Integrand size = 22, antiderivative size = 112 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{5 e^4 (d+e x)^{5/2}}-\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right )}{3 e^4 (d+e x)^{3/2}}+\frac {2 c (3 B d-A e)}{e^4 \sqrt {d+e x}}+\frac {2 B c \sqrt {d+e x}}{e^4} \]
2/5*(-A*e+B*d)*(a*e^2+c*d^2)/e^4/(e*x+d)^(5/2)-2/3*(-2*A*c*d*e+B*a*e^2+3*B *c*d^2)/e^4/(e*x+d)^(3/2)+2*c*(-A*e+3*B*d)/e^4/(e*x+d)^(1/2)+2*B*c*(e*x+d) ^(1/2)/e^4
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (3 a A e^3+a B e^2 (2 d+5 e x)+A c e \left (8 d^2+20 d e x+15 e^2 x^2\right )-3 B c \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )}{15 e^4 (d+e x)^{5/2}} \]
(-2*(3*a*A*e^3 + a*B*e^2*(2*d + 5*e*x) + A*c*e*(8*d^2 + 20*d*e*x + 15*e^2* x^2) - 3*B*c*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3)))/(15*e^4*(d + e*x)^(5/2))
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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{(d+e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {a B e^2-2 A c d e+3 B c d^2}{e^3 (d+e x)^{5/2}}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 (d+e x)^{7/2}}+\frac {c (A e-3 B d)}{e^3 (d+e x)^{3/2}}+\frac {B c}{e^3 \sqrt {d+e x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (a B e^2-2 A c d e+3 B c d^2\right )}{3 e^4 (d+e x)^{3/2}}+\frac {2 \left (a e^2+c d^2\right ) (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac {2 c (3 B d-A e)}{e^4 \sqrt {d+e x}}+\frac {2 B c \sqrt {d+e x}}{e^4}\) |
(2*(B*d - A*e)*(c*d^2 + a*e^2))/(5*e^4*(d + e*x)^(5/2)) - (2*(3*B*c*d^2 - 2*A*c*d*e + a*B*e^2))/(3*e^4*(d + e*x)^(3/2)) + (2*c*(3*B*d - A*e))/(e^4*S qrt[d + e*x]) + (2*B*c*Sqrt[d + e*x])/e^4
3.15.34.3.1 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 = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(\frac {\left (-30 x^{2} \left (-B x +A \right ) c -6 a \left (\frac {5 B x}{3}+A \right )\right ) e^{3}-40 \left (x \left (-\frac {9 B x}{2}+A \right ) c +\frac {B a}{10}\right ) d \,e^{2}-16 c \,d^{2} \left (-15 B x +A \right ) e +96 B c \,d^{3}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(80\) |
gosper | \(-\frac {2 \left (-15 B c \,x^{3} e^{3}+15 A c \,e^{3} x^{2}-90 B \,x^{2} c d \,e^{2}+20 A c d \,e^{2} x +5 B x a \,e^{3}-120 B c \,d^{2} e x +3 A a \,e^{3}+8 A c \,d^{2} e +2 B a d \,e^{2}-48 B c \,d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(101\) |
trager | \(-\frac {2 \left (-15 B c \,x^{3} e^{3}+15 A c \,e^{3} x^{2}-90 B \,x^{2} c d \,e^{2}+20 A c d \,e^{2} x +5 B x a \,e^{3}-120 B c \,d^{2} e x +3 A a \,e^{3}+8 A c \,d^{2} e +2 B a d \,e^{2}-48 B c \,d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(101\) |
derivativedivides | \(\frac {2 B c \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-2 A c d e +B a \,e^{2}+3 B c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 c \left (A e -3 B d \right )}{\sqrt {e x +d}}}{e^{4}}\) | \(102\) |
default | \(\frac {2 B c \sqrt {e x +d}-\frac {2 \left (A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-2 A c d e +B a \,e^{2}+3 B c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 c \left (A e -3 B d \right )}{\sqrt {e x +d}}}{e^{4}}\) | \(102\) |
risch | \(\frac {2 B c \sqrt {e x +d}}{e^{4}}-\frac {2 \left (15 A c \,e^{3} x^{2}-45 B \,x^{2} c d \,e^{2}+20 A c d \,e^{2} x +5 B x a \,e^{3}-75 B c \,d^{2} e x +3 A a \,e^{3}+8 A c \,d^{2} e +2 B a d \,e^{2}-33 B c \,d^{3}\right )}{15 e^{4} \sqrt {e x +d}\, \left (e^{2} x^{2}+2 d e x +d^{2}\right )}\) | \(124\) |
1/15*((-30*x^2*(-B*x+A)*c-6*a*(5/3*B*x+A))*e^3-40*(x*(-9/2*B*x+A)*c+1/10*B *a)*d*e^2-16*c*d^2*(-15*B*x+A)*e+96*B*c*d^3)/(e*x+d)^(5/2)/e^4
Time = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (15 \, B c e^{3} x^{3} + 48 \, B c d^{3} - 8 \, A c d^{2} e - 2 \, B a d e^{2} - 3 \, A a e^{3} + 15 \, {\left (6 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 5 \, {\left (24 \, B c d^{2} e - 4 \, A c d e^{2} - B a e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]
2/15*(15*B*c*e^3*x^3 + 48*B*c*d^3 - 8*A*c*d^2*e - 2*B*a*d*e^2 - 3*A*a*e^3 + 15*(6*B*c*d*e^2 - A*c*e^3)*x^2 + 5*(24*B*c*d^2*e - 4*A*c*d*e^2 - B*a*e^3 )*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 653 vs. \(2 (116) = 232\).
Time = 0.49 (sec) , antiderivative size = 653, normalized size of antiderivative = 5.83 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {6 A a e^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {16 A c d^{2} e}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {40 A c d e^{2} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {30 A c e^{3} x^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {4 B a d e^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {10 B a e^{3} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {96 B c d^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {240 B c d^{2} e x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {180 B c d e^{2} x^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {30 B c e^{3} x^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-6*A*a*e**3/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 16*A*c*d**2*e/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 40*A*c*d* e**2*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x **2*sqrt(d + e*x)) - 30*A*c*e**3*x**2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e **5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 4*B*a*d*e**2/(15*d**2* e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x )) - 10*B*a*e**3*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 96*B*c*d**3/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 240*B*c*d**2*e* x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*s qrt(d + e*x)) + 180*B*c*d*e**2*x**2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e** 5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 30*B*c*e**3*x**3/(15*d** 2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*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**(7/2), True))
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {e x + d} B c}{e^{3}} + \frac {3 \, B c d^{3} - 3 \, A c d^{2} e + 3 \, B a d e^{2} - 3 \, A a e^{3} + 15 \, {\left (3 \, B c d - A c e\right )} {\left (e x + d\right )}^{2} - 5 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {5}{2}} e^{3}}\right )}}{15 \, e} \]
2/15*(15*sqrt(e*x + d)*B*c/e^3 + (3*B*c*d^3 - 3*A*c*d^2*e + 3*B*a*d*e^2 - 3*A*a*e^3 + 15*(3*B*c*d - A*c*e)*(e*x + d)^2 - 5*(3*B*c*d^2 - 2*A*c*d*e + B*a*e^2)*(e*x + d))/((e*x + d)^(5/2)*e^3))/e
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, \sqrt {e x + d} B c}{e^{4}} + \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} B c d - 15 \, {\left (e x + d\right )} B c d^{2} + 3 \, B c d^{3} - 15 \, {\left (e x + d\right )}^{2} A c e + 10 \, {\left (e x + d\right )} A c d e - 3 \, A c d^{2} e - 5 \, {\left (e x + d\right )} B a e^{2} + 3 \, B a d e^{2} - 3 \, A a e^{3}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{4}} \]
2*sqrt(e*x + d)*B*c/e^4 + 2/15*(45*(e*x + d)^2*B*c*d - 15*(e*x + d)*B*c*d^ 2 + 3*B*c*d^3 - 15*(e*x + d)^2*A*c*e + 10*(e*x + d)*A*c*d*e - 3*A*c*d^2*e - 5*(e*x + d)*B*a*e^2 + 3*B*a*d*e^2 - 3*A*a*e^3)/((e*x + d)^(5/2)*e^4)
Time = 10.50 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2\,\left (-48\,B\,c\,d^3-120\,B\,c\,d^2\,e\,x+8\,A\,c\,d^2\,e-90\,B\,c\,d\,e^2\,x^2+20\,A\,c\,d\,e^2\,x+2\,B\,a\,d\,e^2-15\,B\,c\,e^3\,x^3+15\,A\,c\,e^3\,x^2+5\,B\,a\,e^3\,x+3\,A\,a\,e^3\right )}{15\,e^4\,{\left (d+e\,x\right )}^{5/2}} \]