )\right )}{3 e^4 (d+e x)^{3/2}} \]
(-2*(a*A*e^3 + a*B*e^2*(2*d + 3*e*x) - A*c*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2 ) + B*c*(16*d^3 + 24*d^2*e*x + 6*d*e^2*x^2 - e^3*x^3)))/(3*e^4*(d + e*x)^( 3/2))
Time = 0.25 (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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 (d+e x)^{3/2}}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^{5/2}}+\frac {c (A e-3 B d)}{e^3 \sqrt {d+e x}}+\frac {B c \sqrt {d+e x}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4 \sqrt {d+e x}}+\frac {2 \left (a e^2+c d^2\right ) (B d-A e)}{3 e^4 (d+e x)^{3/2}}-\frac {2 c \sqrt {d+e x} (3 B d-A e)}{e^4}+\frac {2 B c (d+e x)^{3/2}}{3 e^4}\) |
(2*(B*d - A*e)*(c*d^2 + a*e^2))/(3*e^4*(d + e*x)^(3/2)) - (2*(3*B*c*d^2 - 2*A*c*d*e + a*B*e^2))/(e^4*Sqrt[d + e*x]) - (2*c*(3*B*d - A*e)*Sqrt[d + e* x])/e^4 + (2*B*c*(d + e*x)^(3/2))/(3*e^4)
3.15.33.3.1 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 = 0.36 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\left (-B \,x^{3}-3 A \,x^{2}\right ) c +a \left (3 B x +A \right )\right ) e^{3}-12 \left (x \left (-\frac {B x}{2}+A \right ) c -\frac {B a}{6}\right ) d \,e^{2}-8 c \,d^{2} \left (-3 B x +A \right ) e +16 B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(82\) |
risch | \(\frac {2 c \left (B e x +3 A e -8 B d \right ) \sqrt {e x +d}}{3 e^{4}}-\frac {2 \left (-6 A c d \,e^{2} x +3 B x a \,e^{3}+9 B c \,d^{2} e x +A a \,e^{3}-5 A c \,d^{2} e +2 B a d \,e^{2}+8 B c \,d^{3}\right )}{3 e^{4} \left (e x +d \right )^{\frac {3}{2}}}\) | \(96\) |
gosper | \(-\frac {2 \left (-B c \,x^{3} e^{3}-3 A c \,e^{3} x^{2}+6 B \,x^{2} c d \,e^{2}-12 A c d \,e^{2} x +3 B x a \,e^{3}+24 B c \,d^{2} e x +A a \,e^{3}-8 A c \,d^{2} e +2 B a d \,e^{2}+16 B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(100\) |
trager | \(-\frac {2 \left (-B c \,x^{3} e^{3}-3 A c \,e^{3} x^{2}+6 B \,x^{2} c d \,e^{2}-12 A c d \,e^{2} x +3 B x a \,e^{3}+24 B c \,d^{2} e x +A a \,e^{3}-8 A c \,d^{2} e +2 B a d \,e^{2}+16 B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(100\) |
derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}-6 B c d \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A c d e +B a \,e^{2}+3 B c \,d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(108\) |
default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}-6 B c d \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A c d e +B a \,e^{2}+3 B c \,d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(108\) |
-2/3*(((-B*x^3-3*A*x^2)*c+a*(3*B*x+A))*e^3-12*(x*(-1/2*B*x+A)*c-1/6*B*a)*d *e^2-8*c*d^2*(-3*B*x+A)*e+16*B*c*d^3)/(e*x+d)^(3/2)/e^4
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (B c e^{3} x^{3} - 16 \, B c d^{3} + 8 \, A c d^{2} e - 2 \, B a d e^{2} - A a e^{3} - 3 \, {\left (2 \, B c d e^{2} - A c e^{3}\right )} x^{2} - 3 \, {\left (8 \, B c d^{2} e - 4 \, A c d e^{2} + B a e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]
2/3*(B*c*e^3*x^3 - 16*B*c*d^3 + 8*A*c*d^2*e - 2*B*a*d*e^2 - A*a*e^3 - 3*(2 *B*c*d*e^2 - A*c*e^3)*x^2 - 3*(8*B*c*d^2*e - 4*A*c*d*e^2 + B*a*e^3)*x)*sqr t(e*x + d)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (114) = 228\).
Time = 0.35 (sec) , antiderivative size = 449, normalized size of antiderivative = 4.01 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=\begin {cases} - \frac {2 A a e^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {16 A c d^{2} e}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {24 A c d e^{2} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {6 A c e^{3} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {4 B a d e^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {6 B a e^{3} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {32 B c d^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {48 B c d^{2} e x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {12 B c d e^{2} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {2 B c e^{3} x^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-2*A*a*e**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 16*A*c*d**2*e/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 24*A*c* d*e**2*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 6*A*c*e**3*x* *2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 4*B*a*d*e**2/(3*d*e **4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 6*B*a*e**3*x/(3*d*e**4*sqrt( d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 32*B*c*d**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 48*B*c*d**2*e*x/(3*d*e**4*sqrt(d + e*x) + 3*e** 5*x*sqrt(d + e*x)) - 12*B*c*d*e**2*x**2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x *sqrt(d + e*x)) + 2*B*c*e**3*x**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt( d + e*x)), Ne(e, 0)), ((A*a*x + A*c*x**3/3 + B*a*x**2/2 + B*c*x**4/4)/d**( 5/2), True))
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} B c - 3 \, {\left (3 \, B c d - A c e\right )} \sqrt {e x + d}}{e^{3}} + \frac {B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3} - 3 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{3}}\right )}}{3 \, e} \]
2/3*(((e*x + d)^(3/2)*B*c - 3*(3*B*c*d - A*c*e)*sqrt(e*x + d))/e^3 + (B*c* d^3 - A*c*d^2*e + B*a*d*e^2 - A*a*e^3 - 3*(3*B*c*d^2 - 2*A*c*d*e + B*a*e^2 )*(e*x + d))/((e*x + d)^(3/2)*e^3))/e
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (e x + d\right )} B c d^{2} - B c d^{3} - 6 \, {\left (e x + d\right )} A c d e + A c d^{2} e + 3 \, {\left (e x + d\right )} B a e^{2} - B a d e^{2} + A a e^{3}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B c e^{8} - 9 \, \sqrt {e x + d} B c d e^{8} + 3 \, \sqrt {e x + d} A c e^{9}\right )}}{3 \, e^{12}} \]
-2/3*(9*(e*x + d)*B*c*d^2 - B*c*d^3 - 6*(e*x + d)*A*c*d*e + A*c*d^2*e + 3* (e*x + d)*B*a*e^2 - B*a*d*e^2 + A*a*e^3)/((e*x + d)^(3/2)*e^4) + 2/3*((e*x + d)^(3/2)*B*c*e^8 - 9*sqrt(e*x + d)*B*c*d*e^8 + 3*sqrt(e*x + d)*A*c*e^9) /e^12
Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2\,B\,c\,{\left (d+e\,x\right )}^3-2\,A\,a\,e^3+2\,B\,c\,d^3+2\,B\,a\,d\,e^2-2\,A\,c\,d^2\,e-6\,B\,a\,e^2\,\left (d+e\,x\right )+6\,A\,c\,e\,{\left (d+e\,x\right )}^2-18\,B\,c\,d\,{\left (d+e\,x\right )}^2-18\,B\,c\,d^2\,\left (d+e\,x\right )+12\,A\,c\,d\,e\,\left (d+e\,x\right )}{3\,e^4\,{\left (d+e\,x\right )}^{3/2}} \]